Question

In the figure, $\angle Q>\angle R$, PA is the bisector of $\angle QPR$ and $PM\perp QR$. Prove that $\angle APM=1/2(\angle Q-\angle R)$ Find a

Answer 1

$In△PAQ$

$Sumoftwooppositeinteriorangles=Exteriorangle$

$\alpha =\theta +\angle \theta -\left(i\right)$

$In△PAR$

$\alpha +\theta +\angle R={180}^{\circ }$

$\alpha ={180}^{\circ }-\theta -\angle R-\left(ii\right)$

$In△PMQ$

$\angle MPQ=180-\angle Q-{90}^{\circ }$

$=90-\angle Q$

$\angle APM=\angle APQ-\angle MPQ$

$=\theta -(90-\angle Q)$

$=\theta +\angle Q-{90}^{\circ }-\left(iii\right)$

$\because \left(i\right)=\left(ii\right)$

$=>\theta +\angle Q=180-\theta -\angle R$

$=>2\theta =180-\angle Q-\angle R$

$=>\theta =90-\frac{1}{2}(\angle Q+\angle R)$

$puttingthisvalueof\theta in\left(iii\right)$

$\angle APM=\angle Q+90-\frac{1}{2}(\angle Q+\angle R)-90$

$=\frac{\angle Q}{2}-\frac{\angle R}{2}$

$=\frac{1}{2}(\angle Q-\angle R)$

$\therefore a=2$