Marks' first lessons in geometry / In two parts. Objectively presented, and designed for the use of primary classes in grammar schools, academies, etc.

PART SECOND. AXIOMS AND THEOREMS.

Bernhard Marks

AXIOMS ILLUSTRATED.

Axiom 1.

The triangle A is equal to the triangle C.

The triangle B is also equal to the triangle C.

What do you think of the two triangles A and B? Why?

If two things are separately equal to the same thing, they are equal to each other.

Axiom 2.

The square A is equal to the square B.

To the rectangle C add the square A, and we have an L pointing in what direction?

To the same rectangle C add the square B, and we have an L pointing in what direction?

Which is larger, the L pointing to the left, or that pointing to the right?

To what same thing did you add two equals?

What two equals did you add to it?

What was the first sum?

The second?

What do you think of the two sums?

If equals be added to the same thing, the sums will be equal.

Axiom 3.

The square A is equal to the square B.

From the inverted T take away the square A, and we have an L pointing in what direction?

From the same Fig. T take away the square B, and we have an L pointing in what direction?

Which is larger, the L pointing to the right, or that pointing to the left?

What two equal things did we take away from the same thing?

From what same thing did we take them away?

What did we find true of the two remainders?

If equals be taken from the same thing, the remainders will be equal.

Axiom 4.

The rectangle 1 2 is equal to the rectangle 1 3.

From the rectangle 1 2 take away the square A, and what rectangle remains?

From the rectangle 1 3 take away the same square A, and what rectangle remains?

Which is greater, the rectangle B, or the rectangle C?

What same thing did we take away from equals?

From what did we first take it?

What remained?

From what did we next take it?

What remained?

What did we find true of the two remainders?

If the same thing be taken from equals, the remainders will be equal.

Axiom 5.

If equals be added to equals, the sums will be equal.

Axiom 6.

If equals be subtracted from equals, the remainders will be equal.

Axiom 7.

If the halves of two things are equal, the wholes will be equal.

Axiom 8.

Every Whole is equal to the sum of all its parts.

Axiom 9.

From one point to another only one straight line can be drawn.

Axiom 10.

A straight line is the shortest distance between two points.

Axiom 11.

If two things coincide throughout their whole extent, they are equal.

THEOREMS ILLUSTRATED.

Diagram 29.

DEVELOPMENT LESSON.

Do the angles Blue, Red, take up all the space on the line a b?

Do the angles Blue, Yellow, Red, take up all the space on the line?

Do the angles Blue, Yellow, Green, Red, take up all the space on the line?

Is there room between any two of the angles to put in another angle?

Then are not the angles Blue, Yellow, Green, Red, equal to all the space on the line a b?

Note.—The word space, as here used, means angular space; and it is indispensable that the teacher impress this fact upon the learner.

By means of former lessons, the pupil has learned positively, that an angle is the difference between the directions of two lines; and, impliedly, that the included space has nothing to do with the size of the angle. There cannot, therefore, be much danger that the pupil will imbibe any erroneous notion from this style of expression, which is very much more simple than to say that the difference of direction of two given lines is equal to the difference of direction of two other given lines, which style will be used somewhat later in these lessons.

Diagram 30.

PROPOSITION I. THEOREM.

DEVELOPMENT LESSON.

Are the adjacent angles Green, Red, equal to all the angular space on the line a b?

Place a paper square corner or right angle on the line a b at the left of c d with its vertex at c.

It will cover all the angle Green and part of the angle Red up to the line c d.

Now place another square corner on the line a b to the right of the line c d, and with its vertex at the point c.

It will cover the remaining part of the angle Red, and two edges of the square corners will meet along the line c d.

Are the two right angles equal to all the angular space on the line a b?

Then if the two adjacent angles Green, Red, are equal to all the angular space on the line a b, and the two right angles are also equal to the same space, what do you infer concerning the adjacent angles and the two right angles?

What axiom do you apply when you say that the adjacent angles are equal to the two right angles?

To what same thing did you find two things separately equal?

What did you first see equal to it?

What did you next see equal to it?

Then what did you find true?

If the angle Red were smaller, and the angle Green larger, would the adjacent angles still be equal to two right angles?

Then,—

Any two adjacent angles are equal to two right angles.

If we draw the straight line c d where the edges of the square corners come together, what kind of angles will a c d, d c b, be?

See now if you can understand the following demonstration:—

DEMONSTRATION.

We wish to prove that

Any two adjacent angles are equal to two right angles.

Let the two straight lines a b, m n, intersect each other in the point c. (Diagram 30.)

Then will any two adjacent angles, as Green, Red, be equal to two right angles?

For, from the point c, draw the straight line c d so as to make the angles a c d, d c b, right angles.

The adjacent angles Green, Red, are equal to all the angular space on the line a b.

The right angles a c d, d c b, are also equal to all the angular space on the line a b.

Therefore the adjacent angles Green, Red, are equal to two right angles.

TEST QUESTIONS.

To what same thing did you find two things equal?

What did you first see equal to it?

What did you next see equal to it?

Then what new thing did you find true?

What axiom did you make use of?

Diagram 31.

TEST LESSON.

