Question

In the adjoining figure, ABCD is a square and ∆EDC is an equilateral triangle. Prove that 1. AE=BE and 2. Angle DAE=15 degree.​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

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

So we get ∠ EDC = ∠ ECD = 60

We know that ABCD is a square

So we get ∠ CDA = ∠ DCB = 90o

Consider △ EDA

We get ∠ EDA = ∠ EDC + ∠ CDA By substituting the values in the above equation ∠ EDA = 60+ 90

So we get ∠ EDA = 150 …….

(1) Consider △ ECB

We get ∠ ECB = ∠ ECD + ∠ DCB

By substituting the values in the above equation

∠ ECB = 60 + 90

So we get ∠ ECB = 150

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 By angle sum property ∠ EDA + ∠ DAE + ∠ DEA = 180

By substituting the values we get 150 + ∠ DAE + ∠ DEA = 180

We know that ∠ DEA = ∠ DAE

So we get 150+ ∠ DAE + ∠ DAE = 180

On further calculation 2 ∠ DAE = 180 – 150

By subtraction 2 ∠ DAE = 30°

By division ∠ DAE = 15 °

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