Question

A triangle ABC is inscribed in a circle. The bisectors of angle BAC, ABC and ACB meet the circumcircle of the triangle at points P, Q and R respectively. Prove that:
(i) $\angle ABC=2\angle APQ,$
(ii) $\angle ACB=2\angle APR,$
(iii) $\angle QPR={90}^0-\frac{1}{2}\angle BAC.$

Answer 1

$Given△ABCisinscribedinC(0,r).$

$Thebisectorsof\angle BAC,\angle ABCand\angle ACBmeetsthecircumcircle$

$of△ABC,inP,Q,Rrespectively.$

$InthefigureJoinRQ,$

$\angle ABQ=\angle APQ-\left(i\right)$

$\left\{Anglesinthesamesegmentofacircleareequal\right\}.$

$\angle ABQ=\angle QBC\left\{BQisthebisectorof\angle ABC\right\}.$

$\therefore \angle QBC=\angle APQ-\left(ii\right)$

$Adding\left(i\right)\&\left(ii\right)$

$\angle ABQ+\angle QBC=\angle APQ+\angle APQ$

$\therefore \angle ABC=2\angle APQ-\left(iii\right)$

$Similarily,\angle ACB=2\angle APR-\left(iv\right)$

$Adding\left(iii\right)\&\left(iv\right)$

$\angle ABC+\angle ACB=2(\angle APQ+\angle APR)$

$\therefore \angle ABC+\angle ACB=2\angle QPR-\left(v\right)$

$In△ABC,$

$\angle ABC+\angle BAC+\angle ACB={180}^{\circ }\left\{Anglesumproperty\right\}$

$\angle ABC+\angle ACB={180}^{\circ }-\angle BAC-\left(vi\right)$

$From\left(v\right)\&\left(vi\right),weget$

${180}^{\circ }-\angle BAC=2\angle QPR$

$\therefore \angle QPR=\frac{1}{2}(180-\angle BAC)$

$\angle QPR=\frac{1}{2}\times {180}^{\circ }-\frac{1}{2}\angle BAC$

$⟹\angle QPR=90-\frac{1}{2}\angle BAC$