Diagonal AC of a parallelogram ABCD bisects ∠A see Figure. Show that (i) it bisects ∠C also, (ii) ABCD is a rhombus.

(i) ∠DAC = ∠BAC …(1) [∵ Given]
∠DAC = ∠BCA …(2) [∵ Alternate angle]
∠BAC = ∠ACD …(3) [∵Alternate angles]
From the equations (1), (2) and (3), we have
∠ACD = ∠BCA …(4)
Hence, Diagonal AC bisects angle C also.
(ii) From the equation (2) and (4), we have
∠ACD = ∠DAC
In ΔADC,
∠ACD = ∠DAC [∵ Prove above]
AD = DC [∵ In a triangle, the sides opposite to equal angle are equal]
A parallelogram whose adjacent sides are equal, is a rhombus. Hence, ABCD is a rhombus.