present the sides of a triangle, the sum of any two must be greater than the third, and the difference of any two must be less than the third.

PROPOSITION VIII. THEOREM.

If from any point within a triangle two straight lines be drawn to the extremities of any side, their sum will be less than that of the two remaining sides of the triangle.

Let be any point within the triangle BAC, and let the lines OB, OC, be drawn to the extremities of any side, as BC:

then will the sum of BO and Ос be less than the sum of the sides BA and AC.

B

A

AD

Prolong one of the lines, as BO, till it meets the side AC in D; then, from Prop. VII., we shall have,

OC < OD + DC ;

adding BO to both members of this inequality, recollecting that the sum of BO and OD is equal to BD, we have (A. 4),

BO + OC < BD + DC.

From the triangle BAD, we have (P. VII.),

BD < BA + AD ;

adding DC to both members of this inequality, recollecting that the sum of AD and DC is equal to AC, we have,

BD + DC <BA + AC.

But it was shown that BO+ OC is less than BD +- DC; still more, then, is BO+ OC, less than BA + AC; which was to be proved.

PROPOSITION IX. THEOREM.

If two triangles have two sides of the one equal to two sides of the other, each to each, and the included angles unequal, the third sides will be unequal; and the greater side will belong to the triangle, which has the greater included angle.

In the triangles BAC and DEF, let AB be equal to DE, AC to DF, and the angle A greater than the angle D: then will BC be greater than EF.

Let the line AG be drawn, making the angle CAG equal to the angle D (Post. 7); make AG equal to DE, and draw GC. Then will the triangles AGC and DEF have two sides and the included angle of the one equal to two sides and the included angle of the other, each to each; consequently, GC is equal to EF (P. V.).

Now, the point G may be without the triangle ABC, it may be on the side BC, or it may be within the triangle ABC. Each case will be considered separately.

GI+IC > GC, and BI+IA > AB;

whence, by addition, recollecting that the sum of BI and IC is equal to BC, and the sum of GI and IA, to GA, we have,

AG + BC > AB + GC.

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

AB + BC > AB + EF.

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

3o. When G is within the triangle ABC.

From Proposition VIII., we have,

BA+ BC > GA + GC;

D

F

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 BU greater than EF, then will the angle BAC be greater than the angle EDF.

For, if not, BAC must either be equal to, or less than, EDF. In the former case, BC 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: then will the triangles be equal in all their parts.

BC would be greater than EF; and if the angle A were 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 be equal to the angle B.

Join the vertex A

BC. Then, AB is common, and BD

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

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

D

Cor. 1. An equilateral triangle is equiangular.

AD

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 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