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 ∆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

Answer:see below

Step-by-step explanation: