Question

the side AB,BC and CA of Δ ABC touch a circle with centre O at P. Q and R respectively. prove that: AB + CQ=AC+BQ.​

Answer 1

AQ = AR, BP = BQ, CP =CR tangents from exterior points.

AQ = AR

AB +BQ = AC +CR

AB + BP = AC + CP. BQ = BP and CR = CP

Perimeter of ∆ABC

AB + BC + CA

= AB + ( BP + PC ) + CA

= AB + BP +(AC + PC)

= 2AB + 2BP

= 2 ( AB + BP )

= 2AQ

$= \frac{1}{2} \: perimeter \: of ∆abc$

hop this will help you...

-3idiots29

Answer 2

AQ = AR, BP = BQ, CP =CR tangents from exterior points.

AQ = AR

AB +BQ = AC +CR

AB + BP = AC + CP. BQ = BP and CR = CP

Perimeter of ∆ABC

AB + BC + CA

= AB + ( BP + PC ) + CA

= AB + BP +(AC + PC)

= 2AB + 2BP

= 2 ( AB + BP )

= 2AQ

$= \frac{1}{2} \: perimeter \: of ∆abc$

hop this will help you...