Question

A circle touches the side BC of ∆ABC at P and touches AB and AC produced at Q and R, respectively, as shown in the given figure. Show that $AQ=\frac{1}{2}\times (\mathrm{perimeter}\mathrm{of}∆ABC).$
Figure

Answer 1

We know that the lengths of tangents drawn from an external point to a circle are equal.
$$\begin{gathered} \therefore AQ=AR\dots \dots \left(\mathrm{i}\right)\mathrm{}\left[\mathrm{Tangents}\mathrm{}\mathrm{from}A\right] \\ BP=BQ\dots \dots ..\left(\mathrm{ii}\right)\left[\mathrm{Tangents}\mathrm{}\mathrm{from}B\right] \\ CP=CR\dots \dots \dots ..\left(\mathrm{iii}\right)\mathrm{}\left[\text{T}\mathrm{angents}\mathrm{}\mathrm{from}C\right] \\ \mathrm{Perimeter}\mathrm{of}∆ABC=AB+BC+AC \\ =AB+BP+CP+AC \\ =AB+BQ+CR+AC[\mathrm{Using}\left(ii\right)\mathrm{and}\left(iii\right)] \\ =AQ+AR \\ =2AQ\left[\mathrm{Using}\left(i\right)\right] \\ \therefore AQ=\frac{1}{2}\left(\mathrm{Perimeter}\mathrm{}\mathrm{of}∆ABC\right) \end{gathered}$$