The diagonal AC of the parallelogram ABCD bisects ∠BAD. Prove that the diagonal AC also bisects ∠BCD.

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Solution

Given ABCD is a parallelogram with diagonals AC and BD.

Also, AC bisects ∠BAD

So, ∠DAC=∠BAC …………..(1)

Since, CD||AB,

∠DCA=∠BAC (alternate interior angles)

And ∠BCA=∠DAC (alternate interior angles)

From (1),

∠DAC=∠BAC

⇒ ∠BCA=∠DCA

Hence, AC bisects ∠BCD

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$[3 M a r k s]$

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