Question

in the adjoining figure ABCD is a square and Triangle EDC is an equilateral triangle prove that AE=BE, angle DAE=15°

Answer 1

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Answer 2

Given: ABCD is a square and EDC is an equilateral triangle.

To prove: AE=BE and $\angle DAE=15^{\circ}$

Proof: In $\triangle ADE \text{ and }\triangle BCE$

           AD=BD             ( Side of same square)

      ∠ADE=∠BCE         (Each angle is 150°)

           DE=BE             ( Side of same equilateral triangle)

So, $\triangle ADE \cong \triangle BCE$ by SAS congruence

Therefore,  AE=BE   (By CPCT )

Hence Proved

In $\triangle ADE$

AD=DE

$\angle DAE=\angle DEC$  (If sides of a triangle are equal then their opposites angles are equal)

In $\triangle ADE$

$\angle DAE+\angle DEC+\angle ADE=180^{\circ}$

Angle sum property of triangle

$2\angle DAE+150^{\circ}=180^{\circ}$

$2\angle DAE=180^{\circ}-150^{\circ}$

$2\angle DAE=30^{\circ}$

$\angle DAE=15^{\circ}$

Hence Proved