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Home / Geometry / In the quadrilateral ABCD, AB = CD, BC = CD and CD>AD. Prove that ∠DAB > ∠BCD

In the quadrilateral ABCD, AB = CD, BC = CD and CD>AD. Prove that ∠DAB > ∠BCD

Solution: General enunciation: In the quadrilateral ABCD, AB = CD, BC = CD and CD>AD. Prove that ∠DAB > ∠BCD.

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Particular enunciation: Given that, In the quadrilateral ABCD, AB = CD, BC = CD and CD>AD. We have to proved that ∠DAB > ∠BCD.

Construction: We join A and C.

Proof: In the triangle ACD, CD>AD

∴∠CAD>∠ACD ————————————-(i)

[the angle opposite to greater side is greater]

Again, in triangle, BC>AB

∴∠BAC>∠ACB ————————————-(ii)

Adding (i) and (ii) we have

∠CAD + ∠BAC>∠ACD + ∠ACB

⟹∠DAB>∠BCD. (Proved)

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