Question

Prove that angle bisector of any angle of a triangle and perpendicular bisector of the opposite side if intersect, they will intersect on the circumcircle of the triangle.

Answer 1

Proof of the statement:

Consider the triangle $∆ABC$

We have to prove that the angle bisector of $\angle A$ and perpendicular bisector of $BC$ intersect on the circumcircle of $∆ABC$

Let the angle bisector of $\angle A$ intersect circumcircle of $∆ABC$ at $P$

Join $BP$ and $CP$

So, $\angle BAP=\angle BCP$ (Angles made by a chord on circumference in same segment are equal)

Since, $AP$ is a bisector of $\angle A$

$\angle BAP=\angle PAC$

$\angle A=\angle BAP+\angle PAC$

So, $\angle BAP=\angle BCP=\frac{1}{2}\angle A$ ……..(i)

Similarly $\angle PAC=\angle PBC=\frac{1}{2}\angle A$ …….(ii)

From (i) and (ii)

$\angle BCP=\angle PBC$

In $∆BPC$

$\angle BCP=\angle PBC$

So, $BP=CP$ (Opposite sides of equal angles are equal in a triangle)

Now, perpendicular bisector of side $BC$ of triangle $∆BPC$ will pass via vertex $P$,

Because $∆BPC$ is an isosceles triangle.

Therefore, angle bisector of $\angle A$ and perpendicular bisector of $BC$ intersect on the circumcircle of $∆ABC$.

Hence,it is proved that angle bisector of any angle of a triangle and perpendicular bisector of the opposite side if intersect, they will intersect on the circumcircle of the triangle.