![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](/thinking-about-geometry/r5081/images/pic1.png)
In the diagram, \(\widehat{ABC}=\widehat{BDC}=90^\circ\).
- 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](/thinking-about-geometry/r5081/images/anglepic.png)
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](/thinking-about-geometry/r5081/images/pic1.png)
- 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 \(CD=\dfrac{49}{10}=\quantity{4.9}{cm}\).