Question : In an isosceles $\triangle ABC$, AD is the median to the unequal side BC meeting at D, DP is the angle bisector of $\angle ADB$ and PQ is drawn parallel to BC meeting AC at Q. Then, the measure of $\angle PDQ$ is:
Option 1: 130°
Option 2: 90°
Option 3: 180°
Option 4: 45°
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Correct Answer: 90°
Solution : Given: In an isosceles $\triangle ABC$, $AB = AC$. The midpoint of side BC is point D. $\angle ADB=90°=\angle ADC$ PD is the internal bisector of $\angle ADB$. So, $\angle PDA=45°$. Since $PQ||BC$, $\angle ADQ=45°$ $\angle PDQ=\angle ADQ+\angle PDA=45°+45°=90°$ Hence, the correct answer is 90°.
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