Pritam drew a qua

A. Draw the given quadrilateral ABCD.

B. Draw the diagonal DB of quadrilateral ABCD.

C. Draw a parallel line through point A to diagonal DB of quadrilateral ABCD which intersects at F produced BC.

D. Join FD, AF||BD

ΔDFC is the required triangle.

Proof:

∆ABD = ∆BFD (on same base DB and between same parallels DB and AF)

∴ ∆ABD = ∆BFD

∆DBC + ∆ABD = ∆BFD + ∆DBC (adding area of ∆DBC on both sides)

∴ quadrilateral ABCD = ∆DFC