Question

Maths class 9 NCERT Solution for Exercise- 7.3 Q1 (iii) and (iv) part.
pls no faltu answers pls​

Answer 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.

(iv) BPD = CPD (by CPCT as ΔBPD ΔCPD)

and BP = CP — (i)

also,

BPD +CPD = 180° (Since BC is a straight line.)

⇒ 2BPD = 180°

⇒ BPD = 90° —(ii)

Now, from equations (i) and (ii), it can be said that

AP is the perpendicular bisector of BC.

I hope it helps you

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