Question

ABCD is an isosceles trapezium in which $\overline{AB}\parallel \overline{CD}$ , If the bisectors of $\angle BAD$ and $\angle ADC$ intersect at P. Find $\angle APD$

• A. ${45}^{\circ }$
• B. ${60}^{\circ }$
• C. ${90}^{\circ }$
• D. ${120}^{\circ }$

Answer 1

The correct option is C ${90}^{\circ }$

Since ABCD is isosceles trapezium

So angle DAB =angle ABC =$x$ (say)

and Angle ADC = angle DCB = $y$ (say)

Sum of all angles of quadrilateral is 360

So $x$ + $x$ +$y$ + $y$ =360

$x$+$y$ =180

Angle ADP = $\frac{y}{2}$ and angle DAP =$\frac{x}{2}$ (Since DP and AP are angle bisectors)

$\frac{y}{2}$+$\frac{x}{2}$ = 90

In triangle APD

$\frac{y}{2}$+$\frac{x}{2}$ + angle DPA =180

So angle DPA = 90

So correact answer is Option C