In the diagram ABC is a straight line, $BC = 10\ \text{cm}$, $CD = 8\ \text{cm}$ and $BD = 6\ \text{cm}$.

1. Check whether triangle $BCD$ is right-angled using the side lengths.

$BD^{2}+CD^{2}=6^{2}+8^{2}=36+64=100$
$BC^{2}=10^{2}=100$
Hence,
$BD^{2}+CD^{2}=BC^{2}.$

2. Conclude the position of the right angle.

By the converse of Pythagoras' theorem, $riangle BCD$ is right-angled at $D$, since $BC$ is the longest side.

So $\angle BDC=90^\circ$.

3. Use the circle theorem to justify that one circle passes through $B$, $C$ and $D$.

A circle drawn with $BC$ as diameter has angle $\angle BDC=90^\circ$ at any point $D$ on the circumference.

Since $\angle BDC=90^\circ$, $D$ lies on the circle with diameter $BC$.

Therefore, a circle passing through $B$, $C$ and $D$ can be drawn.

1. Relate $\angle ABD$ to the angle at $B$ in right-angled triangle $BCD$ and find its cosine.

Since $ABC$ is a straight line, $\angle ABD$ and $\angle CBD$ are supplementary.

In right-angled $\triangle BCD$,
$\cos CBD=\frac{BD}{BC}=\frac{6}{10}=\frac{3}{5}.$
Hence
$\cos ABD=\cos(180^\circ-\angle CBD)=-\cos CBD=-\frac{3}{5}.$