Answer :

Given.

E is any point on side DC of parallelogram ABCD, produced AE meets produced BC at the point F

Formula used.

If 2 triangles are on same base And lies between 2 parallel lines then both triangles are equal.

If triangle and parallelogram are on same base And lies between 2 parallel lines then triangle gets half of parallelogram.

Draw the diagonal BD of parallelogram ABCD

In triangle ADF and triangle ADB

Both lies on same base AD

And AD || BC ∵ opposite sides of parallelogram are parallel

As BF is extended part of BC

Then AD || BF

∴ triangle ADF = triangle ADB = of parallelogram ABCD

∵ Diagonal of parallelogram gives 2 congruent triangles

In triangle ABE and parallelogram ABCD

Both lies on same base AB

And AB || DC ∵ opposite sides of parallelogram are parallel

∴ triangle ABE = of parallelogram ABCD

Both triangle ADF and triangle ABE are half of parallelogram

∴ triangle ADF = triangle ABE

If ∴ triangle ABE = of parallelogram ABCD

⇒ triangle ADE + triangle EBC = Another of parallelogram ABCD

∴ triangle ADF = of parallelogram ABCD

⇒ triangle ADF= triangle ADE+ triangle DEF

→ triangle ADE+ triangle DEF = triangle ADE+ triangle ABE

∴ triangle DEF = triangle ABE

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