Question

If the bisectors of $\angle Q$ and $\angle R$ of a triangle $△PQR$ meet at point $S$, then prove that
$\angle QSR={90}^{\circ }+\frac{1}{2}\angle P$.

Answer 1

given:In$△PQR$,Bisectors of $\angle Q\&\angle R$ meet at point S.

$ToProve:\angle QSR={90}^{\circ }+\frac{1}{2}\angle P$

$Proof:Let\angle PQR=\angle Q;\angle QRP=\angle R;\angle QPR=2P$

$In△PQR$

$\angle P+\angle Q+\angle R={180}^{\circ }$ (sum of angles of a triangle)

$⟹\angle Q+\angle R=180-P-\left(1\right)$

$In△SQR$

$\angle QSR+\frac{\angle Q}{2}+\frac{\angle R}{2}={180}^{\circ }$ (sum of angles of a triangle)

$\frac{\angle Q+\angle R}{2}=180-\angle QSRfrom\left(1\right)$

$\frac{180-\angle P}{2}=180-\angle QSR$

$⟹90-\frac{\angle P}{2}=180-\angle QSR$

$⟹\angle QSR=90+\frac{\angle P}{2}$

Hence,proved