Question

The incircle of a triangle $ABC$ touches the sides $AB,BC$ and $CA$ at the points $P,Q$ and $R$ respectively. Show that $AP+BQ+CR=BP+QC+RA=\frac{1}{2}$$Perimeterof∆ABC$

Answer 1

Step1: Draw an appropriate diagram according to the question

Following figure shows a triangle $\mathrm{ABC}$ and incircle touching the sides $\mathrm{AB},\mathrm{BC}$ and $\mathrm{CA}$ at the points $\mathrm{P},\mathrm{Q}$ and $\mathrm{R}$ respectively.

We know that tangents drawn from a fixed point to a circle are the same length,

$\therefore AR=AP$

$BP=BQ$

and $CQ=CR$

Step2 : Using obtained relation prove the required result

Adding all we get $AR+BP+CQ=AP+BQ+CR$

$\mathrm{Perimeter}\mathrm{of}∆ABC=AB+BC+AC$

$\text{Perimeter of}∆ABC=(AP+PB)+(BQ+QC)+(AR+RC)$

$\text{Perimeter of}∆ABC=(AP+AR)+(BQ+BP)+(CQ+CR)$

$\text{Perimeter of}∆ABC=2AR+2BP+2CQ$

$\text{Perimeter of}∆ABC=2(AR+BP+CQ)$

$\Rightarrow AR+BP+CQ=\frac{1}{2}(\mathrm{P}\text{erimeter of}∆ABC)$

Thus $AP+BQ+CR=BP+QC+RA=\frac{1}{2}$$\left(Perimeterof∆ABC\right)$

Hence Proved.