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# $ \mathrm{ABC} $ and $ \mathrm{DBC} $ are two isosceles triangles on the same base $ BC $ (see Fig. 7.33). Show that $ \angle \mathrm{ABD}=\angle \mathrm{ACD} $.

"

**Given:**

$ABC$ and $DBC$ are two isosceles triangles on the same base $BC$.

**To do: **

We have to show that $\angle ABD=\angle ACD$.

**Solution:**

Let us consider $\triangle ABD$ and $\triangle ACD$

We know that,

Side-Side-Side congruence rule states that if three sides of one triangle are equal to three corresponding sides of another triangle, then the triangles are congruent.

Since, $AD$ is the common side of the two triangles we get,

$AD=DA$

Since $ABC$ and $DBC$ are two isosceles triangles we get,

$AB=AC$ and $BD=CD$

Therefore,

$\angle ABD \cong \angle ACD$

We also know

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

Therefore,

$\angle ABD=\angle ACD$.

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