Question

ABC is an isoceles ∆ in which AB=AC, circumscribed about a circle.Show that BC is bisected at the point of contact.

Answer 1

Sol:

The tangents drawn from an externel point to a circle are equal in length.

∴ AP = AQ  ............ (1)

 BP = BR  ................ (2)

 CQ = CR    .............. (3)

Given that ABC is an isosceles triangle with AB = AC.

Subtract AP on both sides, we obtain

AB – AP = AC – AP

⇒ AB – AP = AC – AQ  (from (1))

∴ BP = CQ.

⇒ BR = CQ  (from (2))

⇒ BR = CR  (from (3))

 ∴BR = CR, it shows that BC is bisected at the point of contact R.
Hope it helps.