Question

ABCD is a square and on the side DC, an equilateral triangle is constructed. Prove that (i) AE = BE and (ii) ​

Answer 1

AD = BC (sides of a square)

DE = CE (sides of equilateral triangle)

ADE = BCE = 90o + 60o = 150o

(BY SAS congruence theorem)

AE = BE

ADE = 90o + 60o = 150o

As AD=DC(sides of a square)

DC=DE

so, AD=DE

ADE=DEA

DAE =

= 15o

Answer 2

Step-by-step explanation:

Given : ABCD is a square and EDC is an equilateral triangle. AD = BC, DE = CE

To Prove : i) AE = BE

               ii) ∠DAE = 15°

Construction : Join A to E and B to E.

Proof :

i) In ΔADE and ΔBCE,

AD = BC (given)

∠ADE = ∠BCE (90° + 60°)

DE = CE (given)

Therefore, ΔADE is congruent to ΔBCE ( SAS rule)

AE = BE (CPCTC)

ii) ∠DAE + ∠ADE + ∠DEA = 180°

150° + ∠DAE + ∠DEA = 180°

∠DAE + ∠DEA = 180° -150°

2 ∠DAE = 30°

∠DAE = 15°

Hence Proved.