Question

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

Answer 1

Given, that $ABC$ and $DBC$ are two isosceles triangles on the same base$.$

Proof that$,\angle ABD=\angle ACD$

proof$:-$

$°.°$ ABC isosceles triangle on the base$.$

.$°.$ $<ABC=<ACB$ ---------$\left(1\right)$

$°.°$ DBC isosceles triangle on the base$.$

$.°.$ $<DBC=<DCB$ ----------$\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$ we get$,$

$<ABC+<DBC=<ACB+<DCB$

$=><ABD=<ACD$