Question
$CDE$ is an equilateral triangle formed on a side $CD$ of a square $ABCD$. Show that $\triangle\text{ADE}\cong\triangle\text{BCE}.$
✓
Answer
We have to prove that $\triangle\text{ADE}\cong\triangle\text{BCE}$
Given $ABCD$ is a square So $\text{AB}=\text{BC}=\text{CD}=\text{AD}$
Now in $\triangle\text{EDC}$ is equilateral triangle.
So $\text{DE}=\text{EC}=\text{CB}$ In $\triangle\text{AED}$ and $\triangle\text{CEB}$
$\text{AD}=\text{BC}$ (Side of triangle) $\text{DE}=\text{CE}$ (Side of equilateral triangle) $\angle\text{ADE}=\text{ADC}+\angle\text{CDE}$
$=90+60$
$=150$ And, $\angle\text{BCE} =\angle\text{BCD}+\angle\text{DCE}$
$90+60$
$=150$ So $\angle\text{ADE}=\angle\text{BCE}$ Hence from $SAS$ congruence $\triangle\text{ADE}\cong\triangle\text{BCE}$ Proved.
Given $ABCD$ is a square So $\text{AB}=\text{BC}=\text{CD}=\text{AD}$
Now in $\triangle\text{EDC}$ is equilateral triangle.
So $\text{DE}=\text{EC}=\text{CB}$ In $\triangle\text{AED}$ and $\triangle\text{CEB}$
$\text{AD}=\text{BC}$ (Side of triangle) $\text{DE}=\text{CE}$ (Side of equilateral triangle) $\angle\text{ADE}=\text{ADC}+\angle\text{CDE}$
$=90+60$
$=150$ And, $\angle\text{BCE} =\angle\text{BCD}+\angle\text{DCE}$
$90+60$
$=150$ So $\angle\text{ADE}=\angle\text{BCE}$ Hence from $SAS$ congruence $\triangle\text{ADE}\cong\triangle\text{BCE}$ Proved.
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