Question

ABC is a triangle. A circle touches sides AB and AC produced and side BC at X, Y and Z respectively. Show that AX=1/2 perimeter of triangle ABC.

Answer 1

Tangents from an exterior point to a circle are equal in length.
 
 ∴ BP = BQ ----- -(i)
   CP = CR ------(ii)
   and, AQ = AR ----- (iii)
 
AQ = AR
 ⇒ AB + BQ = AC + CR
 ⇒ AB + BP = AC + CP...(iv) [From eqn (i) and eqn (ii)]
 
Perimeter of ΔABC
= AB + BC + AC
= AB + (BP + PC) + AC
= (AB + BP) + (AC + PC)
= 2(AB + BP) [From eqn (iv)]
= 2 AQ [From eqn (i)]
 ⇒ AQ = 1/2(Perimeter of ΔABC)

Answer 2

Sol:

Given: A circle touching the side BC of ΔABC at P and AB, AC produced at Q and R respectively.

RTP: AP = 1/2 (Perimeter of ΔABC)

Proof: Lengths of tangents drawn from an external point to a circle are equal.
⇒ AQ = AR, BQ = BP, CP = CR.
Perimeter of ΔABC = AB + BC + CA
= AB + (BP + PC) + (AR – CR)
= (AB + BQ) + (PC) + (AQ – PC) [AQ = AR, BQ = BP, CP = CR]
= AQ + AQ
= 2AQ

⇒ AQ = 1/2 (Perimeter of ΔABC)

∴ AQ is the half of the perimeter of ΔABC.