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# CDE is an equilateral triangle formed on a side CD of a square ABCD (Fig.7.5). Show that ∆ ADE ≅ ∆ BCE.

**Solution:**

Given, CDE is an equilateral triangle formed on the side CD of a square ABCD.

We have to show that the triangles ADE and BCE are congruent.

We know that sides of a square are equal and each angle is equal to 90 degrees.

AB = BC = CD = AD

∠ABC = ∠BCD = ∠CDA = ∠DAB = 90°

We know that in an equilateral triangle all the sides are equal and each angle is equal to 60 degrees.

CD = CE = DE

∠CDE = ∠DCE = ∠DEC = 60°

Considering triangles ADE and BCE,

Sides of a square, AD = BC

Sides of equilateral triangle, DE = CE

∠ADE = ∠ADC + ∠CDE

∠ADE = 90° + 60°

∠ADE = 150°

∠BCE = ∠BCD + ∠DCE

∠BCE = 90° + 60°

∠BCE = 150°

∠ADE = ∠BCE

By SAS criterion, ΔADE ≅ΔBCE

Therefore, the triangles ADE and BCE are congruent.

**✦ Try This:** In the given figure, DE || BC. If DE = 3cm, BC = 6 and ar(ΔADE) = 15 square cm, find the area of Δ ABC.

**☛ Also Check: **NCERT Solutions for Class 9 Maths Chapter 7

**NCERT Exemplar Class 9 Maths Exercise 7.3 Problem 3**

## CDE is an equilateral triangle formed on a side CD of a square ABCD (Fig.7.5). Show that ∆ ADE ≅ ∆ BCE

**Summary:**

CDE is an equilateral triangle formed on a side CD of a square ABCD (Fig.7.5). It is shown that ∆ ADE ≅ ∆ BCE by SAS criterion

**☛ Related Questions:**

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