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} $.
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.
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