By means of Fig. A,—

1. Prove that the adjacent angles Green, Red, are equal to two right angles.

2. Prove that the adjacent angles Blue, Yellow, are equal to two right angles.

By means of Fig. B,—

3. Prove that the adjacent angles Green, Red, are equal to two right angles.

4. Prove that the adjacent angles Yellow, Blue, are equal to two right angles.

By means of Fig. C,—

5. Prove that the adjacent angles Red, Blue, are equal to two right angles.

6. Prove that the adjacent angles Green, Yellow, are equal to two right angles.

7. Give the preceding demonstrations again, but name the angles by their letters instead of by their colors.

Diagram 32.

TEST LESSON.

By means of Fig. A prove,—

1. That the adjacent angles a c m, m c b, are equal to two right angles.

2. That the adjacent angles a c n, n c b, are equal to two right angles.

By means of Fig. B prove,—

3. That the adjacent angles a c n, n c b, are equal to two right angles.

4. That the adjacent angles a c m, m c b, are equal to two right angles.

By means of Fig. C prove,—

5. That the adjacent angles a c m, m c b, are equal to two right angles.

6. That the adjacent angles a c n, n c b, are equal to two right angles.

By means of Fig. D prove,—

7. That the adjacent angles a c n, n c b, are equal to two right angles.

8. That the adjacent angles b c m, m c a, are equal to two right angles.

Diagram 33.

PROPOSITION II. THEOREM.

DEVELOPMENT LESSON.

What kind of angles are P and S?

How do the adjacent angles Yellow, Blue, compare with the right angles P, S?

How do the adjacent angles Blue, Red, compare with the two right angles?

Then if the adjacent angles Yellow, Blue, are equal to two right angles, and the adjacent angles Blue, Red, are also equal to two right angles, what do you think of the two pairs of adjacent angles, Yellow, Blue, and Blue, Red?

If, from the adjacent angles Yellow, Blue, we take away the angle Blue, what remains?

If, from the adjacent angles Blue, Red, we take away the same angle Blue, what remains?

Then, since the same angle Blue has been taken from equal pairs of adjacent angles, what do you think of the two remainders, Yellow, Red?

Suppose the lines a b and m n were so drawn that the angles Yellow, Red, were larger or smaller, would they still be equal to each other?

Then,—

All vertical angles are equal to each other.

Diagram 34.

DEMONSTRATION.

We wish to prove that

All vertical angles are equal to each other.

Let the straight lines a b, m n, intersect each other at the point c, then will any two vertical angles, as Yellow, Red, be equal to each other.

For the adjacent angles Yellow, Blue, are equal to two right angles.^[3]

3.When this comparison is made, let the pupil look at the right angles P and S.

The adjacent angles Blue, Red, are also equal to two right angles.

Therefore the adjacent angles Yellow, Blue, are equal to the adjacent angles Blue, Red.

If, from the adjacent angles Yellow, Blue, we take away the angle Blue, we shall have left the angle Yellow.

If, from the adjacent angles Blue, Red, we take away the same angle Blue, we shall have left the angle Red.

Therefore the vertical angles Yellow, Red, are equal to each other.

TEST QUESTIONS.

When you say that the adjacent angles Yellow, Blue, are equal to two right angles, do you know it because you see it, or because you have proved it?

How do you know that the adjacent angles Blue, Red, are equal to two right angles?

When you say the adjacent angles Yellow, Blue, are equal to the adjacent angles Blue, Red, what axiom do you use?

What same thing do you take away from equals?

From what equals do you take it away?

When you take the angle Blue from the adjacent angles Yellow, Blue, what is the remainder?

When you take the same angle Blue from the adjacent angles Blue, Red, what is the remainder?

What do you find true of the two remainders?

What axiom do you use?

Diagram 35.

OTHER METHODS OF DEMONSTRATION.

The adjacent angles Yellow, Green, are equal to what?

The adjacent angles Green, Red, are equal to what?

Then what do you know of the two pairs of adjacent angles Yellow, Green, and Green, Red?

From the adjacent angles Yellow, Green, take away the angle Green. What remains?

From the adjacent angles Green, Red, take the same angle Green. What remains?

What do you know of the two remainders?

Why?

What axiom do you use?

In the last lesson, when you proved the vertical angles Yellow, Red, equal to each other, you made use of the angle Blue; now prove the same two angles equal by means of the angle Green.

The adjacent angles Blue, Red, are equal to what?

The adjacent angles Red, Green, are equal to what?

Then what do you know of the two pairs of adjacent angles Blue, Red, and Red, Green?

From the adjacent angles Blue, Red, take away the angle Red. What remains?

From the adjacent angles Red, Green, take away the same angle Red. What remains?

Then what do you know of the two remainders, Blue, Green?

Now apply the preceding demonstration to the vertical angles Blue, Green.

Prove the vertical angles Blue, Green, equal to each other by means of the angle Yellow.

Diagram 36.

TEST LESSON.

By means of Fig. A,—

1. Prove that the vertical angles Yellow, Red, are equal to each other, using the angle Green.

2. Prove the same thing, using the angle Blue.

3. Prove that the vertical angles Blue, Green, are equal to each other, using the angle Yellow.

4. Prove the same thing, using the angle Red.

By means of Fig. B,—

5. Prove the vertical angles Yellow, Red, equal to each other, using the angle Green.

6. Prove the same thing, using the angle Blue.

7. Prove the vertical angles Green, Blue, equal by means of the angle Red.

8. Prove the same thing by means of the angle Yellow.

Go through the preceding eight demonstrations again, calling the angles by their letters instead of by their colors.

