Q. 74.8( 8 Votes )

# <span lang="EN-US

A. Draw the given quadrilateral ABCD.

B. Draw the diagonal DB of quadrilateral ABCD.

C. Draw a parallel line through point C to diagonal DB of quadrilateral ABCD which intersects at Q produced AD.

D. CQ||DB, ΔABQ is the required triangle.

Proof:

∆DCB = ∆DBQ (on same base DB and between same parallels DB and CQ)

∆DCB = ∆DBQ

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

quadrilateral ABCD = ∆ABQ

Taking BQ as a base, we draw another ∆BCE with one angle 30⁰ and between the parallels DQ and BC.

∆BCE = quadrilateral ABCD

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