Diagonal AC of a parallelogram ABCD bisects ∠A. Show that

(i) it bisects ∠C also,

(ii) ABCD is a rhombus.

(Ex.8.1Q3.png)

(i) We know that,

ABCD is a parallelogram.

So, opposite sides are parallel and equal.

⇒ ∠DAC = ∠BCA (Alternate interior angles are equal) ……(1)

⇒ ∠BAC = ∠DCA (Alternate interior angles are equal) …….(2)

Given,

⇒ AC bisects ∠A

⇒ ∠DAC = ∠BAC …….(3)

From equations (1), (2), and (3), we get :

⇒ ∠DCA = ∠BAC = ∠DAC = ∠BCA ………(4)

∴ ∠DCA = ∠BCA

∴ AC bisects ∠C.

Hence, proved that AC bisects ∠C.

(ii) In Δ ABC,

⇒ ∠BAC = ∠BCA (Proved above)

⇒ BC = AB (Side opposite to equal angles are equal) …..(5)

⇒ DA = BC and AB = CD (Opposite sides of a parallelogram are equal) ….(6)

From equations (5) and (6), we get :

⇒ AB = BC = CD = DA.

Since, all sides of quadrilateral ABCD are equal.

Hence, proved that ABCD is a rhombus.