By means of Fig. C, prove that

9. a c n equals m c b, by means of a c m.

10. a c n equals m c b, by means of b c n.

11. a c m equals n c b, by means of a c n.

12. a c m equals n c b, by means of m c b.

By means of Fig. D, prove that

13. m c a equals b c n, by means of a c n.

14. m c a equals b c n, by means of m c b.

15. m c b equals a c n, by means of m c a.

16. m c b equals a c n, by means of b c n.

Diagram 37.

PROPOSITION III. THEOREM.

DEVELOPMENT LESSON.

In the above diagram, the lines a b, c d, are parallel, and are intersected by the line e f at the points m and n.

The angle Red measures the difference of direction between the line m b and what other line?

The angle Yellow measures the difference of direction between the line n d and what other line?

Then, as the lines m b and n d are parallel, must there not be the same difference of direction between them and the line e f?

Then can there be any difference between the angles which measure those equal directions?

Then what do you think of the opposite exterior and interior angles Red, Yellow?

DEMONSTRATION.

We wish to prove that

Opposite exterior and interior angles are equal to each other.

Let the straight line e f intersect the two parallel straight lines a b, c d, at the points m and n.

Then will any two opposite exterior and interior angles, as Red, Yellow, be equal to each other.

For the angle Red measures the difference of direction of the lines m b and e f.

And the angle Yellow measures the difference of direction of the lines n d and e f.

But because the lines m b, n d, are parallel, these differences are equal.

Therefore the angles which measure them are equal; that is,

The opposite exterior and interior angles Red, Yellow, are equal to each other.

Diagram 38.

TEST LESSON.

By means of Fig. A,—

1. Prove that the opposite exterior and interior angles Green, Blue, are equal to each other.

2. Prove that the opposite exterior and interior angles Red, Yellow, are equal to each other.

3. Prove the opposite exterior and interior angles c n e, a m n, equal.

4. Prove the opposite exterior and interior angles e n d, n m b, equal.

By means of Fig. B,—

5. Prove the opposite exterior and interior angles e m a, m n d, equal.

6. Prove the opposite exterior and interior angles a m n, d n f, equal.

7. Prove the opposite exterior and interior angles e m b, m n c, equal.

8. Prove the opposite exterior and interior angles b m n, c n f, equal.

Diagram 39.

PROPOSITION IV. THEOREM.

DEVELOPMENT LESSON.

What do you know of the opposite exterior and interior angles Red, Yellow?

What do you know of the vertical angles Red, Green?

Then if the interior alternate angles Green, Yellow, are separately equal to the angle Red, what new fact do you know?

What axiom do you employ?

To what same thing did you find two things equal?

What two things did you find equal to it?

DEMONSTRATION.

We wish to prove that

Any two interior alternate angles are equal to each other.

Let the straight line e f intersect the two parallel straight lines a b, c d, in the points m and n.

Then will any two interior alternate angles, as Green, Yellow, be equal to each other.

For the opposite exterior and interior angles Red, Yellow, are equal.

The vertical angles Red, Green, are also equal.

Then because the interior alternate angles Green, Yellow, are separately equal to the angle Red, they are equal to each other.

Diagram 40.

TEST LESSON.

What do you know of the vertical angles Green, Red, in Fig. A?

What do you know of the opposite exterior and interior angles Red, Yellow?

Then if the interior alternate angles Green, Yellow, are separately equal to the angle Red, what do you infer?

By means of Fig. A,—

1. Prove that the interior alternate angles Green, Yellow, are equal, using the angle Red.

2. Prove the same angles equal, using the angle Blue.

3. Go through the same demonstrations again, calling the angles by their letters instead of by their colors.

By means of Fig. B,—

4. Prove the interior alternate angles Red, Blue, equal, using the angle Yellow.

5. Prove the same angles equal, using the angle Green.

6. Go through the same two demonstrations again, naming the angles by their letters instead of by their colors.

By means of Fig. C,—

7. Prove the interior alternate angles c n m, n m b, equal, using the angle f n d.

8. Prove the same, using the angle a m e.

9. Prove the interior alternate angles a m n, m n d, equal, using the angle e m b.

10. Prove the same, using the angle c n f.

Diagram 41.

PROPOSITION V. THEOREM.

DEVELOPMENT LESSON.

What do you know of the opposite exterior and interior angles Red, Yellow?

What do you know of the vertical angles Yellow, Green?

Then if the exterior alternate angles Red, Green, are separately equal to the angle Yellow, what new thing do you know to be true?

What axiom do you employ?

To what same thing did you know two things to be equal?

What two things did you know to be equal to it?

Then what new thing did you find to be true?

DEMONSTRATION.

We wish to prove that

Any two exterior alternate angles are equal to each other.

Let the straight line e f intersect the two parallel straight lines a b, c d, at the points m and n.

Then will any two exterior alternate angles, as Red, Green, be equal.

For the opposite exterior and interior angles Red, Yellow, are equal to each other.

And the vertical angles Yellow, Green, are also equal to each other.

