Question

From the following figure, prove that :
$AP+BQ+CR=BP+CQ+AR=\frac{1}{2}\times$ $perimeterof△ABC$

Answer 1

Perimeter of $△ABC=AB+BC+CA$

Since the inner circle cuts each side of triangle into 2 equal halves, we have,

$AB=AP+PB$

Similarly, $BC=BQ+QC$ and $CA=CR+RA$

Therefore perimeter of $△ABC=AB+BC+CA$

$=(AP+PB)+(BQ+QC)+(CR+RA)$

$=(AP+BQ+CR)+(PB+QC+RA)$

$\therefore AP+BQ+CR=PB+QC+RA=\frac{1}{2}$ perimeter of $△ABC$