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Proof.
We have, now, two triangles DB C and ACB

(DB C lying on ACB).

B

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In those triangles,
DB is equal to AC (6),
the two sides DB, B C are equal to

the two sides A C, CB (B C being common), and angle D B C is equal to angle A CB (Given a).

, (C) The triangle D B C is equal to the tri

angle ACB. (d) But, the triangle DBC is less than the

triangle A CB (self-evident). ... If A B is not equal to A C, the triangle

DBC is both equal to and less than

the triangle A CB; which is absurd. .. A B is not unequal to AC, that is, A B is equal to A C. Q. E. D.

NOTE.The side of a triangle opposite to any particular angle is said to subtend that angle.

EXERCISES.—I. Given.-AC equal to A D, and BC equal to BD (Fig. 1).

Required.—To prove that triangle A CB is equal to triangle A D B, without using Euc. I. 8. (Join CD, and use Euc. I. 5,

and then I. 4.)

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II. Given.-The two isosceles triangles A CB, ADB (Fig. 2).

Required.—To prove that AE is equal to E B. (Apply Euc. I. 8 to the triangles CÒA, CDB. Then Euc. 1. 13 to the angles at D. Then apply Euc. 1. 4 to the triangles ADE, B D E.)

THEOREM (Euclid 1. 7). Repeat.-—The enunciation of Euc. I. 5 and Axiom 9. General Enunciation. On the same base, and on the same side

of it, there cannot be two triangles having their sides which are terminated at one extremity of the base equal to one another, and likewise those sides which are terminated at the other ex

tremity equal to one another. The same, in tabular form.

GIVEN.

REQUIRED.

Two triangles on the To prove that, same base, and on the the two sides terminated same side of the base. at one end of the base

cannot be equal if those terminated at the other

end of the base are equal. Particular Enunciation (Case I). Given.—The two triangles CAB, DAB, on the

same base A B, and on the same side of it, and

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А.

B having the vertex of each triangle without the

other. Required.—To prove that it is not possible that

the two sides CA, DA should be equal, and, at the same time, the two sides , DB be

equal (see explanation of this on page 74). Construction.

Join CD

Proof.
Let us suppose (a) that C A and D A are equal,

and also that
(6) C B and D B are equal.

A А

B

Then ACD is an isosceles triangle, having side

A C equal to side A D (a). (c) ... by Euc. I. 5, angle ACD is equal to

angle ADC. But angle A C D is greater than angle B C D.

by Axiom 9, angle A D C is greater than angle BCD (c). (d) and angle BDC is still greater than

angle BCD. Again, BDC is an isosceles triangle having side

BD equal to side B C (6). (e) .. by Euc. I. 5, angle BDC is equal to

angle BCD, but we have proved that
angle B D C is greater than angle BCD (d).

if A C and A D are equal, and also B D and B C equal, then angle BDC is both greater than (d) and equal

to (c) angle BCD; But this is impossible: and so it is impossible

for A C and A D to be equal at the same time that B D and DC are equal.

Q. E. D.

Particular Enunciation (CASE II).
Given.—The two triangles CAB, DAB, on

the same base A B, and on the same side of it;

B

but having the vertex of one triangle within the

other triangle. Required. To prove that A C and A D cannot

be equal at the same time that D C and D B

are equal. Construction.

Join CD. (See Fig. on page 98.)

Produce A C to E, and A D to F.
Proof.
Let us suppose, as before,

(a) that A C, A D are equal, and

(6) that B C, B D are also equal. Then ACD is an isosceles triangle having its equal

sides A C, A D (a) produced to E and F. (c) ... by Euc. I. 5, angle ECD is equal to

angle FDC, But angle ECD is greater than angle BCD. .. by (c) angle FDC is greater than angle BCD. (d) Still greater is angle BDC than angle

BCD. Again,
BDC is an isosceles triangle, having
side B C equal to side BD (6).

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