If a diagonal of a parallelogram bisects one of the angles of the parallelogram, prove that it also bisects the second angle and then the two diagonals are perpendicular to each other.

Let ABCD be a parallelogram, and let the diagonal AC bisect ∠BAD.

Since ABCD is a parallelogram:

AB ∥ DC

AD ∥ BC

∠DAC = ∠CAB [AC bisects ∠DAB]

In △ABC and △ADC:

AC = AC (common side)

∠BAC = ∠DCA (Alternate interior angles are equal)

∠DAC = ∠BCA (Alternate interior angles are equal)

△ABC ≅ △ADC (A.S.A. congruence)

So, ∠DCA = ∠BCA (By C.P.C.T.C.)

That means AC also bisects the opposite angle ∠DCB.

Now,

AD = AB (From congruence)

AB = CD and AD = BC (Opposite sides of parallelogram)

∴ AB = BC = CD = DA

Thus, ABCD is a Rhombus.

We know that,

Diagonals of a rhombus are perpendicular to each other.

Thus, AC ⊥ BD.

Hence, proved that the diagonal bisects the second angle and AC ⊥ BD.