In any triangle $\mathrm{ABC}$, if the angle bisector of $\angle \mathrm{A}$ and perpendicular bisector of $\mathrm{BC}$ intersect, prove that they intersect on the circumcircle of the triangle $\mathrm{ABC}$.

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Given:

In a triangle $\mathrm{ABC}$, the angle bisector of $\angle \mathrm{A}$ and perpendicular bisector of $\mathrm{BC}$ intersect.

To do:

We have to prove that they intersect on the circumcircle of the triangle $\mathrm{ABC}$.

Solution:

Let $ABC$ be a triangle in which the angle bisector of $\angle \mathrm{A}$ and perpendicular bisector of $\mathrm{BC}$ intersect at $E$ as shown in the figure.

Join $BE$ and $CE$

$AE$ is the bisector of $\angle BAC$

This implies,

$\angle BAE = \angle CAE$

$arc BE = arc EC$

This implies,

chord $BE =$ chord $EC$

In $\triangle BDE$ and $\triangle CDE$,

$DE = DE$     (Common side)

$BD = CD$     (Given)

$BE = CE$      (Proved)

Therefore, by SSS congruency,

$\triangle BDE \cong \triangle CDE$

This implies,

$\angle BDE = \angle CDE$

$\angle BDE + \angle CDE = 180^o$                   (Linear pair)

$2\angle BDE=180^o$

$\angle BDE=\frac{180^o}{2}$

$\angle BDE=90^o$

$\angle CDE = \angle BDE = 90^o$

Therefore,

$DE \perp BC$

$BE = CE$ and $DE \perp BC$

Point $E$ is equidistant from the points $B$ and $C$. This is only possible when point $E$ lies on the perpendicular bisector of $BC$.

This implies,

$ED$ is the perpendicular bisector of $BC$.

Therefore, the perpendicular bisector of $BC$ and the angle bisector of $\angle A$ meet on the circumcircle of the triangle $\mathrm{ABC}$ at point $E$.

Hence proved.

Updated on 10-Oct-2022 13:46:58