Q62
In an isosceles ΔABC, 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
A.
130°
B.
90°
AnswerC.
180°
D.
45°
Answer: Option B
Solution
Answer: Option B
Solution:

∠PDQ = 45° + 45° = 90°