but the angle APB is greater than the angle PQB;

much more then is the angle APB greater than the angle ACB*.

Q. E. D.

THEOREM 14.

If two triangles have two sides of the one equal to two sides of the other, each to each, but the included angles unequal, their bases are unequal, the base of that which has the greater angle being greater than the base of the other ↑.

Part. En. Let ABC, DEF be two triangles, having

AB=DE,

AC=DF,

but the included angle BAC greater than the included angle EDF;

it is required to prove that the base BC is greater than

the base EF.

Proof. Place the point A on D, and AB along DE;

then because AB = DE,

therefore the point B will fall on the point E;

(Hyp.)

and because the angle BAC is greater than the angle EDF, the line AC will fall outside DF, as DG;

and BC will fall as EG.

(Hyp.)

Let DH be the bisector of the angle FDG, (Ax. 4.) meeting EG in H.

* Euclid, I. 21. + Euclid, I. 24.

Join FH.

Then because in the triangles FDH, GDH, we have

and the included angle FDH= the included angle GDH,

therefore HF=HG;

(Constr.) (Th. 5.)

and therefore EH and HF together are equal to EG. But EH and HF together are greater than EF; (Th. 12.) therefore EG or BC is greater than EF.

THEOREM 15.

Q. E. D.

If two triangles have the three sides of the one equal to the three sides of the other, each to each, then the triangles are identically equal, and of the angles those are equal which are opposite to equal sides*.

Part. En. Let ABC, DEF be two triangles

then shall the triangles be equal in all respects.

Proof. The angle BAC must be either equal to EDF,

or greater than EDF, or less than EDF.

But BAC is not greater than EDF;

for then the base BC would be greater than the base EF.

(Th. 14.)

Nor is BAC less than EDF;

for then the base BC would be less than the base EF:

(Th. 14.)

therefore the angle BAC is equal to the angle EDF.

Alternative Proof.

If the point B were placed on E, and BC were placed along EF, then because BC=EF (Hyp.), therefore the

point C would fall on F: and let the triangles BAC, EDF fall on opposite sides of EF; BA, AC falling as EG, GF, and the angle BAC as EGF. Join DG.

(Constr.)

therefore the whole angle EDF= the whole angle EGF;

and therefore the triangles BAC, EDF are equal in all respects. (Th. 5.)

Q. E. D.

NOTE. The student should examine for himself the cases in which DG passes through an extremity of the base, and passes outside the base.

THEOREM 16.

If two triangles have two sides of the one equal to two sides of the other, each to each, but the base of the one is greater than the base of the other, then the angle contained by the sides of that which has the greater base is greater than the angle contained by the sides of the other.

Part. En. Let BAC, DEF be two triangles which

but the base BC greater than the base DF;

11

it is required to prove that the angle BAC is greater than the angle DEF.

Proof. For the angle BAC must either be equal to the angle DEF, or less than the angle DEF, or greater than the angle DEF

But the angle BAC is not equal to the angle DEF, for then the base BC would be equal to the base DF; but it is not.

(Th. 5.)

Nor is the angle BAC less than the angle DEF, for then the base BC would be less than the base DF;

If two triangles have two angles of the one equal to two angles of the other, each to each, and have the sides opposite to one of the equal angles in each equal, then the triangles are equal in all respects, those sides being equal which are opposite to the equal angles.

*Euclid I. 25.

Part. En. Let the two triangles ABC, DEF have the

two angles

ABC= DEF,

ACB=DFE,

and the side AB = the side DE;

A A

G F

it is required to prove the triangles are equal in all respects.

Proof. Let the point A be placed on the point D,

AB along DE,

then because AB = DE,

therefore the point B will fall on the point E.

and

(Hyp.)

And because the angle ABC is equal to the angle

DEF,

therefore the line BC will lie along EF

(Hyp.)

And the point C will fall on F, for if it fell otherwise as G, then, since the angle A CB is equal to the angle EFD, (Hyp.) the angle EGD would be equal to the angle EFD, the exterior angle equal to the interior and opposite, which is impossible; (Th. 9.)

therefore the triangles would coincide and are equal in all respects, AC being equal to DF, BC to EF, and the angle BAC to the angle EDF*.

* Eucl. 1. 26. Part 2.

Q. E. D.