ABC and DBC are two isosceles triangles on the same base BC (see the given figure). Show that ∠ABD = ∠ACD.
Solution:
Let us join AD.
In $\triangle \mathrm{ABD}$ and $\triangle \mathrm{ACD}$
$\mathrm{AB}=\mathrm{AC}$ (Given)
BD = CD (Given)
$A D=A D$ (Common side)
$\therefore \triangle \mathrm{ABD} \cong \triangle \mathrm{ACD}$ (By SSS congruence rule)
$\Rightarrow \angle \mathrm{ABD}=\angle \mathrm{ACD}(\mathrm{By} \mathrm{CPCT})$
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