Question
$ABC$ and $DBC$ are two isosceles triangles on the same base $BC$. Show that $\angle ABD = \angle ACD.$
✓
Answer
Given : $A B C$ and $D B C$ are two isosceles triangles on the same base $B C$.
To Prove : $\angle ABD =\angle ACD$.
Proof : $A s A B C$ is an isosceles triangle on the base $B C$
$\therefore \angle ABC=\angle ACB \ldots(1)$
As $A B C$ is an isosceles triangle on the base $B C$
$\therefore \angle DBC=\angle DCB \ldots(2)$
Adding the corresponding sides of $(1)$ and $(2)$
$\angle ABC + \angle DBC = \angle ACB + \angle DCB$
$\Rightarrow \angle ABD = \angle ACD$
To Prove : $\angle ABD =\angle ACD$.
Proof : $A s A B C$ is an isosceles triangle on the base $B C$
$\therefore \angle ABC=\angle ACB \ldots(1)$
As $A B C$ is an isosceles triangle on the base $B C$
$\therefore \angle DBC=\angle DCB \ldots(2)$
Adding the corresponding sides of $(1)$ and $(2)$
$\angle ABC + \angle DBC = \angle ACB + \angle DCB$
$\Rightarrow \angle ABD = \angle ACD$
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