A circle touches the side Bc of a triangle ABC at a point P and touches AB and AC when produced at Q and R respectively. Show that
- AQ = 1/2 ( perimeter of triangle ABC)
Here is the answer to your query.
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))