Question

In the adjoining figure, ABCD is a square and ∆EDC is an equilateral triangle. Prove that (i) AE = BE, (ii) ∠DAE = 15°.

Answer 1

Given: ABCD is a square in which AB = BC = CD = DA. ∆EDC is an equilateral triangle in which ED = EC = DC and
$\angle$ EDC = $\angle$ DEC = $\angle$​DCE = 60^∘.
To prove: AE = BE and $\angle$DAE = 15^∘​
Proof: In ∆ADE and ∆BCE, we have:
AD = BC [Sides of a square]

DE = EC​ [Sides of an equilateral triangle]
$\angle$ADE = $\angle$BCE = 90^∘ + 60^∘ = 150^∘​

∴ ∆ADE ≅ ∆BCE
i.e., AE = BE

Now, $\angle$ADE = 150^∘
DA = DC [Sides of a square]
DC = DE [Sides of an equilateral triangle]
So, DA = DE
∆ADE and ∆BCE are isosceles triangles.
i.e., $\angle$DAE = $\angle$DEA = $\frac{1}{2}\left(180°-150°\right)=\frac{30°}{2}=15°$