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(DB C lying on ACB).
In those triangles,
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.)
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.
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
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 DĀ, DB be
equal (see explanation of this on page 74). Construction.
and also that
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
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).
the same base A B, and on the same side of it;
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.
(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