Given : A parallelogram ABCD, in which diagonal AC bisects ∠A, i.e., ∠DAC = ∠BAC.
To Prove :
(i) Diagonal AC bisects
∠C i.e., ∠DCA = ∠BCA
(ii) ABCD is a rhomhus.
Proof :
(i) ∠DAC = ∠BCA [Alternate angles]
∠BAC = ∠DCA [Alternate angles]
But, ∠DAC = ∠BAC [Given]
∴ ∠BCA = ∠DCA
Hence, AC bisects ∠DCB
Or, AC bisects ∠C Proved.
(ii) In ∆ABC and ∆CDA
AC = AC [Common]
∠BAC = ∠DAC [Given]
and ∠BCA = ∠DAC [Proved above]
∴ ∆ABC ≅ ∆ADC [ASA congruence]
∴ BC = DC [CPCT]
But AB = DC [Given]
∴ AB = BC = DC = AD
Hence, ABCD is a rhombus Proved.