Then because the exterior alternate angles Red, Green, are separately equal to the angle Yellow, they are equal to each other.

Diagram 42.

TEST LESSON.

What do you know of the opposite exterior and interior angles Yellow, Red?

What do you know of the vertical angles Red, Blue?

Then if the exterior alternate angles Yellow, Blue, are separately equal to the angle Red, what do you know of them?

By means of Fig. A,—

1. Prove that the exterior alternate angles Yellow, Blue, are equal, using the angle Red.

2. Prove the same thing, using the angle Green.

3. Go through the same demonstrations, calling the angles by their letters.

4. Prove the exterior alternate angles e m b, c n f, equal, using the angle a m n.

5. Prove the same, using the angle m n d.

By means of Fig. B,—

6. Prove that the exterior alternate angles c m e, f n b, are equal, using the angle n m d.

7. Prove the same, using the angle a n m.

8. Prove the exterior alternate angles e m d, a n f, equal, using the angle c m n.

9. Prove the same, using the angle m n b.

Diagram 43.

PROPOSITION VI. THEOREM.

DEVELOPMENT LESSON.

What do you know of the interior alternate angles Yellow, Red?

If to the angle Green you add the angle Yellow, what is the sum?

If to the same angle Green you add the equal angle Red, what is the sum?

Then, having added equals to the same thing, what do you think of the two sums,—the adjacent angles Green, Yellow, and the interior opposite angles Green, Red?

What do you know of the adjacent angles Green, Yellow, and the right angles P, S?

Then if the interior opposite angles Green, Red, and the two right angles P, S, are separately equal to the adjacent angles Green, Yellow, what new thing do you know?

DEMONSTRATION.

We wish to prove that

Any two interior opposite angles are equal to two right angles.

Let the straight line e f intersect the two parallel straight lines a b, c d, in the points m and n.

Then will any two interior opposite angles be equal to two right angles.

For the interior alternate angles Yellow, Red, are equal.

If to the angle Green we add the angle Yellow, we shall have the adjacent angles Green, Yellow.

If to the same angle Green we add the equal angle Red, we shall have the interior opposite angles Green, Red.

Then the adjacent angles Green, Yellow, are equal to the interior opposite angles Green, Red.

But the adjacent angles Green, Yellow, are equal to two right angles.

Then because the interior opposite angles Green, Red, and two right angles, are separately equal to the two adjacent angles Green, Yellow, they are equal to each other.

Diagram 44.

TEST LESSON.

By means of Fig. A,—

1. Prove the interior opposite angles Green, Yellow, equal to two right angles, using the angle Red.

2. Prove the same, using the angle Blue.

3. Prove the same, using the angle e g b.

4. Prove the same, using the angle f h d.

5. Go through the same demonstrations again, naming the angles by their letters instead of by their colors.

6. Prove the interior opposite angles Red, Blue, equal to two right angles, using the angle Yellow.

7. Prove the same, using the angle Green.

8. Prove the same, using the angle e g a.

9. Prove the same, using the angle c h f.

10. Go through the same demonstrations again, calling the angles by their letters instead of by their colors.

By means of Fig. B,—

11. Prove the interior opposite angles a g h, g h c, equal to two right angles, using the angle g h d.

12. Prove the same, using the angle c h f.

13. Prove the same, using the angle a g e.

14. Prove the interior opposite angles b g h, g h d, equal to two right angles, using the angle a g h.

15. Prove the same, using the angle e g b.

16. Prove the same, using the angle f h d.

Compare the angles Yellow, Green, each with its exterior opposite angle, and see if you can prove that the exterior opposite angles e g b, f h d, are also equal to two right angles.

Diagram 45.

PROPOSITION VII. THEOREM.

DEVELOPMENT LESSON.

Suppose we do not know whether the lines a b, c d, are parallel, or not;

But, by measuring, we find that the interior angles Blue, Yellow, on the same side of the secant^[4] line e f, are equal to two right angles:

4.“Secant” means “cutting.”

The adjacent angles Blue, Red, are equal to what?

Then, if the interior angles Blue, Yellow, are equal to two right angles,

And the adjacent angles Blue, Red, are also equal to two right angles,

What do you infer?

From the interior angles Blue, Yellow, take away the angle Blue: what remains?

From the adjacent angles Blue, Red, take away the same angle Blue: what remains?

What do you know of the two remainders?

The angle Red measures the direction of the line g b from what line?

The equal angle Yellow measures the direction of the line h d from what line?

Then if the lines g b, h d, have the same direction from the line e f, what do you call them?

Diagram 46.

DEMONSTRATION.

We wish to prove, that,

If a straight line intersects two other straight lines so that two interior angles on the same side of the intersecting line are equal to two right angles, the two lines are parallel.

Let the straight line e f intersect the two straight lines a b, c d, in the points g and h, so that the angles Red, Blue, are equal to two right angles.

Then will the lines a b, c d, be parallel.

For the angles Red, Blue, are supposed equal to two right angles.

The adjacent angles Red, Green, are known to be also equal to two right angles.

Then the interior angles Red, Blue, are equal to the adjacent angles Red, Green.

If from the interior angles Red, Blue, we take away the angle Red, we have left the angle Blue.

If from the adjacent angles Red, Green, we take the same angle Red, we shall have left the angle Green.

