Question
In the given figure $\triangle\text{CDE}$ is an equilateral triangle formed on a side CD of a square ABCD. Show that $\triangle\text{ADE}\cong\triangle\text{BCE}.$
✓
Answer
Given, $\triangle\text{CDE}$ is an equilateral triangle formes on a side CD of a ABCD.
$\triangle\text{ADE}\cong\triangle\text{BCE}$
In $\triangle\text{ADE}$ and $\triangle\text{BCE},$
$\text{DE}=\text{CE}$
$\angle\text{ADE}=\angle\text{BCE}$
$\text{AD}=\text{BC}$
$\triangle\text{ADE}\cong\triangle\text{BCE}$
$\triangle\text{ADE}\cong\triangle\text{BCE}$
In $\triangle\text{ADE}$ and $\triangle\text{BCE},$
$\text{DE}=\text{CE}$
$\angle\text{ADE}=\angle\text{BCE}$
$\text{AD}=\text{BC}$
$\triangle\text{ADE}\cong\triangle\text{BCE}$
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