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1. Angle BAC is not equal to angle EDF, for if it were (by Euc. I. 4) the base BC would be equal to the base E F, but it is not.

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II. Angle BAC is not less than angle EDF, for if it were (by Euc. I. 24) the base BC would be less than the base E F, but it is not. III. Then as angle BAC is neither equal to (1) nor less than (11) angle EDF, it must be greater than angle EDF (d).

THEOREM (Euclid I. 26).

Q. E. D.

Repeat. The enunciations of Euc. I. 4 and 16, and Axioms I and 9.

General Enunciation.

If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, namely, either the side adjacent to the equal angles, or the side opposite to equal angles in each; then shall the

other sides be equal each to each, and also the third angle of the one equal to the third angle of the other. The same, in tabular form.

GIVEN.

Two triangles, having (a) two angles of the one equal to two angles of the other, each to each; (b) one side equal to one side.

Particular Enunciation.

REQUIRED.

To prove that(a) The other sides are equal, each to each. (b) The third angle of one is equal to the third angle of the other.

Given. The two triangles ABC, DEF, having (a) angle A B C equal to angle DE F. (b) angle A C B equal to angle D F E. (c) and one side equal to one side.

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Now, of the three sides in each of these triangles, one side in each, namely, BC and E F, is said to be adjacent to the equal angles, because it is a side of each angle. The other two sides in each triangle, namely, BA and AC in AB C, and ED and DF in DEF, are said to be opposite to equal angles.

A different construction is required for the proof, according as we have given an adjacent side or an opposite side.

CASE I.-Let the two sides given equal, (d) be the sides BC and EF, adjacent to the equal angles. Required. To prove that

Proof.

(e) AB is equal to DE,
(f) AC is equal to D F,

(g) angle BAC is equal to angle EDF.

Let us suppose that A B is not equal to DE; then, one of the two must be greater than the other.

Let us suppose AB to be greater than DE.
And let us suppose (1⁄2) B G equal to D E.
Join G C.

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Then in the two triangles G BC, DEF,
GB is equal to DE (h), and

BC is equal to EF (Given d),

and angle GBC is equal to angle DEF (Given a). (i) ... (by Euc. I. 4) angle G C B is equal to angle DFE.

Again,

(k) angle ACB is greater than angle GCB, but angle ACB is equal to angle DFE (Given b). And angle DFE is equal to angle G CB (i). (2).. angle AC B is equal to angle G C B.

So that by supposing A B unequal to DE we are able to prove that the whole angle ACB is equal to its part G C B: which is impossible (Axiom 9).

... A B is not unequal to DE, that is, (m) A B is equal to DE, and

BC is equal to EF, and

the included angle ABC is equal to the included angle DEF (Given a).

.. (by Euc. I. 4) the base AC is equal to the base D F, and the remaining angle BAC is equal to the remaining angle

Q. E. D.

EDF. CASE II. Let the sides given equal be the (d) sides A B and DE, opposite to the equal angles ACB and DFE (see Fig. on next page). Required. To prove that

(e) BC is equal to EF,

AC is equal to DF, and

(g) angle BAC is equal to angle EDF. Proof.

Let us suppose that BC is not equal to EF, then one of the two must be greater than the other.

Let us suppose B C to be greater than E F. (h) And let us suppose BH to be equal to EF. Join AH.

Then in the two triangles A B H and DEF, AB is equal to DE (Given d).

BH is equal to EF (h),

and angle ABH is equal to angle DEF; (i). (by Euc. I. 4) angle BHA is equal to angle EF D.

Again,

(k) Angle BHA is greater than angle BCA (Euc. I. 16).

And angle BCA is equal to angle EFD (Given b). (1) .. angle BHA is greater than angle EFD. So that, by supposing B C to be unequal to EF, we are able to prove that angle BHA is

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at the same time both equal to (i) and greater than (1) EFD, which is impossible. .. BC is not unequal to EF; that is to say, BC is equal to EF,

and, AB is equal to DE (Given d),

and angle ABC is equal to angle DE F

(by Euc. I. 4) base AC is equal to base DF, and angle B A C is equal to angle EDF.

Q. E. D.

EXERCISE.-CASE III. Let the sides given equal be the sides AC and DF, opposite to the equal angles ABC and DEF

Required. To prove that

(a) BC is equal to EF.
(b) A B is equal to DE, and

(c) angle B A C is equal to angle EDF.

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