Question

A circle is touching the side BC of triangle ABC at P and touching AB and AC produced Q and R respectively . Prove that AQ = 1/2[Perimeter of triangle ABC.

Answer 1

Tangents drawn from an external point to the circle are equal.
BP = BQ  ….(1)
CP = CR  ….(2)
AQ = AR  ….(3)
Now, AQ = AR
⇒ AB + BQ = AC + CR
⇒ AB + BP = AC + CP  (Using (1) and (2)

Perimeter of ∆ABC = AB + BC + CA
  = AB + (BP + PC) + AC
  = (AB + BP) + (PC + AC)
  = 2(AB + BP)  ( AB + BP = AC + CP)
  = 2(AB + BQ)  (Using (1))
  = 2AQ
hence, AQ = 1/2(perimeter of triangle ABC)