Question

if each diagonal of a quadrilateral separate it into two Triangles of equal area then show that quadrilateral is a parallelogram.


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Answer 1

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Answer 2

In quadrilateral ABCD, AC is a diagonal.

∴ ar $\Delta ABC$ = ar $\Delta ADC$

ar $\Delta AOB$ + ar $\Delta BOC$ = ar $\Delta AOD$ + ar $\Delta DOC.... (i)$

In quadrilateral ABCD, BD is diagonal

∴ ar $\Delta ABD$ = ar $\Delta BCD$

ar $\Delta AOD$ + ar $\Delta AOB$ = ar $\Delta BOC$ + ar $\Delta COD.... (ii)$

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

ar $\Delta AOD$ - ar $\Delta BOC$ = ar $\Delta BOC$ - ar $\Delta AOD$

So,

2ar $\Delta AOD$ = 2ar $\Delta BOC$

$\Delta AOD$ =$\Delta BOC$

ar $\Delta AOD$ + ar $\Delta AOB$ = ar $\Delta AOB$ + ar $\Delta BOC$

ar $\Delta ADB$ = ar $\Delta ABC$

$\Delta ADB$ and $\Delta ABC$ 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.