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)
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