Question

Prove it.......

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Answer 1

As ABCD is parallelogram

So AB= DC

As AB = 2MB ( AM + MB = AB, AM= MB)

and DC = 2CN ( DN+NC= DC, DN= CN)

As AB= DC

So MB= CN

similarly DN= AM

As M and N are mid points

So MN//BC

it can be proved by taking MNBC

join MC

then as MB= NC

BMC= MCN( alternate)

MC= MC

so SAS

therefore MN= BC

So MB= NC and MN= BC

therefore MBCN is parallelogram

Also AM= DN
and AD= MN

therefore AMND is parallelogram

In AMD and BNC

AD = BC ( opp sides of parallelogram are equal)

A = C ( opp angles are equal)

AM= NC( AB= DC and M and N are mid points)

So SAS therefore
DM= BN

In DMBN

DM= BN
MB= DN

so it's parallelogram

In DMO and BON

DM= BN

MN= MN

DMO = BNO ( alternate)

So SAS

therefore OM= ON

DO= OB


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Answer 2

Given-    ABCD is a parallelogram

To prove-  area (BEC) = area (MCBN)

construction- PRODUCE AB TO E AND

JOIN CE

Proof;-
since M and N are the midpoints of the sides AB and CD,

      ar (MCBN) = ar (MDAN)

or, ar (MCBN) = 1/2 ar (ABCD)              ...(i)

and ar 1/2ar(ABCD) = ar(triangleBCD)   ...(ii)    (because diagonal of a llgm divides it into 2 triangles of equal area)

FROM EQUATION 1 AND 2, WE HAVE
           AR (MCBN =BCD)   ...(iii)

NOW,

     AB ll DC
or, AE ll DC
or, BE ll DC ...(iv)
also,BD ll CE ..(v)

so, from equation (iv) and (v), BECD is a llgm, whose diagonal is BC

so BEC=BCD......(vi)
from( v) and( vi)
we get ar BEC =ar MCBN....

hense the proof..

hope this helps u...