In the diagram ABCDE is a regular pentagon. BE and BD are straight lines.

1. Find the angle at $A$ in the regular pentagon.

Since $ABCDE$ is a regular pentagon, each interior angle is
$\frac{(5-2)\times 180^\circ}{5}=108^\circ.$

2. Use triangle $ABE$ to find $\angle ABE$.

In $\triangle ABE$, $AB=AE$ because all sides of a regular pentagon are equal.

So $\triangle ABE$ is isosceles, and
$\angle ABE=\angle BEA.$
Hence
$\angle ABE=\frac{180^\circ-108^\circ}{2}=36^\circ.$

1. Find the angle between $BE$ and $BD$ at $B$.

In regular pentagon $ABCDE$, $\angle ABC=108^\circ$.

In $\triangle BCD$, $BC=CD$, so
$\angle CBD=\angle BDC=\dfrac{180^\circ-108^\circ}{2}=36^\circ$.

Hence
$\angle ABD=108^\circ-36^\circ=72^\circ$.

Using $\angle ABE=36^\circ$ from part (a)(i),
$\angle EBD=72^\circ-36^\circ=36^\circ.$

1. Compare a pair of alternate interior angles formed by transversal $BD$.

From part (a)(ii), $\angle EBD = 36^\circ$.

In a regular pentagon, diagonal $BD$ makes $\angle BDC = 36^\circ$ by the same base-angle reasoning in isosceles triangle $BCD$.

Hence $\angle EBD = \angle BDC$.

Therefore, $BE \parallel CD$ since alternate interior angles are equal.

1. Use angles on a straight line at $E$ to find $\angle DEX$.

In a regular pentagon, each interior angle is $108^\circ$, so $\angle AED = 108^\circ$.

Since $AE$ is produced to $X$, $AEX$ is a straight line.

$\angle DEX = 180^\circ - 108^\circ = 72^\circ$

1. Show that one pair of opposite sides is parallel.

Since $X$ lies on $CD$ produced, $D$, $C$ and $X$ are collinear.

From part (b), $BE \parallel CD$.

Therefore, $BE \parallel DX$.

2. Show that the other pair of opposite sides is parallel.

At $D$,
$\angle BDE = \angle CDE - \angle BDC.$
Since $ABCDE$ is a regular pentagon, $\angle CDE = 108^\circ$.

Also, $\angle BDC = 36^\circ$, so
$\angle BDE = 108^\circ - 36^\circ = 72^\circ.$
From part (c)(i),
$\angle DEX = 72^\circ.$
Thus $\angle BDE = \angle DEX$, so $BD \parallel EX$.

3. Conclude using the definition/property of a parallelogram.

In quadrilateral $BDXE$,
$BE \parallel DX \quad \text{and} \quad BD \parallel EX.$
Since both pairs of opposite sides are parallel, $BDXE$ is a parallelogram.