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Question
In the figure, ∠BCD = ∠ADC and ∠ACB =∠BDA. Prove that AD = BC and ∠A = ∠B.
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Solution
∠BCD = ∠ADC
∠ACB = ∠BDA
∠BCD + ∠ACB = ∠ADC + ∠BDA
⇒ ∠ACD = ∠BDCACD = BDC
In ΔACD and ΔBCD
∠ACD =∠BDCACD = BDC
∠ADC = ∠BCD
ADC = BCD
CD = CD
Therefore, ΔACD ≅ ΔBCD ...(ASA criteria)
Hence, AD = BC and ∠A = ∠B.
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