In a quadrilateral ABCD, there is a point O inside it such that OB = OD. Also, AB = AD and BC = DC. Prove that O lies on AC

In ∆AOB and ∆AOD,

AB = AD

OB=OD

OA=OA

∠1 = ∠2

Similarly, in ∆CBO and ∆CDO,

BC = DC [given]

OB = OD [given]

and OC = OC [common side]

∠3 = ∠4

On adding Eqs. (i) and (ii), we get ∠1+∠3 = ∠2 +∠4

We know that∠1 +∠3+∠2 +∠4 = 360

[∴ sum of all angles around a point is 360°]

∠1 + ∠3+ ∠1 + ∠3 = 360° [from Eqs. (I) and (ii)]

2(∠1+∠3) = 360°

∠1+∠3 = 180°

∠AOC = 180°

Hence, O lies on AC.