In \( \triangle \mathrm{ABC}, \mathrm{AD} \) is the perpendicular bisector of \( \mathrm{BC} \) (see Fig. 7.30). Show that \( \triangle \mathrm{ABC} \) is an isosceles triangle in which \( \mathrm{AB}=\mathrm{AC} \).
![](/assets/questions/media/153848-1659407074.png)
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Given:
In $\triangle ABC, AD$ is the perpendicular bisector of $BC$.
To do:
We have to show that $\triangle ABC$ is an isosceles triangle in which $AB=AC$.
Solution:
Let us consider $\triangle ADB$ and$\triangle ADC$,
We know that,
According to Rule of Side-Angle-Side Congruence:
Triangles are said to be congruent if any pair of corresponding sides and their included angles are equal in both triangles.
Since $AD$ is the common side of both the triangles,
We get,
$AD=DA$
This implies,
$\angle ADB= \angle ADC$
Since $AD$ is a perpendicular bisector of $\triangle ABC$ we get,
$BD=CD$
Therefore,
$\triangle ADB \cong \triangle ADC$
We also know
From corresponding parts of congruent triangles: If two triangles are congruent, all of their corresponding sides must be equal.
Therefore,
$AB=AC$.
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