Question

The perpendicular bisectors of AB and AC of the triangle ABC meet BC at D If the bisectors meet AB at P and AC at Q then find $\angle BDP+\angle CDQ$

• A. ${80}^{\circ }$
• B. ${90}^{\circ }$
• C. ${60}^{\circ }$
• D. ${100}^{\circ }$

Answer 1

Answer: (B)

${90}^{\circ }$

From figure,

$\angle DPA={90}^{\circ }and\angle AQD={90}^{\circ }$ as $PDandDQ$ are perpendicular bisectors of $ABandAC$ respectively

The perpendicular bisectors from two sides of the triangle meet on the third side.

They are perpendicular to each other.

$\therefore \angle PQD={90}^{\circ }$

$BDC$ is a straight line.

$\angle BDP+\angle PDQ+\angle QDC={180}^{\circ }$

$\angle BDP+{90}^{\circ }+\angle QDC={180}^{\circ }$

$\angle BDP+\angle QDC={180}^{\circ }-{90}^{\circ }={90}^{\circ }$

Option B