$CDE$ is an equilateral triangle formed on side $CD$ of a square $ABCD$ (see figure). Show that $\triangle ADE \cong \triangle BCE$.

(N/A) Given: An equilateral triangle $CDE$ is formed on side $CD$ of square $ABCD$.
To prove: $\triangle ADE \cong \triangle BCE$.
Proof: In square $ABCD$,we have:
$AD = BC$ (Sides of a square are equal) ... $(1)$
$\angle ADC = \angle BCD = 90^{\circ}$ (Angles of a square) ... $(2)$
In equilateral triangle $CDE$,we have:
$DE = CE$ (Sides of an equilateral triangle are equal) ... $(3)$
$\angle CDE = \angle DCE = 60^{\circ}$ (Angles of an equilateral triangle) ... $(4)$
Adding $(2)$ and $(4)$:
$\angle ADC + \angle CDE = \angle BCD + \angle DCE$
$\angle ADE = \angle BCE$ ... $(5)$
Now,in $\triangle ADE$ and $\triangle BCE$:
$AD = BC$ (From $1$)
$\angle ADE = \angle BCE$ (From $5$)
$DE = CE$ (From $3$)
Therefore,by $SAS$ congruence criterion,$\triangle ADE \cong \triangle BCE$.