(i) From the figure we know that △ EDC is an equilateral triangle.

So we get ∠ EDC = ∠ ECD = 60^o

We know that ABCD is a square

So we get ∠ CDA = ∠ DCB = 90^o

Consider △ EDA

We get

∠ EDA = ∠ EDC + ∠ CDA

By substituting the values in the above equation

∠ EDA = 60^o + 90^o

So we get

∠ EDA = 150^o ……. (1)

Consider △ ECB

We get

∠ ECB = ∠ ECD + ∠ DCB

By substituting the values in the above equation

∠ ECB = 60^o + 90^o

So we get

∠ ECB = 150^o

So we get ∠ EDA = ∠ ECB …… (2)

Consider △ EDA and △ ECB

From the figure we know that ED and EC are the sides of equilateral triangle

So we get

ED = EC

We also know that the sides of square are equal

DA = CB

By SAS congruence criterion

△ EDA ≅ △ ECB

AE = BE (c. p. c. t)

(ii) Consider △ EDA

We know that

ED = DA

From the figure we know that the base angles are equal

∠ DEA = ∠ DAE

Based on equation (1) we get ∠ EDA = 150^o

By angle sum property

∠ EDA + ∠ DAE + ∠ DEA = 180^o

By substituting the values we get

150^o + ∠ DAE + ∠ DEA = 180^o

We know that ∠ DEA = ∠ DAE

So we get

150^o + ∠ DAE + ∠ DAE = 180^o

On further calculation

2 ∠ DAE = 180^o – 150^o

By subtraction

2 ∠ DAE = 30^o

By division

∠ DAE = 15^o