Question

$△ABC$ and $△DBC$ are two isosceles triangles on the same base $BC$ (in figure). Show that $\angle ABD=\angle ACD$.

Answer 1

$△ABC$ is isosceles

So, $AB=AC.........\left(1\right)$

$△DBC$ is isosceles

So, $DB=DC.........\left(2\right)$

Let us join AD.

In $△ABD$ and $△ACD$

$AB=AC$ [From (1)]

$BD=CD$ [From (2)]

$AD=AD$ [Common]

$\therefore △ABD\cong △ACD$ [SSS congruency]

$\therefore \angle ABD=\angle ACD$ [CPCT}

Hence proved.