(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.