In the adjoining figure, ABCD is a square and is an equilateral triangle. Prove that: 1. AE = BE, 2. =15^ .

Given: ABCD is a square in which AB = BC = CD = DA. is an equilateral triangle in which ED = EC = DC and = = =60^ . To prove: AE = BE and =15^ Proof: in and , we have: AD = BC [Sides of a square] DE = EC [Sides of an equilateral triangle] = =90^ +60^ =150^ i.e., AE = BE Now, =150^ DA = DC [Sides of a square] DC = DE [Sides of an equilateral triangle] So, DA = DE and are isosceles triangle. i.e., = =1 2 (180^ -150^ )= 30^ 2 =15^

In the adjoining figure, ABCD is a square and is an equilateral t…

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Question

In the adjoining figure, ABCD is a square and $\triangle\text{EDC}$ is an equilateral triangle. Prove that:

AE = BE,
$\angle\text{DAE}=15^{\circ}.$

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Answer

Given: ABCD is a square in which AB = BC = CD = DA. $\triangle\text{EDC}$ is an equilateral triangle in which ED = EC = DC and
$\angle\text{EDC}=\angle\text{DEC}=\angle\text{DCE}=60^{\circ}.$
To prove: $\text{AE = BE}$ and $\angle\text{DAE}=15^{\circ}$
Proof: in $\triangle\text{ADE}$ and $\triangle\text{BCE},$ we have:
$\text{AD = BC}$ [Sides of a square]
$\text{DE = EC}$ [Sides of an equilateral triangle]
$\angle\text{ADE}=\angle\text{BCE}=90^{\circ}+60^{\circ}=150^{\circ}$
$\therefore\triangle\text{ADE}\cong\triangle\text{BCE}$
i.e., $\text{AE = BE}$
Now, $\angle\text{ADE}=150^{\circ}$
$\text{DA = DC}$ [Sides of a square]
$\text{DC = DE}$ [Sides of an equilateral triangle]
So, $\text{DA = DE}$
$\triangle\text{ADE}$ and $\triangle\text{BCE}$ are isosceles triangle.
i.e., $\angle\text{DAE}=\angle\text{DEA}=\frac{1}{2}(180^{\circ}-150^{\circ})=\frac{30^{\circ}}{2}=15^{\circ}$