hey abhi bhai , here is your answer..

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explanations:

Let us consider EF, side of ∥gm BCFE meets AC

at G. We know that,

E is the mid-point and EF∥BC

G is the mid-point of AC.

So, AG=GC

Now, in △AEG and ΔCFG,

The alternate angles are:

∠EAG,∠GCF

Vertically opposite angles are:

∠EGA=∠CGF

So,AG=GC

Proved.

∴ΔAEG≅ΔCFG

ar(ΔAEG)=ar(ΔCFG)

Now,

ar (∥gm EBCF )

=arBCGE+ ar (ΔCFG)

=arctgE+ar(ΔAEG)

=ar(ΔABC)

We know that, ar (ΔABC)=25sq. units

Hence, ar (∥gm EBCF )=25 sq. units

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miss you abhi bro❤❤......

plzz mark me as brainleist........

Let us consider EF, side of ∥gm BCFE meets AC at G. We know that,

E is the mid-point and EF∥BC

G is the mid-point of AC.

So, AG=GC

Now, in △AEG and ΔCFG,

The alternate angles are:

∠EAG,∠GCF

Vertically opposite angles are:

∠EGA=∠CGF

So,AG=GC

Proved.

∴ΔAEG≅ΔCFG

ar(ΔAEG)=ar(ΔCFG)

Now,

ar (∥gm EBCF )

=arBCGE+ ar (ΔCFG)

=arctgE+ar(ΔAEG)

=ar(ΔABC)

We know that, ar (ΔABC)=25sq. units

Hence, ar (∥gm EBCF )=25 sq. units