In the diagram, there are two right-angled triangles ΔABD and ΔCEB such that AD = 12 cm, BD = 15 cm and BE = 20 cm.

1. Find the length of $AB$ in right-angled triangle $ABD$.

In $\triangle ABD$, the right angle is at $A$, so $BD$ is the hypotenuse.

$AB^2 + AD^2 = BD^2$

$AB^2 + 12^2 = 15^2$

$AB^2 = 225 - 144 = 81$

$AB = 9\text{ cm}$

2. Use the cosine ratio for angle $ABD$.

For $\angle ABD$, the adjacent side is $AB$ and the hypotenuse is $BD$.

$\cos \angle ABD = \frac{AB}{BD} = \frac{9}{15} = \frac{3}{5}$

1. Find the length of $AB$ in right-angled triangle $ABD$.

In $\triangle ABD$,
$AB^2+AD^2=BD^2$
$AB^2+12^2=15^2$
$AB^2=225-144=81$
$AB=9\text{ cm}$

2. Use similarity to find the corresponding side $BC$.

Since $\triangle ABD$ and $\triangle CEB$ are similar, corresponding sides are proportional.

The hypotenuse $BD$ corresponds to $BE$, and $AB$ corresponds to $BC$.
$\frac{BC}{AB}=\frac{BE}{BD}$
$\frac{BC}{9}=\frac{20}{15}$
$BC=9\times \frac{20}{15}=12\text{ cm}$