At the point C (Prop. X.) erect the perpendicular CE. The sum of the angles ACE and ECB is equal to two right angles (Def. 10). But the sum of the angles BCD, DCE, and A ACE is equal to the sum of the an

E

B

gles ACE and ECB. Therefore the sum of the angles BCD, DCE, and ACE is equal to two right angles (Ax. 1). But the angles DCE and ACE are together equal to ACD (Ax. 8); therefore the sum of the angles BCD and ACD is equal to two right angles.

PROPOSITION XVI.-THEOREM.

If two straight lines meet a third straight line, making the sum of the adjacent angles equal to two right angles, these two straight lines will form one and the same straight line.

Let two straight lines, AB and BC, meet a third straight line, BD, making the sum of the adjacent angles, ABD and CBD, equal to two right angles, then will these two lines form A one and the same straight line.

B

D

E

For, suppose that AB and BC do not form one and the same straight line,* then some other line, as BE, must be the continuation of AB; then, if ABE be a straight line, the sum of the angles ABD and EBD is equal to two right angles (Prop. XV.). But (by hypothesis) the sum of the angles ABD and CBD is also equal to two right angles. Hence the sum of the angles ABD and EBD is equal to the sum of the angles ABD and CBD; but the angle ABD is common to both; take it away, and there will remain the angle EBD equal to CBD, which is absurd (Ax. 9). Therefore BE is not a continuation of

* This method of demonstration is called indirect, or the reductio ad absurdum.

AB, and in a similar manner it can be demonstrated that no other line than BC can be the continuation of AB.

PROPOSITION XVII.-THEOREM.

If one straight line intersect another straight line, the opposite or vertical angles are equal.

D Let the straight line AB inter.sect the straight line CD in the point G; then will the angle AGD be equal to the angle CGB, and AGC equal to BGD.

B

For the sum of the angles AGD and AGC is equal to two right angles (Prop. XV.), and the sum of the angles BGC and AGC is also equal to two right angles (Prop. XV.). Therefore the sum of the angles AGD and AGC is equal to the sum of the angles BGC and AGC. Take away the common angle AGC (Ax. 3), and the remaining angle AGD will be equal to the remaining angle BGC, and in the same manner it can be demonstrated that the angle AGC is equal to the angle BGD.

Corollary. It is manifest that the sum of all the an gles made by straight lines meeting at a common point is equal to four right angles.

PROPOSITION XVIII.-THEOREM.

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

A

B

G

D

E

Let the triangle ABC have one of its sides, AB, produced to D; then will the exterior angle CBD be greater than either of the two interior and opposite angles, CAB and BCA.

Bisect BC at the point F (Prop. IX.); join the points F and A; produce AF until FE is equal to it; join B and E.

In the two triangles AFC and EFB, the sides CF and FB are equal (const.), the sides AF and FE are also equal (const.), and the angles AFC and EFB are equal (Prop. XVII.). Therefore the 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, and are every way equal (Prop. IV.); the angle EBF is equal to the angle FCA. But DBC is greater than EBF (Ax. 9), and therefore greater than FCA. In the same manner, by producing BC and bisecting AB, it can be demonstrated that ABG, or its equal, DBC, is greater than САВ.

Cor. Any two angles of a triangle are together less than two right angles. Because the exterior angle is greater than either of the interior and opposite angles; to this inequality add the adjacent angle, and the sum of the exterior angle and the adjacent angle will be greater than the sum of the adjacent angle, and either of the other two. But the sum of the adjacent angle and the exterior angle is equal to two right angles (Prop. XV.). Therefore the sum of the two angles of the triangle is less than two right angles.

PROPOSITION XIX.-THEOREM.

If two angles of a triangle be equal, the sides opposite to them are also equal.

Let ABC be a triangle having the angle BAC equal to the angle ABC; then will the side BC be equal to the side AC.

For, suppose AC and BC are not equal, then one of them must be the greater; let BC be the greater; then A

B

from BC (Prop. III.) cut off a part, BD, equal to the less, AC. In the two triangles BAC and ABD, the side AC is equal to BD (const.), the side AB common, and the

angles BAC and ABD are equal (hypothesis.) There fore the 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. Hence (Prop. IV.) they are equal in all their parts, which is impossible (Ax. 9). Therefore BD is not equal to AC; and in like man-◄ ner it can be proved that no other line than BC is equal o AC.

PROPOSITION XX.-THEOREM.

The greater side of every triangle has the greater angle opposite to it.

A

D

B

Let ABC be a triangle having the side AC greater than the side BC; then will the angle ABC be greater than the angle CAB.

CD

From the greater line, AC, cut off equal to CB (Prop. III.). Join the points D and B. Because the triangle DCB is isosceles, the angle CDB is equal to the angle DBC (Prop. VI.). But the angle CDB is greater than the angle A (Prop. XVIII.), and hence the angle DBC must also be greater than the angle A. But the angle ABC is greater than DBC (Ax. 9), therefore ABC is much greater than A.

PROPOSITION XXI.-THEOREM.

The greater angle of every triangle has the greater side opposite to it.

A

B

Let ABC be a triangle having the angle ABC greater than the angle BAC; then will the side AC be greater than the side BC.

For if AC be not greater than BC, it must be equal to, or less than it. It can not be equal to it; for then the angle ABC would be equal to the angle BAC (Prop. VI.), which is con

trary to the hypothesis. It can not be less, for then (Prop. XX.) ABC would be less than BAC, which is also contrary to the hypothesis. Since AC is neither equal to, nor less than BC, it must be greater.

PROPOSITION XXII.-THEOREM.

If, from a point within a triangle, two straight lines be drawn to the extremities of one side, these two lines will be less than the other two sides of the triangle.

Let ABC be a triangle, and the two lines AE and EB drawn from the point E to the extremities A and B; then will the sum of AE and EB be less than the sum of AC and CB.

Produce AE to D. In the triangle BDE the side BE is less than A

C

B

the sum of BD and DE (Prop. XII.). To this inequality add AE; then the sum of BE and AE (Ax. 4) is less than the sum of BD, DE, and AE, or BD and DA. In the triangle ACD, the side AD is less than the sum of AC and CD (Prop. XII.). To this inequality add BD; then the sum of AD and DB is less than the sum of AC, CD, and DB, or than the sum of AC and CB. But it was before proved that the sum of AE and EB is less than the sum of AD and DB; hence the sum of AE and EB is much less than the sum of AC and CB.

Cor. The two lines drawn from the point E to the extremities of the side AB will contain a greater angle than the other two sides of the triangle.

For the angle AEB is greater than the angle EDB; and EDB is greater than the angle ACD (Prop. XVIIL). Hence AEB is much greater than ACB.

B 2