Question

$\bf \huge {\underline {\underline \red{QuEsTiOn}}}$

Diagonals AC and BD of a trapezium ABCD with AB||DC intersect each other at O. Prove that ar (AOD) = ar (BOC)​

Answer 1

NOTE:-

Triangle with same base and between same parallels are equal in area.

Given,

A trapezium ABCD where AB|| DC And diagonals AC AND BD intrest each other at O.

We need to prove that,

ar [AOD] , ar [BOC]      

Now, we need to proof it that,(triangle=<)

<ADS and <BDC Lie on the base of DC and between parallels AB and CD.

Therefore, Area= <ADC, <BDC

Now, subtracting ar (DOC) "Both of sides"

Area=(<ADC)- AREA=(<DOC)

=Area (<AOD) = Area (<BOC)

HOPE ITS HELPFUL MARK BRAINLIST ALSO FOLLOW ME.....

Answer 2

$\huge\mathfrak{\color{orange}{\underline {\underline{Answer♡}}}}$

Given

↪Diagonals of trapezium is AC and BD

↪AB || DC

To Prove

↪ ar (AOD) = ar (BOC)

Solution

△ACD and △BCD are on same base DC and between same parallels AB and DC.

( Theorem - Two triangles on the same base and between the same parallels are equal in area. )

So, △ACD = △BCD

☞ △AOD - △DOC = △BOC - △DOC

☞ ar (AOD) = ar (BOC)

$\sf \pink{hence \: proved...}$

Itz ur Adi❤