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)

Hi!

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))

= 2AQ

Cheers!

**
**