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# In quadrilateral $ A C B D $,

$ \mathrm{AC}=\mathrm{AD} $ and $ \mathrm{AB} $ bisects $ \angle \mathrm{A} $ (see Fig. 7.16). Show that $ \triangle \mathrm{ABC} \cong \triangle \mathrm{ABD} $.

What can you say about $ \mathrm{BC} $ and $ \mathrm{BD} $ ?

"64391"

**Given:**

In quadrilateral $ABCD, AC=AD$ and bisects $\angle A$.

**To do:**

We have to show that $\triangle ABC\cong ABCD$ and say about $BC$ and $BD$.

**Solution:**

Let us consider $\triangle ABC$ and $\triangle ABD$.

Given,

$AC=AD$

The line segment $AB$ bisects $\angle A$.

Therefore,

$\angle CAB=\angle DAB$

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.

Therefore,

$\triangle ABC\cong ABCD$

We also know that,

From corresponding parts of congruent triangles: If two triangles are congruent, all of their corresponding angles and sides must be equal.

This implies,

$BC=BD$.

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