Solution: General enunciation: In the quadrilateral ABCD, AB = CD, BC = CD and CD>AD. Prove that ∠DAB > ∠BCD.
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
[the angle opposite to greater side is greater]
Again, in triangle, BC>AB
Adding (i) and (ii) we have
∠CAD + ∠BAC>∠ACD + ∠ACB