$ \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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