Triangle ABC and DBC are on the same base BC with A and D on opposite sides of line BC such that $area\left( {\Delta ABC} \right) = $$area\left( {\Delta DBC} \right)$. Show that BC bisects AD.

Hint: Best way of solving this question is to draw a suitable diagram of what is given in the question. Relations between areas of both triangles are given so if you have to use this relation. construct altitudes to use the area of triangles.

Complete step-by-step answer:

Let BC cut AD at O.
Draw perpendicular on BC from A as AM and from D as DN.
As given $ar\left( {\Delta ABC} \right)$$ = $$ar\left( {\Delta DBC} \right)$ so,
$ \Rightarrow \dfrac{1}{2}\left( {{\text{AM}} \times {\text{BC}}} \right) = \dfrac{1}{2}\left( {{\text{DN}} \times {\text{BC}}} \right)$
$ \Rightarrow {\text{AM = DN}}$
$ \Rightarrow \angle AMN = \angle DNM = {90^0}$
$ \Rightarrow \angle AOM = \angle DON$ (vertically opposite angle)
So triangle AMO congruent with DNO by AAS criteria.
Hence AO=DO (by corresponding sides of corresponding triangles)
$\because $ O lies on BC and AO+DO=AD (from figure)
Hence we can say that BC bisects AD.

Note: Whenever you get this type of question the key concept of solving is you have to first draw a diagram and use what is given in question to proceed further. you have to use congruence also and also knowledge of vertically opposite angles.