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, $\angle ADE = \angle AED$.

The sum of the angles in a triangle is $180^\circ$. Thus, in $\triangle ADE$, we have

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

$54^\circ + m\angle ADE + m\angle ADE = 180^\circ$

$2 \cdot m\angle ADE = 180^\circ - 54^\circ = 126^\circ$

$m\angle ADE = \frac{126^\circ}{2} = 63^\circ$

Since $CB || DA$, $\angle CDE$ and $\angle ADE$ are supplementary.

Therefore, $m\angle CDE + m\angle ADE = 180^\circ$

$m\angle CDE = 180^\circ - m\angle ADE = 180^\circ - 63^\circ = 117^\circ$

Since $BA || CE$, $\angle ABC$ and $\angle BCE$ are supplementary. Also, $CB || DA$, so $ABCD$ is a parallelogram. Thus, $\angle BCD = \angle BAD$.

Since $BA || CE$ and $CB || DA$, the quadrilateral $ABCD$ is a parallelogram. Thus, opposite angles are equal, so $\angle BCD = \angle BAD$. Also, consecutive angles are supplementary, so $\angle ABC + \angle BCD = 180^\circ$.

Since $CB || DA$, $m\angle ADC + m\angle BCD = 180^\circ$.

Since $BA || CE$, $m\angle ABC + m\angle BCE = 180^\circ$.

Since $CB || DA$, $\angle CDA$ and $\angle BCD$ are supplementary, so $m\angle CDA + m\angle BCD = 180^\circ$. Also, $\angle ADC$ and $\angle ADE$ are supplementary, and we know $m\angle ADE = 63^\circ$. Therefore, $m\angle ADC = 180^\circ - 63^\circ = 117^\circ$. So we have $117^\circ + m\angle BCD = 180^\circ$, and $m\angle BCD = 180^\circ - 117^\circ = 63^\circ$.

Consider the quadrilateral $ABCD$. Since $BA || CE$ and $CB || DA$, $ABCD$ is a parallelogram. Opposite angles are equal, so $\angle BCD = \angle DAB$. We are given $\angle DAE = 54^\circ$.

Since $DA || CB$, $m\angle BCD + m\angle CDA = 180^\circ$.

Also, $\angle ADC + \angle ADE = 180^\circ$.

We have $\angle ADE = 63^\circ$, so $\angle ADC = 180^\circ - 63^\circ = 117^\circ$.

Then $m\angle BCD + 117^\circ = 180^\circ$, so $m\angle BCD = 180^\circ - 117^\circ = 63^\circ$.

Since $ABCD$ is a parallelogram, $\angle BAD = \angle BCD$. Since $BC || AD$, $\angle BCE = \angle ABC$.

Since $BA || CE$, $m\angle ABC + m\angle BCE = 180$.

Consider the angles around point $C$.

$m\angle BCD = 180^\circ - m\angle CDA$ since $AD || BC$.

Since $\angle ADE = 63^\circ$, $\angle CDA = 180^\circ - 63^\circ = 117^\circ$.

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

3. Final Answer

$m\angle BCD = 63^\circ$