Triangle ABC is isosceles in which AB=AC circumscribed about a circle. Prove that base is bisected by the point of contact.
@Studious: You had provided the correct answer. Keep Posting!!
Still the detailed solution is as:
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.