Question

can anyone solve this
solution=?

Answer 1

(i) Since ABCD is a square and ∆DCE is an equilateral triangle.
∠ADC = 90°
∠EDC = 60°
∠ADC + ∠EDC = 90° + 60°
∠ADE = 150°
Similarly, we have
∠BCE = 150°

(ii) Thus in ∆ADE and ∆BCE, we have
AD = BC (Sides of a square)
∠ADE = ∠BCE = 150° (Proved)
DE = CE (Sides of a triangle)
So by SAS criterion, we have
∆ADE ≅ ∆BCE

(iii) Thus, AE = BE (by CPCT)

HENCE PROVED!

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