**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*

∴∠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)**