Two triangles $$ABC$$ and $$DBC$$ have common base $$BC$$. Suppose $$AB= DC $$ and $$\angle ABC = \angle BCD $$, then prove that $$AC = BD$$.
In $$\bigtriangleup ABC $$ and $$\bigtriangleup DBC $$
$$AB = DC $$ (Given)
$$\angle ABC = \angle BCD $$ (Given)
$$BC = BC $$ (common side)
$$\therefore \bigtriangleup ABC \cong \bigtriangleup DBC $$ (By $$SAS$$ postulate)
$$\implies AC = BD $$ ( Corresponding sides of congruent triangles)