Question

डेवलप a डायलॉग बिटवीन यू एंड योर मदर ऑन द सेलिब्रेशन ऑफ रिपब्लिक डे इन योर स्कूल​

Answer 1

Answer:

D and E are mid-points of sides BC and AC respectively.

So, DE∥BA⟹DE∥BF

Similarly, FE∥BD. So, BDEF is a parallelogram.

Similarly, DCEF and AFDE are parallelograms.

Now, DF is a diagonal of parallelogram BDEF.

Therefore,

ar(△BDF)=ar(△DEF) .....(1)

DE is a diagonal of parallelogram DCEF.

So, ar(△DCE)=ar(△DEF) .....(2)

FE is a diagonal of parallelogram AFDE.

ar(△AFE)=ar(△DEF) ....(3)

From 1,2 &3, we have,

ar(△BDF)=ar(△DCE)=ar(△AFE)=ar(△DEF)

But, ar(△BDF)+ar(△DCE)+ar(△AFE)+ar(△DEF)=ar(△ABC)

So, 4ar(△DEF)=ar(△ABC)

ar(△DEF)=

4

1

ar(△ABC)

Now, ar(∥

gm

BDEF)=2ar(△DEF)

ar(∥

gm

BDEF)=2×

4

1

ar(△ABC)=

2

1

ar(△ABC)