# <span lang="EN-US

A. Draw the given square ABCD.

B. Draw the diagonal DB of square ABCD.

C. Draw a parallel line through point C to diagonal DB of square ABCD which intersects at E produced AB.

D. CE||DB. ΔADE is the required triangle.

Proof:

∆DCB = ∆DBE (on same base DB and between same parallels DB and CE)

∆DCB = ∆DBE

∆ABD + ∆DCB = ∆DBE + ∆ABD (adding area of ∆ABD on both sides)

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