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°
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
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°.
Related Questions
Know More about
Staff Selection Commission Combined Grad ...
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Get Updates BrochureYour Staff Selection Commission Combined Graduate Level Exam brochure has been successfully mailed to your registered email id “”.