A circle is touching the side `BC` of `DeltaABC` at `P` and touching `AB` and `AC` produced at `Q` and `R` respectively. Prove that `AQ=(1)/(2)("perimeter of"DeltaABC)`.

Explanation

We know that the lengths of tangents drawn from an external point to a circle are equal.
`:.AQ=AR`, ……….`(i)` [tangents from `A`]
`BP=BQ`, ………..`(ii)` [tangents from `B`]
`CP=CR`. ………..`(iii)` [tangents from `C`]
Perimeter of `DeltaABC`
`=AB+BC+AC`
`=AB+BP+CP+AC`
`=AB+BQ+CR+AC` [using `(ii)` and `(iii)`]
`=AQ+AR`
`=2AQ`[ using `(i)` ]
`:.AQ=(1)/(2)("perimeter of"DeltaABC)`.