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  • In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.

Q 10. In any triangle ABC, if the angle bisector of $\angle A$ and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the $\triangle ABC$.

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Let $A E$ be the angle bisector of $\angle A$.
To prove: ED is the perpendicular bisector of BC .
$\angle B A E=\angle C A E$-------- (1) [Since, $A E$ is the angle bisector of $\angle A]$
Now, $\angle E B C=\angle C A E$-------(2) [Angles subtended by the same arc EC]
Also, $\angle \mathrm{ECB}=\angle \mathrm{BAE}$------(3) [Angles subtended by the same arc BE]
But we know that, $\angle B A E=\angle C A E$ From equation (1)]
Hence, $\angle E B C=\angle E C B$ [From equations (2) and (3)]
Therefore, $\mathrm{BE}=\mathrm{EC}$ [Sides opposite to equal angles are equal]

The point E is equidistant from the points B and C .
Therefore, ED is the perpendicular bisector of BC.
The perpendicular bisector of BC and the angle bisector of $\angle A$ meet the circumcircle of $\triangle A B C$ at E.

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