Can we use similar triangles to find this side?

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In the diagram, \(\widehat{ABC}=\widehat{BDC}=90^\circ\).

1. Write down an angle equal to \(\widehat{CBD}\).

The angle \(\widehat{BAD}\) is equal to \(\widehat{CBD}\).

The angle \(\widehat{CBD}=90^\circ-\widehat{BCD}\), since the triangle \(BCD\) is right-angled.

Since the triangle \(ABC\) is right-angled, we also have that \(\widehat{BAD}=90^\circ-\widehat{BCD}=\widehat{CBD}\).

2. Given that \(AC= \quantity{10}{cm}\) and \(BC=\quantity{7}{cm}\), use similar triangles to calculate \(CD\).

The two right-angled triangles \(BCD\) and \(ABC\) are similar, since their angles are equal.

So the ratios of the corresponding side lengths are equal, giving \[ \frac{CD}{7}=\frac{7}{10}, \] and so \(CD=\dfrac{49}{10}=\quantity{4.9}{cm}\).