Question

please answer. .......

Answer 1

In quadrilateral ABCD, AC is a diagonal.

∴ ar ABC = ar ADC

ar AOB + arBOC = arAOD + ar DOC

In quadrilateral ABCD, BD is diagonal

∴ ar ABD = ar BCD

ar AOD + arAOB = arBOC + ar COD

From equation (i) and (ii),
We have ;

arAOD - ar BOC = arBOC - ar AOD

So,

2ar AOD = 2ar BOC

= tri.AOD = tri.BOC

ar AOD + ar AOB = arAOB + ar BOC

arADB = ar ABC

ADB and ABC having common base AB and line between two lines AB and DC.

∴ AB || DC

Similarly we can prove that AD || BC.

∴ ABCD is a parallelogram.