$ABC$ and $DBC$ are two isosceles triangles on the same base $BC$ (see figure). Show that $\angle ABD = \angle ACD$.
(N/A) In $\Delta ABC$,we have
$AB = AC$ (Given,since $\Delta ABC$ is an isosceles triangle)
Since angles opposite to equal sides are equal,we have:
$\angle ABC = \angle ACB$ .......... $(1)$
In $\Delta BDC$,we have
$BD = CD$ (Given,since $\Delta BDC$ is an isosceles triangle)
Since angles opposite to equal sides are equal,we have:
$\angle CBD = \angle BCD$ .......... $(2)$
Adding equations $(1)$ and $(2)$,we get:
$\angle ABC + \angle CBD = \angle ACB + \angle BCD$
$\Rightarrow \angle ABD = \angle ACD$
Hence proved.
$AB = AC$ (Given,since $\Delta ABC$ is an isosceles triangle)
Since angles opposite to equal sides are equal,we have:
$\angle ABC = \angle ACB$ .......... $(1)$
In $\Delta BDC$,we have
$BD = CD$ (Given,since $\Delta BDC$ is an isosceles triangle)
Since angles opposite to equal sides are equal,we have:
$\angle CBD = \angle BCD$ .......... $(2)$
Adding equations $(1)$ and $(2)$,we get:
$\angle ABC + \angle CBD = \angle ACB + \angle BCD$
$\Rightarrow \angle ABD = \angle ACD$
Hence proved.
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