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Question
In the figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Prove that BC = DE.
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Solution
In ΔADE and ΔBAC
AE = AC
AB = AD
∠BAD = ∠EAC
∠DAC = ∠DAC = DAC ...(common)
⇒ ∠BAC = ∠EAD = EAD
Therefore, ΔADE ≅ ΔBAC ...(SAS criteria)
Hence, BC = DE.
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