In the given figure the sides AB, BE and CA of triangle ABC touch a circle with centre O and radius r at P,Q and R respectively.
Prove that : `(i) AB+CQ=AC+BQ`
(ii) Area `(triangleABC)=(1)/(2)("Perimeter of"triangleABC)xxr`

Explanation

We know that the lengths of tangents from an exterior point to a circle are equal.
`:. AP=AR`, ……….`(i)` [tangents from `A`]
`BP=BQ`, ………`(ii)` [tangents from `B`]
`CQ=CR` ………….`(iii)` [tangents from `C`]
`(a) AB+CQ=AP+BP+CQ`
`=AR+BQ+CR` [using `(i)`, `(ii)` and `(iii)`]
`=(AR+CR)+BQ=AC+BQ`.
`(b)` Join `OA`, `OB` and `OC`.
Area `(DeltaABC)="area"(DeltaOAB)+"area"(DeltaOBC)+"area"(DeltaOCA)`
`=((1)/(2)xxABxxOP)+((1)/(2)xxBCxxOQ)+((1)/(2)xxCAxxOR)`
`=((1)/(2)xxABxxr)+((1)/(2)xxBCxxr)+((1)/(2)xxCAxxr)`
`=(1)/(2)(AB+BC+CA)xxr`
`=(1)/(2)("perimeter of"DeltaABC)xxr`.