Question

A quadrilateral ABCD is such that diagonals BD divides its area into two equal parts. Prove that BD bisects AC.

Answer 1

Join AC. Suppose AC and BD intersect at O. Draw AL and CM perpendicular to BD.
ar(ΔABD) = ar(ΔBDC)
Thus ΔABD and ΔABC are on the same base AB and have equal area.
Therefore, their corresponding altitudes are equal i.e. AL = CM.
Now,
In ΔALO and ΔCMO,
∠1 = ∠2             ...(vertically opposite angles)
∠ALO = ∠CMO ...(right angles)
AL = CM
Therefore,
ΔALO ≅ ΔCMO   ...(AAS axiom)
⇒ AO = OC
⇒ BD bisects AC.