Diagonal AC of a parallelogram ABCD bisects ∠A (see Fig. 8.19). Show that

(i) It bisects ∠ C also,

(ii) ABCD is a rhombus.

(i) ABCD is a parallelogram.

∠DAC = ∠BCA (Alternate interior angles) ... (1)

And,

∠BAC = ∠DCA (Alternate interior angles) ... (2) However, it is given that AC bisects A

∠DAC = ∠BAC ... (3)

From equations (1), (2), and (3), we obtain

∠DAC = ∠BCA = ∠BAC = ∠DCA ... (4)

∠DCA = ∠BCA

Hence, AC bisects C

From equation (4), we obtain

∠DAC = ∠DCA

DA = DC (Side opposite to equal angles are equal)

However,

DA = BC and AB = CD (Opposite sides of a parallelogram)

AB = BC = CD = DA

Hence, ABCD is a rhombus