Question

ABC is an isoceles triangle in which AB = AC, circumscribed about a circle
Prove that base is bisected by point of contact ...
Pls provide me with the answer ASAP

Answer 1

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

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

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

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

 

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

 

Subtract AP from both sides, we get

AB – AP = AC – AP

⇒ AB – AP = AC – AQ  (Using (1))

BP = CQ

 

⇒ BR = CQ  (Using (2))

⇒ BR = CR  (Using (3))

 

So BR = CR, shows that BC is bisected at the point of contact.

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