Question
Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar(ΔAOD) = ar(ΔBOC). Prove that ABCD is a trapezium.
Solution
Given: Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar(ΔAOD) = ar(ΔBOC)
To Prove: □ABCD is a trapezium.
Proof: ar(ΔAOD) = ar(ΔBOC)
⇒ ar(ΔAOD) + ar(ΔAOB)
= ar(ΔBOC) + ar(ΔAOB)
| Adding the same areas on both sides
⇒ ar(ΔABD) = ar(ΔABC)
But ΔABD amd ΔABC are on the same base
AB.
∴ ΔABD and ΔABC will have equal corresponding altitudes.
ΔABD and ΔABC will lie between the same parallels.
∴ AB || DC
∴ [□ABCD is a trapezium.
A quadrilateral is a trapezium if exactly one pair of opposite sides is parallel
