Question

$\text{In parallelogram ABCD, AR and DP are}\text{the angle bisectors of}\angle A\text{and}\angle D\text{respectively. If}\angle ADS=45°,\text{then find}\text{the value of}\angle PSR.$

• A. 75°
• B. 135°
• C. 45°
• D. 90°

Answer 1

The correct option is D 90°

Given figure,

$\text{Given:}\angle ADS=45°$
$\text{AR and DP are angle bisectors.}$

$\angle ADS=45°\text{and PD is the angle}\text{bisector.}\text{Hence,}\angle SDC={45}^{\circ }$

$\angle BAD+\angle ADC={180}^{\circ }$
$\text{(Adjacent angles in a}\text{parallelogram are supplementary)}$
$\angle ADC={180}^{\circ }-{90}^{\circ }\angle ADC={90}^{\circ }...\left(i\right)$

$\angle A=\angle DAS+\angle SAB\text{AR is the angle bisector.}\text{Hence, from (i)}\angle DAS=\angle SAB={45}^{\circ }$

$\text{In triangle ASD,}\angle DSA+\angle SAD+\angle ADS=180°\text{(Angle sum property of triangle}ASD)$
$\angle DAS+{45}^{\circ }+{45}^{\circ }={180}^{\circ }\angle DAS={180}^{\circ }-{90}^{\circ }\angle DAS={90}^{\circ }$

$\angle PSR\text{and}\angle DSA\text{are vertically}\text{opposite angles.}$
$\therefore \angle DSA=\angle PSR=90°$