Question 11.4.4

Let’s see what happens when we apply Ptolemy’s theorem (see Question 10.6.3) to a cyclic quadrilateral with one of its diagonals forming a diameter of the circle.

1. Given that $AC$ is a diameter of the circle through $A$, $B$, $C$, and $D$, find $\angle ABC$ and $\angle ADC$. Show answer

   Angles at the circumference subtended by a diameter are $90^\circ$ (see Question 10.4.5).

   

2. Find $BD$ in terms of $\theta$, $\phi$, and the diameter, $AC$, where $\theta = \angle ACB$ and $\phi = \angle ACD$. Show hint

   Consider $\angle BOD$ where $O$ is the centre of the circle.

   Show answer

   You may already be familiar with the use of the sine rule (Section 11.4.4: Sine and Cosine Rules) for finding the diameter of the circumscribed circle of a triangle, in which case you get \begin{align*} \frac{BD}{\sin(\angle BCD)} &= AC\\ BD &= AC\sin(\theta + \phi). \end{align*} Otherwise, consider that $\angle BOD = 2(\theta + \phi)$ (angle at the centre is twice the angle at the circumference, standing on the same arc, see Question 10.4.4). Then $BD$ is the chord length corresponding to that angle (Question 11.4.1): \begin{align*} BD &= AC\sin\frac{\angle BOD}{2}\\ &= AC\sin(\theta + \phi). \end{align*}

   

3. Now apply Ptolemy’s theorem, \[AC\cdot BD = AB\cdot CD + BC\cdot AD.\] Show hint

   $\Delta ABC$ and $\Delta ADC$ are right-angled triangles, so we can express their sides in terms of $AC$, $\theta$, and $\phi$.

   Show answer

   Since $\Delta ABC$ is a right-angled triangle, $AB = AC\sin\theta$ and $BC = AC\cos\theta$. Similarly, $AD = AC\sin\phi$ and $CD = AC\cos\phi$. Substituting these in to Ptolemy’s theorem, we get: \begin{align*} AC\cdot BD &= AB\cdot CD + BC\cdot AD\\ AC\cdot AC\sin(\theta + \phi) &= AC\sin\theta \cdot AC\cos\phi + AC\cos\theta \cdot AC\sin\phi\\ \sin(\theta + \phi) &= \sin\theta\cos\phi + \cos\theta\sin\phi \end{align*} So now if we know the sines and cosines of two angles, we can find the sine of their sum.