#### In Fig. 6.17, $\angle Q > \angle R$, PA is the bisector of $\angle QPR$ and $PM \perp QR$. Prove that  $\angle APM = \frac{1}{2} (\angle Q -\angle R)$.

Solution:

Given : In $\triangle PQR$, $\angle Q > \angle R$

PA is the bisector of $\angle QPR$  and $PM \perp QR$.

To prove : $\angle APM =\left ( \frac{1}{2} \right ) (\angle Q -\angle R)$
Proof: Since PA is the bisector of $\angle P$, we have

$\angle APQ = \left ( \frac{1}{2} \right )\angle P$                        … (i)
In right angled $\triangle PMQ$ we have

$\angle Q + \angle MPQ + 90^{\circ} = 180^{\circ}$ (Angle sum property)

$\Rightarrow \angle MPQ = 90^{\circ} - \angle Q$              … (ii)
Now
$\angle APM = \angle APQ - \angle MPQ$
$\angle APM =\frac{1}{2} \angle P - (90 - \angle Q)$         using (i) & (ii)
$\angle APM = \frac{1}{2} \angle P - 90^{\circ} + \angle Q$

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

Hence proved