Question

if each diagonal of a quadrilateral divides it into two triangles of equal areas then of prove that quadrilateral is a parelletogram

Answer 1

since AC diagonal
ar(tri.ABC)=ar(tri.ACD)
we know that sum of triangle is equal to quadrilateral
so
ar(tri.ABC)+ar(tri.ACD)=ar(quad.ABCD)
ar(tri.ABC)+ar(tri.ABC)=ABCD
2ar(ABC)=ABCD
ar(ABC)=1/2ABCD (eq-1 )
since BD diagonal
ar(tri.ABD)=ar(tri.BDC)
we know that sum of triangle is equal to quadrilateral
ar(ABD)+ar(BDC)=arABCD
ar(ABD)+ar(ABD)=arABCD
2ar(ABD)=arABCD
ar(ABD)=1/2arABCD (eq2)
when RHS are equal of eq1&eq2
so LHS are also equal
ar(ABC)=ar(ABD)
AB is base in both triangle
longitude are also equal
AD=BC& AD||BC
since
opposite side are parallel &diagonal are equal
so we can say that ABCD is parallelogram

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.