Q

# Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (BOC). Prove that ABCD is a trapezium.

Q: 15    Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that $\small ar(AOD)=ar(BOC)$. Prove that ABCD is a trapezium.

Views

We have,
ABCD is a quadrilateral and diagonals AC and BD intersect at O such that ar($\Delta$AOD) = ar ($\Delta$BOC) ...........(i)

Now, add ar ($\Delta$BOA) on both sides, we get

ar($\Delta$AOD) + ar ($\Delta$BOA)  = ar ($\Delta$BOA) + ar ($\Delta$BOC)
ar ($\Delta$ABD) = ar ($\Delta$ABC)
Since the $\Delta$ABC and $\Delta$ABD lie on the same base AB and have an equal area.
Therefore, AB || CD

Hence ABCD is a trapezium.

Exams
Articles
Questions