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In an isosceles ∆ABC, AD is the median to the unequal side meeting BC at D. DP is the angle bisector of ∠ADB and PQ is drawn parallel to BC meeting AC at Q. Then the measure of ∠PDQ is :
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- 130°
- 90°
- 180°
- 45°
- 130°
Correct Option: B
On the basis of given in question , we draw a figure of an isosceles triangle ABC ,
Given AB = AC
Point D is the mid-point of side BC.
∴ ∠ADB = 90° = ∠ADC
PD is internal bisector of ∠ADB.
∴ ∠PDA = 45°
PQ || BC
∴ ∠ADQ = 45°
∴ PDQ = ∠ADQ + ∠PDA = 45° + 45° = 90°
