Solution

A picture of the triangle A B C, where A B C is a right-angle and B C is the hypotenuse of another triangle B C D  where d lies on A C

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 \(\widehat{CBD}\).

A picture of the triangle A B C with the angles marked

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 \(\widehat{BAD}=90^\circ-\widehat{BCD}=\widehat{CBD}\).

A picture of the triangle A B C, where A B C is a right-angle and B C is the hypotenuse of another triangle B C D where d lies on A C
  1. 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 \(CD=\dfrac{49}{10}=\quantity{4.9}{cm}\).