The diagram shows triangle ABD and BCD is a straight line. It is also known that $AB = 7$ cm, $BC = 5$ cm and $\angle BAC = \angle ADC$.

1. Identify the equal angle at $B$ using the straight line $BCD$.

Since $B$, $C$ and $D$ are collinear, $BC$ and $BD$ lie on the same straight line.

Hence
$\angle ABC = \angle DBA.$

2. Use the given angle equality and conclude similarity by AA.

It is given that
$\angle BAC = \angle ADC.$
Since $B$, $C$ and $D$ are collinear, $\angle ADC = \angle BDA$.

So
$\angle BAC = \angle BDA.$

Therefore, two pairs of corresponding angles are equal, so
$\triangle ABC \sim \triangle DBA \quad \text{(AA similarity).}$

1. Use corresponding sides of the similar triangles to find $BD$.

From part (a), $\triangle ABC \sim \triangle DBA$.

Corresponding sides are proportional:
$\frac{BC}{BA}=\frac{AB}{BD}$
Substitute $BC=5$ and $AB=BA=7$:
$\frac{5}{7}=\frac{7}{BD}$
$5BD=49$
$BD=9.8\text{ cm}$

2. Find $CD$ using the straight-line length $BD=BC+CD$.

Since $B$, $C$ and $D$ are collinear,
$CD=BD-BC$
$CD=9.8-5=4.8\text{ cm}$