# Q : 10    Diagonals AC and BD of a trapezium ABCD with  $\small AB\parallel DC$  intersect each other at O. Prove that  $\small ar(AOD)=ar(BOC)$.

We have a trapezium ABCD such that AB || CD and it's diagonals AC and BD intersect each other at O
$\Delta$ABD and $\Delta$ ABC are on the same base AB and between same parallels AB and CD
$\therefore$ ar($\Delta$ABD) = ar ($\Delta$ABC)

Now, subtracting $\Delta$AOB from both sides we get

ar ($\Delta$ AOD) = ar ($\Delta$ BOC)

Hence proved.

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