We are given that $AD \cong AE$, $BA || CE$, $CB || DA$, and $m\angle DAE = 54^\circ$. We need to find $m\angle BCD$.

1. Problem Description

We are given that $AD \cong AE$, $BA || CE$, $CB || DA$, and $m\angle DAE = 54^\circ$. We need to find $m\angle BCD$.

2. Solution Steps

Since $AD \cong AE$, $\triangle ADE$ is an isosceles triangle. Therefore, $\angle ADE \cong \angle AED$. Let $x = m\angle ADE = m\angle AED$. The sum of the angles in $\triangle ADE$ is $180^\circ$.

$m\angle DAE + m\angle ADE + m\angle AED = 180^\circ$

$54^\circ + x + x = 180^\circ$

$2x = 180^\circ - 54^\circ$

$2x = 126^\circ$

$x = \frac{126^\circ}{2} = 63^\circ$

So, $m\angle ADE = m\angle AED = 63^\circ$.

Since $CB || DA$, $ABCD$ is a trapezoid, but we don't know if it's an isosceles trapezoid. Since $BA || CE$, $ABCE$ is a trapezoid. Since $CB || DA$, $\angle BCD$ and $\angle ADC$ are supplementary angles. Also, since $BA || CE$, $\angle ABC$ and $\angle BCE$ are supplementary angles.

$m\angle BCD + m\angle ADC = 180^\circ$

$m\angle ADC = m\angle ADE = 63^\circ$.

Since $CB || DA$, the quadrilateral $ABCD$ is a trapezoid. Furthermore, we know $BA || CE$. Thus, $\angle ABC$ and $\angle BCE$ are supplementary, so their sum is $180^\circ$. Also, $AD || BC$, so $\angle ADC$ and $\angle BCD$ are supplementary, so their sum is $180^\circ$.

Since $m\angle ADC = 63^\circ$, we have

$m\angle BCD + 63^\circ = 180^\circ$

$m\angle BCD = 180^\circ - 63^\circ = 117^\circ$

3. Final Answer

$m\angle BCD = 117^\circ$