Question
In the given figure, ABC and DBC are two triangles on the same base BC such that $AB = AC$ and $DB = DC$. Prove that $\angle ABD =\angle ACD$,
✓
Answer
In $\triangle ABC$
$AB = AC [$ Given $]$
$\therefore \angle ABC =\angle ACB$ [angles opposite to equal side are equals]
Similarly in, $\triangle DBC , DB = DC$ [Given] ...(1)
$\therefore \angle DBC=\angle DCB \ldots(2)$
Adding (1) and (2)
$\angle ABC+\angle DBC=\angle ACB+\angle DCB$
or $\angle ABD =\angle ACD$
$AB = AC [$ Given $]$
$\therefore \angle ABC =\angle ACB$ [angles opposite to equal side are equals]
Similarly in, $\triangle DBC , DB = DC$ [Given] ...(1)
$\therefore \angle DBC=\angle DCB \ldots(2)$
Adding (1) and (2)
$\angle ABC+\angle DBC=\angle ACB+\angle DCB$
or $\angle ABD =\angle ACD$
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