In the adjoining figure, ABCD is a square and ∆EDC is an equilateral triangle.

Question:

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

 

Solution:

Given: $A B C D$ is a square in which $A B=B C=C D=D A . \triangle E D C$ is an equilateral triangle in which $E D=E C=D C$ and

$\angle E D C=\angle D E C=\angle D C E=60^{\circ}$.

To prove: $A E=B E$ and $\angle D A E=15^{\circ}$

Proof: In $\triangle A D E$ and $\triangle B C E$, we have:

$A D=B C \quad$ [Sides of a square]

$\mathrm{DE}=\mathrm{EC} \quad$ [Sides of an equilateral triangle]

$\angle \mathrm{ADE}=\angle \mathrm{BCE}=90^{\circ}+60^{\circ}=150^{\circ}$

$\therefore \triangle \mathrm{ADE} \cong \triangle \mathrm{BCE}$

i.e., $\mathrm{AE}=\mathrm{BE}$

Now, $\angle \mathrm{ADE}=150^{\circ}$

$\mathrm{DA}=\mathrm{DC} \quad$ [Sides of a square]

$\mathrm{DC}=\mathrm{DE} \quad$ [Sides of an equilateral triangle]

So, $D A=D E$

$\triangle A D E$ and $\triangle B C E$ are isosceles triangles.

i.e., $\angle \mathrm{DAE}=\angle \mathrm{DEA}=\frac{1}{2}\left(180^{\circ}-150^{\circ}\right)=\frac{30^{\circ}}{2}=15^{\circ}$

 

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