Then the angle Blue is equal to the angle Green.

But the angle Blue measures the direction of the line h d from the line e f.

And the angle Green measures the direction of the line g b from the line e f.

Then the lines g b, h d, have the same direction, and are parallel.

TEST LESSON.

1. Prove the same without the colors.

2. Prove the same, using the angle f h d.

3. Prove the same, supposing the angles a g h, g h c, equal to two right angles, and using the angle a g e.

4. Prove the same, using the angle c h f.

See Note E, Appendix.

The following demonstration is very easy. Read it once, and see if you can go through it without a second reading:—

DEMONSTRATION.

We wish to prove that

The sum of any two sides of a triangle is greater than the third side.

Let the figure a b c be a triangle, then will the sum of any two sides, as a c, c b, be greater than the third side a b.

For the straight line a b is the shortest distance between the two points a and b, and is therefore less than the broken line a c b.

PROPOSITION IX. PROBLEM.

The following solution is so easy that you will understand it at once:—

We wish

To construct an equilateral triangle on a given straight line.

SOLUTION.

Let a b be the given line.

With the point a as a centre, and a b as a radius, draw the circumference of the circle, or a part of one.

With the point b as a centre, and the same radius a b, draw another circumference, or a part of one.

From the point c, in which the circumferences or arcs intersect, draw the straight lines a c and b c.

Now, because the lines a b and a c are radii of the same circle, they are equal.

And, because the lines a b and b c are radii of the same circle, they are also equal.

Then, because the two lines a c, b c, are separately equal to the line a b, they are equal to each other, and the triangle is equilateral.

PROPOSITION X. THEOREM.

DEVELOPMENT LESSON.

Let the figure a b c be a triangle.

Produce the side a c to d.

We have now another angle, b c d, and we wish to find out if it is equal to any of the angles of the triangle.

From the point c draw the line c e parallel to a b.

Because the straight line a d intersects the two parallels a b, c e, the angle a is equal to what other angle?

Because the straight line b c intersects the two parallels a b, c e, the angle b is equal to what other angle?

Then the angles a and b are equal to what two angles?

How does the angle b c d compare with the angles b c e, e c d?

Then, if the angles a and b, on the one hand, and the angle b c d, on the other, are separately equal to the angles b c e, e c d,

What have you found out?

What axiom have you just employed?

To what same thing have you found two other things equal?

What two things did you find equal to it?

DEMONSTRATION.

We wish to prove, that,

If any side of a triangle be produced, the new angle formed will be equal to the sum of the angles that are not adjacent to it.

Let a b c be a triangle.

Produce the side a c to d; then will the new angle b c d be equal to the sum of the angles a and b.

For from the point c draw c e parallel to a b.

Then, because the straight line a d intersects the two parallels a b, c e, in the points a and c,

The opposite exterior and interior angles a and e c d are equal to each other.

And because the straight line b c intersects the same parallels in the points b and c,

The interior alternate angles b and b c e are equal.

Then the angles a and b of the triangle are equal to the angles b c e and e c d.

But the new angle b c d is equal to the angles b c e, e c d.

Then because the new angle b c d, and the angles a and b are separately equal to the angles b c e, e c d, they are equal to each other.

PROPOSITION XI. THEOREM.

DEVELOPMENT LESSON.

Let the figure a b c be a triangle.

Produce the side a c to d.

By the last theorem, the angle b c d is equal to what angles of the triangle?

What angle must we add to these angles to make up the three angles of the triangle?

If we add the same angle to the angle b c d, what adjacent angles do we get?

Then the three angles of the triangle, a, b, and c, are equal to what two angles?

But the adjacent angles a c b and b c d are equal to what?

Then, because the three angles of the triangle, a, b, and c, and two right angles, are separately equal to the two adjacent angles c and b c d.

What new thing have you found out?

DEMONSTRATION.

We wish to prove that

The three angles of any triangle are equal to two right angles.

Let the figure a b c be a triangle; then will the sum of the angles a, b, and c, be equal to two right angles.

For, produce the side a c to d.

The new angle b c d is equal to the sum of the angles a and b.

If to the angles a and b we add the angle c, we shall have the three angles of the triangle.

If to the angle b c d we add the same angle c, we shall have the adjacent angles c and b c d.

Then the three angles of the triangle a, b, c, are equal to the adjacent angles c and b c d.

But the adjacent angles c and b c d are equal to two right angles.

Then, because the three angles of the triangle are equal to the adjacent angles c and b c d, they are equal to two right angles.

PROPOSITION XII. THEOREM.

DEVELOPMENT LESSON.

Let the Fig. A B C D be a parallelogram.

Produce the side C D to F.

Because the straight line B D intersects the parallels A B and C F, the angle B is equal to what other angle?

Because the straight line C F intersects the parallels A C and B D, the angle C is equal to what other angle?

Then what follows from this?

To what angle did you find two others equal?

What two angles did you find equal to it?

What axiom do you think of?

See if you can go through the demonstration without reading it even once.

DEMONSTRATION.

We wish to prove that

The opposite angles of a parallelogram are equal to each other.

Let the Fig. A B C D be a parallelogram.

Then will any two opposite angles, as B and C, be equal to each other.

For produce the line C D to F.

Because the straight line B D meets the two parallels A B and C F,

The interior alternate angles B and E are equal to each other.

