Question

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

Answer 1

Given :- $\Delta ABC$ is isosceles

$AB=AC$ __(1)

$\Delta DBC$ is isosceles

$DB=DC$ __(2)

To prove :- $\angle ABD=\angle ACD$

Proof :- let us join $AD$

ln $\Delta ABD$ and $\Delta ACD$,

$AB=AC$ [from eq. (1)]

$BD=CD$ [ from eq. (2)]

$AD=AD$ [common]

$\Delta ABD\cong \Delta ACD$ (sss rule)

$\angle ABD=\angle ACD$ (CPCT)

Hence proved