1. 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
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By: anil on 05 May 2019 01.50 am
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 = $$90^circ$$ $$ecause$$ 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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Solution : The median of an isosceles triangle bisects the opposite side at right angle, => $$angle$$ ADC = $$90^circ$$ $$ecause$$ 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)