Question

in the adjoining figure, ABCD is a square and ∆EDC is an equilateral triangle . prove that AE = BE and Angle DAE = 15°..​

Answer 1

Answer:

I)

∠ CDA = ∠ DCB = 90o

Consider △ EDA

We get ∠ EDA = ∠ EDC + ∠ CDA

∠ EDA = 60o + 90o So we get ∠ EDA = 150o Consider△ECB

We get ∠ECB = ∠ECD + ∠ DCB

∠ ECB = 60o + 90o So we get ∠ ECB = 150o

So we get ∠ EDA = ∠ ECB

Consider △ EDA and △ ECB

So we get ED=EC

DA=CB

By SAS congruence criterion

△ EDA ≅ △ ECB

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

Hence proved

2) Consider △ EDA

We know that

ED = DA

∠ DEA = ∠ DAE

∠ EDA = 150o

By angle sum property

∠ EDA + ∠ DAE + ∠ DEA = 180o

150o + ∠ DAE + ∠ DEA = 180o

We know that

∠ DEA = ∠ DAE

So we get

150o + ∠ DAE + ∠ DAE = 180o

2 ∠ DAE = 180o – 150o

By subtraction

2 ∠ DAE = 30o

By division

∠ DAE = 15o (Hence proved)

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