In an isosceles ΔABC, AD is the median to the unequal side meeting BC at D. DP is the angle disector of ∠ADB and PQ is drawn parallel to BC meeting AC at Q. Then the maeasure of ∠PDQ is

Given : ABC is an isosceles triangle and AD is the median and PD is the angle bisector.

To find : $$\angle$$ PDQ = ?

Solution : The median of an isosceles triangle bisects the opposite side at right angle, => $$\angle$$ ADC = $$90^\circ$$

$$\because$$ PD is angle bisector, => $$\angle$$ PDR = $$\frac{90}{2}=45^\circ$$ ------------(i)

and DQ will bisect $$\angle$$ RDC

=> $$\angle$$ RDQ = $$45^\circ$$ ----------(ii)

Adding equations (i) and (ii), we get :

=> $$\angle PDR + \angle RDQ = 45^\circ+45^\circ$$

=> $$\angle PDQ = 90^\circ$$

=> Ans - (B)