Question

4. In the adjoining figure, ABCD is a square and
AEDC is an equilateral triangle. Prove that
(i) AE = BE, (ii) ZDAE=15°​

Answer 1

Answer:

4. In the adjoining figure, ABCD is a square and AEDC is an equilateral triangle. Prove that (i) AE BE, (ii) ZDAE 15°.

Given: ABCD is a square in which AB = BC = CD = DA. ∆EDC is an equilateral triangle in which ED = EC = DC and

EDC = DEC = DCE = 60∘.

To prove: AE = BE and DAE = 15∘

Proof: In ∆ADE and ∆BCE, we have:

AD = BC [Sides of a square]

DE = EC [Sides of an equilateral triangle]

ADE = BCE = 90∘ + 60∘ = 150∘

∴ ∆ADE ≅ ∆BCE

i.e., AE = BE

Now, 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., DAE = DEA = 1/2(180°-150°)=30/2=15°

Hope, it will help you..