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Question
In the following figure, AB = EF, BC = DE and ∠B = ∠E = 90°.
Prove that AD = FC.
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Solution
Given that, BC = DE
⇒ BC + CD = DE + CD ....( Adding CD on both sides )
⇒ BD = CE ....(i)
Now, in ΔABD and ΔFEC,
AB = EF ....(given)
∠ABD = ∠FEC ....(Each 90°)
BD = CE ....[ From (i) ]
⇒ ΔABD ≅ ΔFEC ...(by SAS congruence criterion)
⇒ AD = FC ...(c.p.c.t.)
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