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# ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig. 7.39). If AD is extended to intersect BC at P, show that (iii) AP bisects A as well as D. Q1(iii)

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Give me the best solution of this question.This is very important question of ncert class 9th of chapter triangles. How I solve the best solution of exercise 7.3 question number 1(iii). Please help me to solve this in a easy and best way.ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig. 7.39). If AD is extended to intersect BC at P, show that (iii) AP bisects A as well as D.

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1. (iii) PAB = PAC by CPCT as ΔABD ΔACD.

AP bisects A. — (i)

Also, ΔBPD and ΔCPD are similar by SSS congruency as

PD = PD (It is the common side)

BD = CD (Since ΔDBC is isosceles.)

BP = CP (by CPCT as ΔABP ΔACP)

So, ΔBPD ΔCPD.

Thus, BDP = CDP by CPCT. — (ii)

Now by comparing (i) and (ii) it can be said that AP bisects A as well as D.

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