Question

If diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar(AOD)= ar(BOC), then ABCD is a:

Answer 1

Answer:

it must be a parallelogram because by this o will be the mid point of diagonals

if this help mark me as brainliest

Answer 2

It's a TRAPEZIUM.

Given -

a quad. ABCD ,

ar( AOD ) = ar ( BOC ).

To prove- ABCD is a trapezium

Proof -

ar(ADC)= ar ( AOD ) +ar(DOC)

similarly

ar(BDC) = ar(BOC) + ar(DOC)

ar( AOD ) = ar ( BOC )

DOC IS COMMON HERE

Therefore,ar(ADC) = ar(BDC)

These two triangles are on the same base DC and their areas are also equal.

Therefore, they lie between the same parallels AB and DC.

AB || DC

Therefore, ABCD is a parallelogram with one pair of opposite sides parallel.

Therefore, ABCD is a trapezium.

I HOPE IT HELP YOU.