Question

In the adjoining figure, ΔCDE is an equilateral triangle and ABCD is a square. Show that:

1. ΔΑDE ≅ ΔBCE
2. ΔАЕВ is an isosceles triangle.

Answer 1

Given:

• ABCD is a square.
• ΔCDE is an equilateral triangle.

To Prove:

1. ΔADE ≅ ΔBCE.
2. ΔAEB is an isosceles triangle.

Proof [Step-wise]:

1. From ABCD being a square:

AD = BC

And ∠ADC = ∠BCD = 90°   ...(Each corner of a square)

2. From ΔCDE equilateral: 

CD = DE = CE

And ∠CDE = ∠DCE = 60°

3. Compute the included angles:

∠ADE is the angle between AD and DE.

Moving from AD to DE can be done via AD → DC.   ...(90°)

Then DC → DE.   ...(60°)

So, ∠ADE = 90° + 60° = 150°.

Similarly, ∠BCE is the angle between BC and CE:

BC → CD   ...(90°) 

Then CD → CE   ...(60°)

So, ∠BCE = 90° + 60° = 150°.

Hence, ∠ADE = ∠BCE.

4. In triangles ADE and BCE we now have:

AD = BC   ...(Sides of the square)

DE = CE   ...(Sides of the equilateral triangle)

Included angle ∠ADE = ∠BCE   ...(Proved above) 

Therefore, by SAS, ΔADE ≅ ΔBCE.

5. From the congruence (CPCTC) corresponding sides AE and BE are equal. 

Therefore, AE = BE.

So, ΔAEB has two equal sides and is an isosceles triangle.

i. ΔADE ≅ ΔBCE   ...(By SAS)

ii. ΔAEB is isosceles   ...(AE = BE, from congruence)