Question

15. 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.

answer correctly earn points thanks in advance. person to explain by writing on a page will get a brainliest tag.

thanks.​

Answer 1

So here is the answer for the question you asked.

Hope it help you.

Answer 2

hi mate,

Solution : 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 helps you.