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Question
The diagonals AC and BC of a quadrilateral ABCD intersect at O. Prove that if BO = OD, then areas of ΔABC an ΔADC area equal.
Sum
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Solution

In ΔABD,
BO = OD
⇒ O is the mid-point of BD
⇒ AO is a median.
⇒ ar(ΔAOB) = ar(ΔAOD) ..........(i)
In ΔCBD, O is the mid-point of BD
⇒ CO is a median.
⇒ ar(ΔCOB) = ar(ΔCOD) ..........(ii)
Adding (i) and (ii)
ar(ΔAOB) 6 ar(ΔCOB) = ar(ΔAOD) + ar(ΔCOD)
Therefore, ar(ΔABC) = ar(ΔADC).
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Diagonal Properties of Different Kinds of Parallelograms
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