two right angles; then will BC and BD be in the same straight line.

For if BD be not in the same line

with CB, suppose BE to be in the same line with CВ.

Then (I. 15) ABC+ABE = 2R; but

(Hyp.) ABC+ABD = 2R.

... (Ax. 1) ABC+ABE=ABC+ABD.

A

E

C

B

D

Taking from both ABC, then ABE = ABD, the less to the

greater, which is impossible. Therefore, BE is not straight with CB, and no line but BD is. Therefore, CBD is a straight line.

Theorem. If two straight lines cut each other, the vertical or opposite angles are equal.

C

B

E

Let AB and CD cut each other at E; then will AEC be equal to BED, and CEB to AED. Because CEA, CEB are adjacent angles, they are together equal to two right angles (I. 15); for the same reason CEB, BED are to

gether equal to two right angles;

A

D

therefore (Ax. 1), CEA and CEB are equal to CEB and BED. Take away from both the common angle CEB, and the remainders AEC, BED (Ax. 3) are equal.

In the same manner it may be proved that CEB is equal to AED.

(I. 15) AEC+CEB=2R(I. 15) DEB+CEB=2R

ANALYSIS.

AEC+CEB=DEB+CЕВ.
СЕВ-СЕВ

{ (Αx.3)

AEC DEВ.

Corollary 1. Hence, all the angles at the intersection of two

straight lines, are equal to four right angles.

Corollary 2. Also, all the angles made by any number of lines meeting in a point, are equal to four right angles.

Proposition 18.

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

Let the side BC of the triangle ABC be produced to D; then will ACD be greater than either ABC or BAC.

By producing AC to G, we may prove similarly that BCG (or ACD) is greater than AВС.

Proposition 19.

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

Let ABC be a triangle; any two of its angles are together less than two right angles.

Produce BC to D.

Then, because ACD is the exterior angle

A

of the triangle ABC, it is greater than ABC. To each add ACB; then

In the same way the theorem may be proved of any two

angles of the triangle.

Proposition 20.

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

Let ABC be a triangle, in which AC is greater than AB;

then will ABC be greater than ACB.

Cut off (I. 1) from AC, AD equal to

A

AB, and join DB.

D

Because ADB is an exterior angle of the triangle DBC, it is greater (I. 18) than

the opposite interior angle DCB. But

(I. 13) ADB =ABD; ... ABD>ACB; therefore still more ABC>ACB.

Proposition 21.

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

Let ABC be a triangle in which the angle ABC is greater than the angle ACB; then will AC be

Theorem. If two triangles have two sides of one equal to two sides of the other, each to each, but the angles contained by these sides unequal, the third side of that which has the greater angle will be greater than the third side of the other.

Let ABC, DEF be two triangles, having the sides BA, AC equal to ED, DF, each to each, and the angle BAC greater than EDF; then will BC be greater than EF.

angles DHF, DHG, DF is equal to DG, DH common, and the angle FDH equal to GDH, therefore (I. 7) FH is equal to GH.

Now (I. 2), EH+HF>EF, or EH+HG>EF;

...

EG or BC is greater than EF.

Theorem. If two triangles have two sides of one equal to two sides of the other, each to each, but the third sides unequal, the angle contained by the two sides of the one which has the greater third side, will be greater than the angle contained by the two sides of the other.

Let ABC, DEF be two triangles having AB, AC equal to DE, DF, each to each, and BC greater than EF; then will BAC be greater than EDF.

For, if not, it must be either equal to it or less. It is not equal, for then BC would be equal to (I. 7) EF. It is not less, for then BC would be less than (I. 22). EF. But BC is greater than EF

(Hyp.), therefore BAC is greater than EDF.

Proposition 24.

F

Theorem. If two triangles have two angles and an opposite side of one, equal to two angles and an opposite side of the other, each to each, the equal sides being opposite equal angles, the triangles will be equal in all their parts.

Let ABC, DEF be two triangles which have ABC, ACB and AB of one, equal to DEF, DFE and DE of the other, each to each; then will the triangles be equal in all their parts.

C