The diagonal AC of a parallelogram ABCD bisects ∠A (see the figure). Show that

(i) it bisects ∠C also.

(ii) ABCD is a rhombus.

(i) Given, diagonal AC of a parallelogram ABCD

bisects ∠A.

Here, AB || CD and AC is transversal.

∴ ∠ DCA = ∠ BAC [pair of alternate angle]… (ii)

Similarly, as BC || AD and AC is a transversal.

∴ ∠ BCA = ∠DAC [pair of alternate angle]… (iii)

From Eqs. (i), (ii) and (iii), we get

∠DAC = ∠BCA = ∠BAC = ∠DCA … (iv)

Thus, diagonal AC also bisects ∠C.

(ii) Now, In ∆ ADC, we have

∠DAC =∠DCA [from Eq. (iv)]

=> CD = AD [∵sides opposite to equal angles are equal]

Also, we have AB =CD and AD = BC

[opposite sides of parallelogram]

∴ AB = CD = AD = BC

Hence, ABCD is a rhombus.