Construction : In quadrilateral ABCD, from A to C. Now, in ΔABC
∵ AB=BC....(Given) ∴ ∠BAC=∠BCA (angles opposite to equal side) In ΔADC, ∵ CD>AD ∴ ∠DAC>∠DCA (since in a triangle, angle opposite to greater side is bigger than the angle opposite to smaller side) On adding eqs. (l) and (ii), we get ∠BAD>∠BCD