Because the straight line C F meets the two parallels B D and A C,

The opposite exterior and interior angles C and E are equal to each other.

Then, because the angles B and C are separately equal to the angle E, they are equal to each other.

1. Prove the same by producing the line A B towards the left.

2. Prove the same by producing the line B D downwards.

3. Prove the angles A and D equal to each other by producing the line C D towards the left.

4. Prove the same by producing the line D B upwards.

5. See if you can prove the same by drawing a diagonal through the points A and D.

PROPOSITION XIII. THEOREM.

DEVELOPMENT LESSON.

In these two triangles we have tried to make the side a b of the one equal to the side d e of the other; the side a c of the one equal to the side d f of the other; and the included angle b a c of the one equal to the included angle e d f of the other.

We now wish to find out if the third side b c of the one is equal to the third side e f of the other, and if the two remaining angles b and c of the one are equal to the two remaining angles e and f of the other.

Suppose we were to cut the triangle d e f out of the page, and place it upon the triangle a b c, so that the line d e should fall upon the line a b, and the point d upon the point a.

As the line d e is equal to the line a b, upon what point will the point e fall?

If the angle e d f were less than the angle b a c, would the line d f fall within or without the triangle?

If the angle e d f were greater than the angle b a c, where would the line d f fall?

Since the angle a is equal to d, where, then, must the line d f fall?

As the line d f is equal to the line a c, upon what point will the point f fall?

Then, if the point e falls upon the point b, and the point f upon the point c, where will the line e f fall?

Now, because the three sides of the triangle d e f exactly fall upon the three sides of the triangle a b c, we say the two magnitudes coincide throughout their whole extent, and are therefore equal.

What three parts of the triangle a b c did we suppose to be equal to three corresponding parts of the triangle d e f before we placed one upon the other.

What line of the one do we find equal to a line in the other?

What two angles of the one do we find equal to two angles in the other?

What do you think of the areas of the triangles?

DEMONSTRATION.

We wish to prove, that,

If two triangles have two sides, and the included angle of the one equal to two sides and the included angle of the other, each to each, the two triangles are equal in all respects.

Let the triangles a b c and d e f have the side a b of the one equal to the side d e of the other; the side a c of the one equal to the side d f of the other; and the included angle b a c of the one equal to the included angle e d f of the other, each to each; then will the two triangles be equal in all their parts.

For, place the triangle d e f upon the triangle a b c, so that the line d e shall fall upon the line a b, with the point d upon the point a.

Because the line d e is equal to the line a b, the point e will fall upon the point b.

Because the angle e d f is equal to the angle b a c, the line d f will fall upon the line a c.

Because the line d f is equal to the line a c, the point f will fall upon the point c.

Then, because the point e is on the point b, and the point f on the point c, the line e f will coincide with the line b c, and the two triangles will be found equal in all their parts;

That is, the angle e is found to be equal to the angle b, the angle f to the angle c, the line e f to the line b c, and the area of the triangle a b c to the area of the triangle d e f.

PROPOSITION XIV. THEOREM.

DEVELOPMENT LESSON.

In these two triangles we have tried to make the angle b of the one equal to the angle e of the other; the angle c of the one equal to the angle f of the other; and the included side b c of the one equal to the included side e f of the other.

We now wish to find out if the remaining angle a of the one is equal to the remaining angle d of the other, and if the two remaining sides a b and a c of the one are equal to the two remaining sides d e and d f of the other.

Suppose we were to cut the triangle d e f out of the page and place it upon the triangle a b c, so that the line e f shall fall upon the line b c, with the point e upon the point b.

Because the line e f is equal to the line b c, upon what point will the point f fall?

Because the angle e is equal to the angle b, where will the line e d fall?

Because the angle f is equal to the angle c, where will the line d f fall?

Then, if the line d e falls upon the line a b and the line d f upon the line a c, where will the point d fall?

Now because the three sides of the triangle d e f exactly fall upon the three sides of the triangle a b c, we say the two magnitudes coincide throughout their whole extent, and are therefore equal.

Suppose the angle e were greater than the angle b, would the line e d fall within or without the triangle?

If it were less, where would the line fall?

Why does the line d e fall exactly upon the line a b?

DEMONSTRATION.

We wish to prove that,

If two triangles have two angles, and the included side of the one equal to two angles and the included side of the other, each to each, the two triangles are equal to each other in all respects.

Let the triangles a b c and d e f have the angle b of the one equal to the angle e of the other; the angle c of the one equal to the angle f of the other; and the included side b c of the one equal to the included side e f of the other, each to each; then will the two triangles be equal in all their parts.

For place the triangle d e f upon the triangle a b c, so that the line e f shall fall upon the line b c, with the point e upon the point b.

Because the line e f is equal to the line b c the point f will fall upon the point c.

Because the angle e is equal to the angle b, the line e d will fall upon the line b a, and the point d will be somewhere in the line b a.

Because the angle f is equal to the angle c, the line f d will fall upon the line c a, and the point d will be somewhere in the line c a.

Then, because the point d is in the two lines, b a and c a, it must be in their intersection, or upon the point a.

And, as the two triangles coincide throughout their whole extent, they are equal in all their parts.

