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Question
CDE is an equilateral triangle formed on a side CD of a square ABCD. Show that ΔADE ≅ΔBCE.
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Solution
We have to prove that ΔADE ≅ ΔBCE
Given ABCDis a square
So AB = BC = CD = AD
Now in ΔEDC is equilateral triangle.
So DE = EC = CB
In ΔAED and ΔCEB
AD = BC (Side of triangle)
DE = CE (Side of equilateral triangle)
∠ADE = ∠ADC + ∠CDE
= 90 + 60
= 150
And,
∠BCE = ∠BCD + ∠DCE
= 90 + 60
= 150
So ∠ACE = ∠BCDE
Hence from SAS congruence ΔADE ≅ ΔBCE Proved.
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