在图中，ABC 和 AED 是直线，且 BE 平行于 CD。

1. State the common angle at $A$.

Since $ABC$ and $AED$ are straight lines, $A, B, C$ are collinear and $A, E, D$ are collinear.

Therefore,
$\angle BAE = \angle CAD.$

2. Use parallel lines to find another pair of equal corresponding angles and conclude similarity.

Since $BE \parallel CD$ and $AB$ lies on the same straight line as $AC$,
$\angle ABE = \angle ACD$
(corresponding angles).

Hence, two pairs of corresponding angles are equal, so
$\triangle ABE \sim \triangle ACD$
(AA similarity).

1. Use similarity to write a proportion between corresponding sides.

From part (a), $\triangle ABE \sim \triangle ACD$.

Therefore, corresponding sides are proportional:
$\frac{AB}{AC}=\frac{AE}{AD}.$

2. Substitute whole side lengths and rearrange to obtain the required ratio.

Since $A,B,C$ are collinear and $A,E,D$ are collinear,
$AC=AB+BC, \qquad AD=AE+ED.$
So
$\frac{AB}{AB+BC}=\frac{AE}{AE+ED}.$
Cross-multiplying,
$AB(AE+ED)=AE(AB+BC).$
Expanding,
$AB\cdot AE+AB\cdot ED=AE\cdot AB+AE\cdot BC.$
Hence
$AB\cdot ED=AE\cdot BC,$
and therefore
$\frac{BC}{ED}=\frac{AB}{AE}.$

1. Compare the given angle sum with the cyclic quadrilateral angle property.

If $E$ were on the circle, then $B,C,D,E$ would form a cyclic quadrilateral, so opposite angles would be supplementary:
$\angle BCD+\angle BED=180^\circ.$
For a point outside the circle, the angle subtended by chord $BD$ at $E$ is smaller than the angle on the circumference, so the sum would be less than $180^\circ$.
Since
$\angle BCD+\angle BED>180^\circ,$
$E$ must lie inside the circle.