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

In the triangles ABC and DEF, let AB be equal to DE, AC to DF, and BC to EF: then are the triangles equal in all respects.

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 were less than

B

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 respects (P. V.); which was to be proved.

Scholium. In triangles, equal in all respects, 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 the angle C is equal to the ingle B.

Join the vertex A and the middle point D of the base BC. Then, AB is equal to AC, by hypothesis, AD com

mon, 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 which was to be proved.

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 straight line drawn from the vertex of an isosceles triangle to the middle of the base, bisects the angle at the vertex, and is perpendicular to the base.

PROPOSITION XII. THEOREM.

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

isosceles.

In the triangle ABC, let the angle ABC be equal to the angle ACB: then

is AC equal to AB, and consequently, the triangle is isosceles.

not equal,

B

For, if AB and AC are 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 respects (P. V.). But this is impossible, because a part can not be equal to the whole (A. 8): hence, the hypothesis that AB and AC are unequal, is false. They must, therefore, be equal; which was to be proved.

Cor. An equiangular triangle is equilateral.

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 the side AB is greater than the side AC.

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 the angle ACB is 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 conclusions contradict

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

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 from A.

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 have AB equal to BF, by construction, 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.

If the given point is on the given line, the proposition is equally true. For, if from 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

PROPOSITION XV. THEOREM.

If from a point without a straight line a perpendicular is let fall on the line, and oblique lines are drawn to different points of it:

1o. The perpendicular is shorter than any oblique line.

2°. Any two oblique lines that meet the given line at points equally distant from the foot of the perpendicular, are equal.

3°. Of two oblique lines that meet the given line at points unequally distant from the foot of the perpendicular, the one which meets it at the greater distance is the longer.

Let A be a given point, DE a given straight line, AB a perpendicular to DE, and AD, AC, AE oblique lines, BC being equal to BE, and BD greater than BC. Then AB is less than any of the oblique lines, AC is equal to AE, and AD greater than AC.

Prolong AB until BF is equal to AB, and FC, FD.

B

E

draw

1o. In the triangles ABC, FBC, we have the side AB equal to BF, by construction, the side BC common, and the included angles ABC and FBC equal, because both are right angles: hence, FC is equal to AC (P. V.). But, AF is shorter than ACF (A. 12): hence, AB, the half of AF, is shorter than AC, the half of ACF; which was to be proved.

2o. In the triangles ABC and ABE, we have the side BC equal to BE, by hypothesis, the side AB common, and the included angles ABC and ABE equal, because both are