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