That is, the angle a is found to be equal to the angle d; the side b a to the side e d; the side c a to the side f d; and the area of the triangle a b c to the area of the triangle d e f.

PROPOSITION XV. THEOREM.

DEMONSTRATION.

We wish to prove that

The opposite sides of any parallelogram are equal.

Let the figure a b c d be a parallelogram; then will the sides a b and c d be equal to each other; likewise the sides a d and b c.

For, draw the diagonal b d.

Because the figure is a parallelogram, the sides a b and d c are parallel, and the interior alternate angles n and o are equal.

Because the figure is a parallelogram, the interior alternate angles r and m are equal.

Then the two triangles a d b, b d c, have two angles and the included side of the one equal to two angles and the included side of the other, each to each, and are therefore equal;

And the side a b opposite the angle m is equal to the side c d opposite the equal angle r;

And the side a d opposite the angle n is equal to the side b c opposite the equal angle o.

TEST.

Prove the same by drawing a diagonal from a to c.

PROPOSITION XVI. THEOREM.

DEVELOPMENT LESSON.

Suppose A B to be a straight line, and C any point out of it.

From the point C draw a perpendicular C F to A B.

Let us see if this perpendicular is not shorter than any other line we can draw from the same point to the same line.

Draw any other line from C to A B as C E.

Now, as C E is any line whatever other than a perpendicular, if we find that the perpendicular C F is shorter than it we must conclude that it is the shortest line that can be drawn from C to A B.

Produce C F until F D is equal to C F, and then join E and D.

In the triangles E F C, E F D, what two sides were drawn equal?

What line is a side to each?

How great an angle is C F E?

What is a right angle?

Then how do the angles C F E and E F D compare with each other?

If the two triangles E F C, E F D, have the side C F of the one equal to the side F D of the other, the side E F common to both, and the included angle E F C of the one equal to the included angle E F D of the other, each to each, what do you infer?

Then what third side of the one have you found equal to a third side of the other?

C E is what part of the broken line C E D?

C F is what part of the line C D?

Which is shorter, the straight line C D, or the broken line C E D?

Then how does the half of C D or C F compare with the half of C E D or C E?

If C E is any line whatever other than a perpendicular, what may we now say of the perpendicular from the point C to the straight line A B?

DEMONSTRATION.

We wish to prove that

A perpendicular is the shortest distance from a point to a straight line.

Let A B be a straight line, and C A point out of it; then will the perpendicular C E be the shortest line that can be drawn from the point to the line.

For draw any other line from C to A B, as C F.

Produce C E until E D equals C E, and join F D.

The two triangles F E C, F E D, have the side C E of the one equal to the side E D of the other, the side F E common, and the included angle F E C of the one equal to the included angle F E D of the other, they are therefore equal, and the side C F equals the side F D.

But the straight line C D is the shortest distance between the two points C D; therefore it is shorter than the broken line C F D.

Then C E, the half of C D, is shorter than C F, the half C F D.

And, as C F is any line other than a perpendicular, the perpendicular C E is the shortest line that can be drawn from C to A B.

PROPOSITION XVII. THEOREM.

DEMONSTRATION.

We wish to prove that

A tangent to a circumference is perpendicular to a radius at the point of contact.

Let the straight line A B be tangent at the point D to the circumference of the circle whose centre is C.

Join the centre C with the point of contact D, the tangent will be perpendicular to the radius C D.

For draw any other line from the centre to the tangent, as C F.

As the point D is the only one in which the tangent touches the circumference, any other point, as F, must be without the circumference.

Then the line C F, reaching beyond the circumference, must be longer than the radius C D, which would reach only to it; therefore C D is shorter than any other line which can be drawn from the point C to the straight line A B; therefore it is perpendicular to it.

PROPOSITION XVIII. THEOREM.

DEMONSTRATION.

We wish to prove that

In any isosceles triangle, the angles opposite the equal sides are equal.

Let the triangle A B C be isosceles, having the side A B equal to the side A C; then will the angle B, opposite the side A C, be equal to the angle C, opposite the equal side A B.

For draw the line A D so as to divide the angle A into two equal parts, and let it be long enough to divide the side B C at some point as D.

Now the two triangles A D B, A D C, have the side A B of the one equal to the side A C of the other, the side A D common to both, and the included angle B A D of the one equal to the included angle C A D of the other; therefore the two triangles are equal in all respects, and the angle B, opposite the side A C, is equal to the angle C, opposite the side A B.

PROPOSITION XIX. THEOREM.

DEMONSTRATION.

We wish to prove that,

If two triangles have the three sides of the one equal to the three sides of the other, each to each, they are equal in all their parts.

Let the two triangles A B C, A D C, have the side A B of the one equal to the side A D of the other; the side B C of the one equal to the side D C of the other, and the third side likewise equal; then will the two triangles be equal in all their parts.

For place the two triangles together by their longest side, and join the opposite vertices B and D by a straight line.

Because the side A B is equal to the side A D, the triangle B A D is isosceles, and the angles A B D, A D B, opposite the equal sides are equal.

Because the side B C is equal to the side D C, the triangle B C D is isosceles, and the angles C B D, C D B, opposite the equal sides are equal.

If to the angle A B D we add the angle D B C, we shall have the angle A B C.

