For, if B D be not in the same straight line with B C,

let B E be in the same straight line with B C. Then, because A B meets the straight line CBE; therefore the adjacent angles CBA, ABE are equal to two right angles (I. 13).

But the angles C B A, therefore the angles C

A B D are equal to two right angles (hyp.)
B A, A B E are equal to the angles C B A,
ABD (Ax. 1).

From these equals take away the common angle CBA, therefore the remaining angle A B E is equal to the remaining angle A B D (Ax. 3),

the less to the greater, which is impossible. Therefore B E is not in the same straight line with B C. In the same way it may be demonstrated

that no other can be in the same straight line with it but B D. Therefore B D is in the same straight line with B C. Therefore, if, at a point in a straight line, two other, etc.

Proposition XV. Theorem.

Q. E. D.

If two straight lines cut one another, the vertical, or opposite, angles shall be equal.

Let the two straight lines A B, CD cut one another in the point E.

Then the angle A E C shall be equal to the angle D E B, and the angle CEB to the angle A E D.

Because the straight line A E makes with CD at the point E the adjacent angles CEA, AED,

these angles are together equal to two right angles (I. 13).

Again, because the straight line DE makes with AB at the

point E

the adjacent angles A E D, DE B,

these angles also are equal to two right angles.

Therefore the angles CEA, A E D are equal to the angles AED, DE B.

Take away the common angle A E D,

and the remaining angle CE A is equal to the remaining angle DEB (Ax. 3).

In the same way it may be demonstrated

that the angle CEB is equal to the angle A ED. Therefore, if two straight lines cut one another, etc.

Q. E. D. COR. 1. From this it is manifest, that if two straight lines cut each other, the angles which they make at the point where they cut, are together equal to four right angles.

COR. 2. And, consequently, that all the angles made by any number of lines meeting in one point, are together equal to four right angles.

Proposition XVI. Theorem.

If one side of a triangle be produced, the exterior angle is greater than either of the interior opposite angles.

Let A B C be a triangle, and let the side B C be produced to D. Then the exterior angle A CD shall be greater than either of the interior opposite angles CBA, BAC.

F

G

Bisect A C in E (I. 10).
Join B E.

Produce B E to F, making E F equal to BE (I. 3).
Join F C.

Because A E is equal to E C and B E to E F,
A E, E B are equal to E C, E F, each to each,
And the angle A E B is equal to the angle C EF,

because they are opposite vertical angles (I. 15),
Therefore the base A B is equal to the base C F (I. 4)
and the triangle A E B to the triangle CE F,

and the remaining angles to the remaining angles, each to each, to which the equal sides are opposite.

Therefore the angle B A E is equal to the angle E CF. But the angle A CD is greater than the angle ECF therefore the angle A C D is greater than the angle B A E. In the same way, if B C be bisected, and A C be produced to G, may be demonstrated that the angle B C G, that is, the angle A C D (I. 15), is greater than the angle A B C. Therefore, if one side of a triangle be produced, etc.

it

Proposition XVII. Theorem.

Q. E. D.

Any two angles of a triangle are together less than two right angles.

Let A B C be a triangle.

Then any two of its angles are together less than two right angles.

B

4

C

Produce B C to D.

Then the exterior angle A CD of the triangle A B C is greater than the interior opposite angle A B C (I. 16).

To each of these add the angle AC B,

therefore the angles A CD, A C B are greater than the angles АВС, А С В.

But the angles AC D, A C B are equal to two right angles (I. 13). therefore the angles A B C, A C B are less than two right angles. In like manner it may be demonstrated

that the angles BA C, A C B are less than two right angles, and that the angles CA B, A B C are less than two right angles. Therefore, any two angles of a triangle, etc.

Proposition XVIII. Theorem.

Q. E. D.

The greater side of every triangle is opposite to the greater angle.

Let A B C be a triangle, of which the side A C is greater than the side A B.

Then the angle A B C shall be greater than the angle B C A.

B

From A C, the greater, cut off A D equal to A B, the less (I. 3).

Join B D.

Then, because in the triangle A B D, A D is equal to A B, therefore the angle A D B is equal to the angle A B D (I. 5). But A D B is the exterior angle of the triangle B D C. Therefore the angle A D B is greater than the angle D C B. But the angle A D B is equal to the angle A B D,

therefore the angle A B D is greater than the angle D C B. Much more then is the angle A B C greater than the angle A CB. Therefore, the greater side of every triangle, etc.

Q. E. D.

Proposition XIX. Theorem.

The greater angle of every triangle is subtended by the greater side, or has the greater side opposite to it..

Let A B C be a triangle, of which the angle A B C is greater than the angle B C A.

Then the side A C shall be greater than the side A B.

B

For, if A C be not greater than A B,

A C must be either equal to A B or less than A B.
If A C were equal to A B,

then the angle A B C would be equal to the angle A C B (I. 5); but it is not (hyp.),

Therefore the side A C is not equal to the side A B.
Again, if A C were less than A B,

then the angle A B C would be less than the angle A C B (I. 18); but it is not (hyp.).

Therefore the side A C is not less than the side A B. Therefore A C is neither equal to nor less than A B. Therefore A C is greater than A B. Therefore, the greater angle of every triangle, etc.

Proposition XX. Theorem.

Q. E. D.

Any two sides of a triangle are together greater than the third side.

Let A B C be a triangle.

Then any two sides of it are together greater than the third side, namely; B A, A C greater than B C; A B, B C greater than A C; and B C, C A greater than A B.