Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Given: The diagonals AC and BD of a quadrilateral ABCD are equal and bisect each other at right angles.
To Prove: Quadrilateral ABCD is a square.
Proof: In ∆OAD and ∆OCB,
OA = OC    | Given
OD = OB    | Given
∠AOD = ∠COB
| Vertically Opposite Angles
∴ ∆OAD ≅ ∆OCB
| SAS Congruence Rule

∴ AD = CB    | C.P.C.T.
∠ODA = ∠OBC    | C.P.C.T.
∴ ∠BDA = ∠DBC
∴ AD || BC
Now, ∵ AD = CB and AD || CB
∴ Quadrilateral ABCD is a || gm.
In ∆AOB and ∆AOD,
AO = AO    | Common
OB = OD    | Given
∠AOB = ∠AOD
| Each = 90° (Given)
∴ ∆AOB ≅ ∆AOD
| SAS Congruence Rule
∴ AB = AD
Now, ∵ ABCD is a parallelogram and
∴ AB = AD
∴ ABCD is a rhombus.
Again, in ∆ABC and ∆BAD,
AC = BD    | Given
BC = AD
| ∵ ABCD is a rhombus
AB = BA    | Common
∴ ∆ABC ≅ ∆BAD
| SSS Congruence Rule
∴ ∆ABC = ∆BAD    | C.P.C.T.
AD || BC
| Opp. sides of || gm ABCD and transversal AB intersects them.
∴ ∠ABC + ∠BAD = 180°
| Sum of consecutive interior angles on the same side of a transversal is 180°
∴ ∠ABC = ∠BAD = 90°
Similarly, ∠BCD = ∠ADC = 90°
∴ ABCD is a square.