In the given figure, the incircle of `DeltaABC` touches the sides `BC`, `CA` and `AB` at `P`, `Q` and `R` respectively. Prove that
`(AR+BP+CQ)=(AQ+BR+CP)`
`=(1)/(2)("perimeter of"DeltaABC)`.

Explanation

We know that the lengths of tangents from an exterior point to a circle are equal.
`:. AR=AQ`, ……….`(i)` [tangents from `A`]
`BP=BR`, ………`(ii)` [tangents from `B`]
`CQ=CP` ………….`(iii)` [tangents from `C`]
`:. (AR+BP+CQ)=(AQ+BR+CP)=k` (say).
Perimeter of `DeltaABC=(AB+BC+CA)`
`=(AR+BR)+(BP+CP)+(CQ+AQ)`
`=(AR+BP+CQ)+(AQ+BR+CP)`
`=(k+k)=2k`
`impliesk=(1)/(2) ("perimeter of " DeltaABC)` .
`:. (AR+BP+CQ)=(AQ+BR+CP)`
`=(1)/(2)("perimeter of" DeltaABC)`.