Question
$\text{ABCD}$ is a square and $\text{DEC}$ is an equilateral triangle. Prove that $\text{AE = BE.}$
✓
Answer
In $\text{DEDA}$ and $\text{DECB,}$
$\text{DE = CE} \ldots \ldots [$ Sides of an equilateral triangle $]$
$\text{AD = BC} \ldots \ldots [$ Sides of a square $]$
$\angle EDA =\angle ECB \ldots[$ As $\angle EDC =\angle ECD$ and $\angle ADC =\angle BCD ]$
$\angle EDC +\angle ADC =\angle ECD +\angle BCD \ldots[$ By addition $]$
$\Rightarrow \angle EDA =\angle ECB$
$\therefore DEDA \cong DECB \ldots[$ By $\text{SAS}$ property $]$
$\therefore AE = BE \ldots \text { [c.p.c.t. }]$
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