Question

In any triangle ABC, if the angle bisector of $\angle A$ and perpendicular bisector of BC intersect, prove that they intersect on the circum circle of the triangle ABC.

Answer 1

Let angle bisector of $\angle A$ intersect circumcircle of $\Delta ABC$ at D.
Join DC and DB.
$\angle BCD=\angle BAD$
[Angles in the same segment]
$\Rightarrow \angle BCD=\angle BAD=\frac{1}{2}\angle A$
[AD is bisector of $\angle A$] ...(i)
Similarly, $\angle DBC=\angle DAC=\frac{1}{2}\angle A$ ...(ii)
From (i) and (ii), $\angle DBC=\angle BCD$
$\Rightarrow BD=DC$ [sides opposite to equal angles are equal]
$\Rightarrow$ D lies on the perpendicular bisector of BC.
Hence, angle bisector of $\angle A$ and the perpendicular bisector of BC interset on the circumcircle of $\Delta ABC$.