Question

In any triangle ABC inscribed in a circle, if the angle bisector of $\angle A$ and perpendicular bisector of BC intersect, then which of the following options is correct.

• A. BP = BQ
• B. BP = AB
• C. BO = AB
• D. BP=CP

Answer 1

The correct option is D BP=CP

Given. ABC is a triangle and O is the centre of its circumcircle. P is a point on the circle such that. AP is the internal bisector of $\angle BAC$ and M is the mid point of BC.
Join BP and CP $\because$ AE is the bisector of $\angle BAC$
$\therefore \angle BAE=\angle CAE$
$\Rightarrow arcBP\cong arcCP$ [$\because$ Equal angles subtended at the circumference of a circle by congruent arcs of circle]
$\Rightarrow chordBP=chordCP$
$\begin{array}{cc}BP=CP& \left[Provedabove\right]\\ BM=CM& [\because Misthemidpoint]\\ MP=MP& \left[Commonside\right]\end{array}$
$\therefore \Delta BMP\cong \Delta CMP$
[By SSS criterion]
$\therefore \angle BMP=\angle CMP$ [CPCT]
Again $\angle BMP+\angle CMP={180}^{\circ }$
[Angles of a linear pair]
$\Rightarrow \angle BMP=\angle CMP={90}^{\circ }$
Hence MP is the right bisector of BC