And if to the equal of A B D, that is, A B D, we add the equal of D B C, that is, B D C, we shall have the angle A D C.

Therefore the angle A B C is equal to the angle A D C.

Then the two triangles A B C, A D C, have two sides, and the included angle of the one equal to two sides and the included angle of the other, each to each, and are equal in all their parts; that is, the three angles of the one are equal to the three angles of the other, and their areas are equal.

PROPOSITION XX. THEOREM.

DEMONSTRATION.

We wish to prove that

An angle at the circumference is measured by half the arc on which it stands.

Let B A D be an angle whose vertex is in the circumference of the circle whose centre is C; then will it be measured by half the arc B D.

For through the centre draw the diameter A E, and join the points C and B.

The exterior angle E C B is equal to the sum of the angles B and B A C.

Because the sides C A, C B, are radii of the circle, they are equal, the triangle is isosceles, the angles B and B A C opposite the equal sides are equal, and the angle B A C is half of both.

Then, because the angle B A C is half of B and B A C, it must be half of their equal E C B.

But E C B, being at the centre, is measured by B E; then half of it, or B A C, must be measured by half B E.

In like manner, it may be proved that the angle C A D is measured by half E D.

Then, because B A C is measured by half B E, and C A D by half E D, the whole angle B A D must be measured by half the whole arc B D.

SECOND CASE.

Suppose the angle were wholly on one side of the centre, as F A B.

Draw the diameter A E and the radius B C as before.

Prove that the angle B A E is measured by half the arc B E.

Draw another radius from C to F, and prove that F A E is measured by half the arc F E.

Then, because the angle F A E is measured by half the arc F E, and the angle B A E is measured by half the arc B E,

The difference of the angles, or F A B, must be measured by half the difference of the arcs, or half of F B.

PROPOSITION XXI. THEOREM.

DEMONSTRATION.

We wish to prove that

Parallel chords intercept equal arcs of the circumference.

Let the chords A B, C D, be parallel; then will the intercepted arcs A C and B D be equal.

For draw the straight line B C.

Because the lines A B and C D are parallel, the interior alternate angles A B C, B C D, are equal.

But the angle A B C is measured by half the arc A C;

And the angle B C D is measured by half the arc B D:

Then, because the angles are equal, the half arcs which measure them must be equal, and the whole arcs themselves must be equal.

PROPOSITION XXII. THEOREM.

DEMONSTRATION.

We wish to prove that

The angle formed by a tangent and a chord meeting at the point of contact is measured by half the intercepted arc.

Let the tangent C A B and the chord A D meet at the point of contact A; then will the angle B A D be measured by half the intercepted arc A D.

For draw the diameter A E F.

Because A B is a tangent, and A E a radius at the point of contact, the angle B A F is a right angle, and is measured by the semicircle A D F.

Because the angle F A D is at the circumference, it is measured by half the arc D F.

Then the difference between the angles B A F and D A F, or B A D, must be measured by half the difference of the arcs A D F and D F, or A D;

That is, the angle B A D is measured by half the arc A D.

PROPOSITION XXIII. THEOREM.

DEMONSTRATION.

We wish to prove that

A tangent and chord parallel to it intercept equal arcs of the circumference.

Let A B be tangent to the circumference at the point D, and let C F be a chord parallel to the tangent; then will the intercepted arcs C D and D F be equal.

For from the point of contact D, draw the straight line D C.

Because the tangent and chord are parallel, the interior alternate angles A D C and D C F are equal.

But the angle A D C, being formed by the tangent D A and the chord D C, is measured by half the intercepted arc D C;

And the angle D C F, being at the circumference, is measured by half the arc on which it stands, D F:

Then, because the angles are equal, the half arcs which measure them are equal, and the arcs themselves are equal.

PROPOSITION XXIV. THEOREM.

DEMONSTRATION.

We wish to prove that

The angle formed by the intersection of two chords in a circle is measured by half the sum of the intercepted arcs.

Let the chords A B and C D intersect each other in the point E; then will the angle B E D or A E C be measured by half the sum of the arcs A C, B D.

For from the point C draw C F parallel to A B.

Because the chords A B and C F are parallel, the arcs A C, B F, are equal.

Add each of these equals to B D, and we have B D plus A C equal to B D plus B F; that is, the sum of the arcs B D, A C, is equal to the arc F D.

Because the chords A B, C F, are parallel, the opposite exterior and interior angles D E B, D C F, are equal.

But D C F is an angle at the circumference, and is therefore measured by half the arc F D.

Then the equal angle D E B must be measured by half of the arc F D, or its equal B D, plus A C.

We wish to prove that

The angle formed by two secants meeting without a circle is measured by half the difference of the intercepted arcs.

Let the secants A B, A C, intersect the circumference in the points D and E; then will the angle B A C be measured by half the difference between the arcs B C and D E.

For from the point D draw the chord D F parallel to E C.

Because A C and D F are parallel, the opposite exterior and interior angles B D F and B A C are equal.

Because the chords D F, E C, are parallel, the arcs D E and F C are equal.

If from the arc B C we take the arc D E, or its equal F C, we shall have left the arc B F;

But the angle B D F, being at the circumference, is measured by half the arc B F:

Then the equal of B D F, or B A C, must be measured by half the arc B F, or half the difference between the intercepted arcs B C and D E.

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