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

(i) it bisects ∠C also, 

(ii) ABCD is a rhombus

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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.

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