Solve both the qns plis

Dear Student,

Q15.

We know that the tangents drawn from an exterior point to a circle are equal in length.

∴ AR = AQ          (Tangents from A)  ..... (1)

 BR = BP           (Tangents from B)  ..... (2)

 CQ = CP             (Tangents from C)  ..... (3)

Now, the given triangle is isosceles, so given AB = AC

Subtract AR from both sides, we get

AB – AR = AC – AR

⇒ AB – AR = AC – AQ                (Using (1))

BR = CQ

⇒ BP = CQ              (Using (2))

⇒ BP = CP                  (Using (3))

So BP = CP, shows that BC is bisected at the point of contact.

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