The diagram below shows part of a regular 18-sided polygon, ABCDEF... The lines BC and ED are extended to meet at the point X such that XC = XD.

1. Find the exterior angle at a vertex of the regular 18-sided polygon

For a regular 18-sided polygon,
$\text{exterior angle}=\frac{360^\circ}{18}=20^\circ.$
Since $BC$ is extended through $C$ to $X$, $\angle XCD$ is this exterior angle.

1. Use the isosceles triangle property and angle sum of a triangle to find $\angle CXD$

Since $XC=XD$, $\triangle XCD$ is isosceles.

So $\angle XDC=\angle XCD=20^\circ$.

$\angle CXD=180^\circ-20^\circ-20^\circ=140^\circ$

1. Identify equal corresponding lengths from the regular polygon and the given condition

In a regular 18-sided polygon, all sides are equal, so
$BC=DE$.
It is also given that
$XC=XD$.

2. Express $BX$ and $EX$ as sums of equal parts

Since $B,C,X$ are collinear and $E,D,X$ are collinear,
$BX=BC+CX$
and
$EX=ED+DX$.

As $BC=DE$ and $CX=DX$,
$BX=EX$.

1. Identify a pair of equal angles at X

Since $B,C,X$ are collinear and $E,D,X$ are collinear,

From part (b), both are $140^\circ$.

2. Show another pair of corresponding angles are equal, then conclude similarity

Since $XC=XD$, $\triangle XCD$ is isosceles, so

From part (c), $BX=EX$, so $\triangle XBE$ is isosceles. Hence

Also $\angle BXE=140^\circ$, so

Thus

and

Therefore,

by AA similarity.

1. Use corresponding equal angles from the similar triangles to identify parallel lines

From part (d), $\triangle XCD \sim \triangle XBE$.

Hence the corresponding angles are equal, so

These are corresponding angles formed by the transversal $XB$ with the lines $CD$ and $BE$.

Therefore,