In the diagram, ACD is a triangle. B is the point on AC such that AB = 10 cm and BC = 4 cm. E is the point on AD where AE = 7 cm and ED = 13 cm.

1. Find the full lengths of the corresponding sides in the larger triangle.

$AC=AB+BC=10+4=14\text{ cm}$

$AD=AE+ED=7+13=20\text{ cm}$

2. Show two pairs of sides are in the same ratio and use the included angle.

Compare the sides about angle $A$:

$\frac{AC}{AE}=\frac{14}{7}=2$

$\frac{AD}{AB}=\frac{20}{10}=2$

Also, $\angle CAD=\angle EAB$ since both are the same angle at $A$.

Therefore, $\triangle ACD \sim \triangle AEB$ by SAS similarity.

1. Relate the area ratio to the ratio of the bases on the same straight line.

Since $F$ lies on $AB$, triangles $\triangle AEF$ and $\triangle AEB$ have the same height from $E$ to line $AB$.

Hence,
$\frac{\text{Area of }\triangle AEF}{\text{Area of }\triangle AEB}=\frac{AF}{AB}.$

2. Use the given ratio and $AB=10$ cm to find $AF$.

Given
$\frac{\text{Area of }\triangle AEF}{\text{Area of }\triangle AEB}=\frac14,$
so
$\frac{AF}{10}=\frac14.$
Therefore,
$AF=\frac14\times 10=2.5.$