Answer :

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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