P. G.

Proposition V. The angles at the base of an isosceles triangle are equal.

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

တ

PROPOSITION XX. THEOREM.

Any two sides of a triangle are together greater than the third side.

GIVEN the triangle ABC;

IT IS REQUIRED TO PROVE that BA and AC are together greater than CB; and AB and BC together greater than CA; and BC and CA together greater than AB.

Because AD is equal to AC;

therefore the angle ADC is equal to the angle ACD.

Because the angle BCD is greater than the angle ACD,
and the angle ACD is equal to the angle ADC;
therefore the angle BCD is greater than the angle ADC.

Because the angle BCD is greater than the angle BDC,
therefore BD is greater than BC.

Because AD is equal to AC,

let BA be added to each;

Post. 1.

Constr.

I. 5.

Axiom 9.

Proved.

u.

Proved.

1. 19.

Constr.

therefore BD is equal to BA and AC.

Axiom 2.

Proved.

Proved.

t.

Because BD is equal to BA and AC,

and BD is greater than BC;

therefore BA and AC are greater than BC.

In the same manner it may be shewn that AB and BC are greater

than CA, and BC and CA greater than AB.

Therefore, any two sides &c.

Q. E. D.

Postulate 2. A terminated straight line may be produced to any length in a straight line.

Postulate 1. A straight line may be drawn from any one point to any other point.

Proposition III. From the greater of two given straight lines a part may be cut off equal to the less.

Proposition V. The angles at the base of an isosceles triangle are equal to one another.

Axiom 9. The whole is greater than its part.

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

Axiom 2. If equals be added to equals the wholes are equal.

PROPOSITION XXI. THEOREM.

If from the ends of the side of a triangle there be drawn two straight lines to a point within the triangle, these shall be less than the other two sides of the triangle, but shall contain a greater angle.

In the triangle ABC;

GIVEN that D is a point within it, and BD and DC are joined;

IT IS REQUIRED TO PROVE that BD and DC are together less than BA and AC, and that the angle BDC is greater than the angle BAC.

B

E

Produce BD,

and let the part produced meet AC at E.

Because BA and AE are together greater than BE,
let EC be added to each;

therefore BA and AC are together greater than BE and EC.

Because DE and EC are together greater than DC,

let DB be added to each;

therefore BE and EC are together greater than BD and DC.

Post. 2.

y.

I. 20.

W.

I. 20.

W.

Because BA and AC are together greater than BE and EC, Proved. and BE and EC are together greater than BD and DC; Proved. therefore BA and AC are together greater than BD and DC.

Because the angle BDC is greater than the angle DEC,
and the angle DEC is greater than the angle BAE;
therefore the angle BDC is greater than the angle BAE.

Therefore, if from the ends &c.

V.

I. 16.

I. 16.

V.

Q. E. D.

Postulate 2. A terminated straight line may be produced to any length in a straight line.

Proposition XX. Any two sides of a triangle are together greater than the third side.

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