Join the vertex A

BC. Then, AB common, and D

is

it

AD

and the middle point "D of the base equal to AC, by hypothesis, 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

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

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

They

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.

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.

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 bave (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 augle 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 conclusions 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 then can no other perpendicular to 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 will have AB equal to BF, by construction, CB common, and the included. angles ABC and FBC equal, because both are right an gles: 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 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. s).

D

-E

A

:

PROPOSITION XV. THEOREM.

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

1°. The perpendicular will be 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, will be 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 will be 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 will AB be less than any of the oblique lines, AC will be equal to AE, and AD greater

than A C.

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

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 and ABE equal,

because both are right angles: hence, AC is equal to AE; which was to be proved.

3o. It may be shown, as in the first case, that AD is equal to DF. Then, because the point triangle ADF, the sum of the lines AD

Clies within the

and DF will be greater than the sum of the lines AC and CF (P. VIII.): hence, AD, the half of ADF, is greater than AC, the half of ACF; which was to be proved.

Cor. 1. The perpendicular is the shortest distance from a point to a line.

Cor. 2. From a given point to a given straight line, only two equal straight lines can be drawn; for, if there could be more, there would be at least two equal oblique lines on the same side of the perpendicular; which is impossible.

If

a perpendicular be drawn to a given straight line at its middle point:

1°. Any point of the perpendicular will be equally distant from the extremities of the line:

2o. Any point, without the perpendicular, will be unequally distant from the extremities.

Let АВ be a given straight line, C its middle point, and EF the perpendicular. Then will any point of EF be equally distant from A and B; and any point without EF, will be unequally distant from A and B. A

1o. From any point of EF, as D, draw the lines DA and DB. Then will DA and DB be equal (P. XV.): hence, D is

D

B

equally distant from 4 and B; which was to be proved