Question
In the adjoining figure,$\text{ABCD}$ is a square and $\triangle\text{EDC}$ is an equilateral triangle. Prove that:
$i. AE = BE$,
$ii. \angle\text{DAE}=15^{\circ}.$
$i. AE = BE$,
$ii. \angle\text{DAE}=15^{\circ}.$
✓
Answer
Given: $\text{ABCD}$ 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.,$\angle\text{DAE}=\angle\text{DEA}=\frac{1}{2}(180^{\circ}-150^{\circ})=\frac{30^{\circ}}{2}=15^{\circ}$
$\triangle\text{EDC}$ 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.,$\angle\text{DAE}=\angle\text{DEA}=\frac{1}{2}(180^{\circ}-150^{\circ})=\frac{30^{\circ}}{2}=15^{\circ}$
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