Question

CDE is an equilateral triangle formed on a side CD of a square ABCD. Show that Δ ADE ≅ Δ BCE.

Answer 1

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

Δ ADE ≅ Δ BCE by SAS theorem.

Given,

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

To prove: ∆ ADE ≅ ∆ BCE.

Proof:

Since all sides of equilateral triangle are equal

we have,  CD=DE=EC  ....... (1)

and all angles are of 60°

∠ EDC = ∠ DCE = ∠ CED = 60°  ..........(2)

Since all sides of square are congruent

we have, AB=BC=CD=DA  ........ (3)

and all angles are of 90°

∠ ABC = ∠ BCD = ∠ CDA = ∠ DAB = 90°  ..........(4)

∴ ∠ ADE = ∠ ADC + ∠ CDE = 90° + 60° = 150°

∠ BCE = ∠ BCD + ∠ DCE = 90° + 60° = 150°

In ∆ ADE and ∆ BCE

AD = BC    (from eqn  3)

DE = EC     (from eqn 1)

∠ ADE = ∠ BCE    (as each angles equal 150°)

Therefore, by SAS theorem ∆ ADE ≅ ∆ BCE.

Answer 2

Given:

CDE is an equilateral triangle formed on a side CD of a square ABCD

To show:

Δ ADE ≅ Δ BCE

Proof:

In ΔADE and ΔBCE

DE = CE → [sides of the equilateral triangle]

∠ADE = ∠BCE

∵ ∠ADC = ∠BCD = 90°

∵ ∠EDC = ∠ECD = 90°

∵ ∠ADE = 90° + 60° = 150°

∵ ∠BCE = 90° + 60° = 150°

AD = BC → [sides of square]

ΔADE ≡ ΔBCE [by SAS congruence rule]

Hence proved.