Given: ABCD is a quadrilateral in which AB=AD and BC=DC

To prove: AC bisects ∠A and ∠C, and AC is the perpendicular bisector of BD

Proof:

In ∆ABC and ∆ADC, we have

AB = AD …given

BC = DC … given

AC = AC … common side

Thus by SSS property of congruence,

∆ABC ≅ ∆ADC

Hence, we know that, corresponding parts of the congruent triangles are equal

∠BAC = ∠DAC

∴ ∠BAO = ∠DAO …(1)

It means that AC bisects ∠BAD ie ∠A

Also, ∠BCA = ∠DCA … cpct

It means that AC bisects ∠BCD, ie ∠C

Now in ∆ABO and ∆ADO

AB = AD …given

∠BAO = ∠DAO … from 1

AO = AO … common side

Thus by SAS property of congruence,

∆ABO ≅ ∆ADO

Hence, we know that, corresponding parts of the congruent triangles are equal

∠BOA = ∠DAO

But ∠BOA + ∠DAO = 180°

2∠BOA = 180°

∴ ∠BOA = = 90°

Also ∆ABO ≅ ∆ADO

So, BO = OD

Which means that AC = BD