Question

1. ABCD is a square and A CDE is an equilateral triangle. Prove that AE = BE and find AED​

Answer 1

$\boxed{\bf{\red{Given}}}$

• ABCD is square
• ∆ CDE is an equilateral triangle

$\boxed{\bf{\red{To\:find}}}$

• AE = BE
• AED = ?

$\boxed{\bf{\red{Proof}}}$

In ∆ADE and ∆BCE

• AD = BC........(given)
• Angle ADE = Angle BCE
• SE = CE..........(given)

.°. ∆ADE ~ ∆BCE.......(SAS rule)

.°. AE = BE....... (PCT)...... (➊ Proved)

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As ABCD is a square,

therefore, angle CDA = angle DCB = 90°

Consider, ∆ EDA

So we get,

Angle EDA = 90° + 60° = 150°

Now,

ED = DA (from figure we know that ED and EC are the sides of equilateral triangle )

From the figure we also know that, the base angles are equal.

Angle DEA = angle DAE

By angles sum property:-

⟹ Angle EDA + Angle DAE + Angle DEA = 180°

⟹ 150° + Angle DAE + Angle DEA = 180°

We know that, Angle DEA = DAE

So we get ,

⟹ 150° + angle DEA + angle AED = 180°

⟹ 2angleAED = 180 – 150°

⟹ 2 angle AED = 30°

⟹ Angle AED = 15°...........(➋ Answer)

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Answer 2

Answer:

Step-by-step explanation: