=

Or, since AG AB, and GC EF, we have,

AB+ BC > AB + EF.

Taking away the common part AB, there remains (A. 5),

3°. When G is within the triangle ABC.

From Proposition VIII., we have,

BA+ BC > GA + GC;

Hence, in each case, BC is greater than EF; which was to be proved.

Conversely: If in two triangles ABC and DEF, the side AB is equal to the side DE, the side AC to DF, and BC greater than EF, then will the angle BAC be greater than the angle EDF

For, if not, BAC must either EDF In the former case, BC

be equal to, or less than,

would be equal to EF

(P. V.), and in the latter case, BC would be less than EF; either of which would be contrary to the hypothesis hence, BAC must be greater than EDF.

PROPOSITION X. THEOREM.

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

In the triangles ABC and DEF, let AB be equal to DE, AC to DF, and BC to EF: EF: then will the tri

angles be equal in all their parts.

B

A

D

E

F

were

For, since the sides AB, AC, are equal to DE, DF, each to each, if the angle A were greater than D, it would follow, by the last Proposition, that the side BC would be greater than EF; and if the angle A less than D, the side BC would be less than EF. But BC is equal to EF, by hypothesis; therefore, the angle A can neither be greater nor less than D: hence, it must be equal to it. The two triangles have, therefore, two sides and the included angle of the one equal to two sides and the included angle of the other, each to each; and, consequently, they are equal in all their parts (P. V.); which was to be proved. Scholium. In triangles, equal in all their parts, the equal sides lie opposite the equal angles; and conversely.

PROPOSITION XI. THEOREM.

In an isosceles triangle the angles opposite the equal sides are

equal.

Let BAC be an isosceles triangle, having the side AB equal to the side AC: then will the angle C be equal to the angle B.

A

AD

Join the vertex A and the middle point D of the base BC. Then, AB AB is equal to AC, by hypothesis, common, and BD equal to DC, by construction: hence, the triangles BAD, and DAC, have the three sides of the one equal to those of the other, each to each; therefore, by the last Proposition, the angle B is equal to the angle C; which was to be proved.

B

C

D

Cor. 1. An equilateral triangle is equiangular.

Cor. 2. The angle BAD is equal to DAC, and BDA to CDA hence, the last two are right angles. Consequently, a line drawn from the vertex of an isosceles triangle to the middle of the base, bisects the vertical angle, and is perpendicular to the base.

PROPOSITION XII. THEOREM.

If two angles of a triangle are equal, the sides opposite to them are also equal, and consequently, the triangle is isosceles.

In the triangle ABC, let the angle ABC be equal to the angle ACB: then will AC be equal to AB, and consequently, the triangle will be isosceles.

B

D

A

For, if AB and AC are not equal, suppose one of them, as AB, to be the greater. On this, take BD equal to AC (Post. 3), and draw DC. Then, in the triangles ABC, DBC, we have the side BD equal to AC, by construction, the side BC common, and the included angle ACB equal to the included angle DBC, by hypothesis: hence, the two triangles are equal

in all their parts (P. V.). But this is impossible, because a part cannot be equal to the whole (A. 8): hence, the hypothesis that AB and AC are unequal, is false. must, therefore, be equal; which was to be proved.

Cor. An equiangular triangle is equilateral.

They

PROPOSITION XIII. THEOREM.

In any triangle, the greater side is opposite the greater angle; and, conversely, the greater angle is opposite the greater side.

In the triangle ABC, let the angle ACB be greater than the angle ABC: then will the side AB be greater than the side AC.

C

A

For, draw CD, making the angle BCD equal to the angle B (Post. 7): then, in the triangle DCB, we have the angles DCB and DBC equal: hence, the opposite sides DB and DC are equal (P. XII.). In the triangle ACD, we have (P. VII.),

AD + DC > AC;

or, since DC = DB, and AD + DB = AB, we have,

which was to be proved.

AB > AC;

Conversely: Let AB be greater than AC: then will the angle ACB be greater than the angle ABC.

For, if ACB were less than ABC, the side AB would be less than the side AC, from what has just been proved; if ACB were equal to ABC, the side AB would be equal to AC, by Prop. XII.; but both conditions are contrary

to the hypothesis: hence, ACB can neither be less than, nor equal to, ABC; it must, therefore, be greater; which was to be proved.

PROPOSITION XIV. THEOREM.

From a given point only one perpendicular can be drawn to a given straight line.

Let A be a given point, and AB · a perpendicular to DE: then can no other perpendicular to DE be drawn through A.

F

For, suppose a second perpendicular AC to be drawn. Prolong AB till BF is equal to AB, and draw CF Then, the triangles ABC and FBC will have AB equal to BF, by construction, CB common, CB common, and the included angles ABC and FBC equal, because both are right angles hence, the angles ACB and FCB are equal (P. V.) But ACB is, by a hypothesis, a right angle: hence, FCB must also be a right angle, and consequently, the line ACF must be a straight line (P. IV.). But this is impossible (A. 11). The hypothesis that two perpendiculars can be drawn is, therefore, absurd; consequently, only one such perpendicular can be drawn; which was to be proved.

B

C

If the given point is on the given line, the proposition is equally true. For, if through A two perpendiculars AB and AC could be drawn to DE, we should have BAE and CAE each equal to a right angle; and consequently, equal to each other; which is absurd (A. 8).

D

E