Question

In following fig., a circle is touching the side BC of Δ ABC at P and AB and AC produced at Q and R respectively. Prove that AQ is half the perimeter of Δ ABC. 

Answer 1

To prove:· AQ = `1/2` (Perimeter of Δ ABC)

Proof : BQ = BR = 5 - r   ....(1)  PC = CR = 12 - r    .....(2)  ] (1)
(Lengths of tangents drawn from an external point to a circle are equal)

Perimeter of 6 ABC = AB + BC+ AC 

= AB + BP + PC + AC 

= AB+BQ+CR+AC           Using (1) 

= AQ+AR 

= 2 AQ 

2 AQ = Perimeter of Δ ABC 

AQ = `1/2` (Perimeter of Δ ABC)