Question

AB,BC,CA are sides of triangle ABC touch the circle at P,Q and R respectively. Prove that AB+CQ=AC+BQ.

Answer 1

hope this helps you with ur problem

Answer 2

Since the lengths of tangents drawn from an exterior point to a circle are equal.
∴ AP = AR ,
   BP = BQ,
   CQ = CR
Now, AB+CQ = AP+PB+CQ
                      = AR+BQ+CQ  (∵ AP=AR and PB=BQ)
                      =(AR+CR)+BQ (∵CQ=CR)
                      = AC+BQ
∴ AB+ CQ = AC+BQ
Hence, proved.
Hope it helps you ^_^