Question

Show that the diagonals of a parallelogram divide it into
four triangles of equal area.
7​

Answer 1

Step-by-step explanation:

To prove:- ar(△AOB)=ar(△BOC)=ar(△COD)=ar(△AOD)

Proof:-

Let ABCD be a parallelogram with diagonals AC and BD

intersecting at O. Since the diagonals of a parallelogram bisect each other at the point of intersection.

Therefore,

AO=OC and BO=OD

We know that the median of a triangle divides it into two equal parts.

Now,

In △ABC,

∵BO is median.

ar(△AOB)=ar(△BOC).....(1)

In △BCD,

∵CO is median.

ar(△BOC)=ar(△COD).....(2)

In △ACD,

∵DO is median.

ar(△AOD)=ar(△COD).....(3)

From equation (1),(2)&(3), we get

ar(△AOB)=ar(△BOC)=ar(△COD)=ar(△AOD)

Hence proved.

Answer 2

Answer:

Step-by-step explanation:

To prove:- ar(△AOB)=ar(△BOC)=ar(△COD)=ar(△AOD)

Proof:-  

Let ABCD be a parallelogram with diagonals AC and BD  

intersecting at O. Since the diagonals of a parallelogram bisect each other at the point of intersection.

Therefore,   AO=OC and BO=OD

We know that the median of a triangle divides it into two equal parts.

Now,

In △ABC,  

∵BO is median.

∴ar(△AOB)=ar(△BOC).....(1)

In △BCD,

∵CO is median.

∴ar(△BOC)=ar(△COD).....(2)

In △ACD,

∵DO is median.

∴ar(△AOD)=ar(△COD).....(3)

From equation (1),(2)&(3), we get

ar(△AOB)=ar(△BOC)=ar(△COD)=ar(△AOD)

Hence proved.