Question 10.6.3

Let’s think about cyclic quadrilaterals, that is, quadrilaterals whose four corners lie on a circle. These were of special interest to ancient astronomers such as Ptolemy, studying the angles between stars and planets on the celestial sphere.

1. If we draw in the diagonals $AC$ and $BD$ of the cyclic quadrilateral $ABCD$, can you see any congruent angles? Show hint

   Angles standing on the same arc are congruent.

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   There is a pair of congruent angles standing on each arc $AB$, $BC$, $CD$, and $DA$. \begin{align*} AB: \angle ADB &\cong \angle ACB\\ BC: \angle BAC &\cong \angle BDC\\ CD: \angle CAD &\cong \angle CBD\\ DA: \angle DBA &\cong \angle DCA \end{align*} Let the intersection of $AC$ and $BD$ be $E$. Then $\angle AEB \cong \angle CED$ and $\angle BEC \cong \angle DEA$ (vertically opposite angles).

   

2. Can you see any similar triangles? Show answer

   $\Delta AEB \sim \Delta DEC$ and $\Delta BEC \sim \Delta AED$.

   

3. Find $AC\div BD$ in terms of $AB$, $BC$, $CD$, and $DA$. This gives Ptolemy’s second theorem. You can skip this part if you like. Show hint

   The similar triangles give you a few equations, and also $AC = EA + EC$ and $BD = EB + ED$. This is a lot of simultaneous equations, so it may take some work to distil them down.

   Show answer

   From the similar triangles we get: \begin{align*} \frac{EA}{ED} &= \frac{AB}{CD}\tag{$1$}\\[5pt] \frac{EB}{EC} &= \frac{AB}{CD}\tag{$2$}\\[5pt] \frac{EA}{EB} &= \frac{DA}{BC}\tag{$3$}\\[5pt] \frac{ED}{EC} &= \frac{DA}{BC}\tag{$4$} \end{align*} And we can write $AC$ and $BD$ in terms of their parts: \begin{align*} AC &= EA + EC\\ BD &= EB + ED \end{align*} If we want to end up with an equation just involving $AC$, $BD$, $AB$, $BC$, $CD$, and $DA$, then we need to get rid of $EA$, $EB$, $EC$, and $ED$. This is a little daunting, but doable. I’ll start by trying to get $EA$, $EB$, and $EC$ in terms of $ED$. We can find $EA$ and $EC$ from equations 1 and 4: \begin{align*} \frac{EA}{ED} &= \frac{AB}{CD}\\[5pt] EA &= \frac{ED\cdot AB}{CD}\\\\ \frac{ED}{EC} &= \frac{DA}{BC}\\[5pt] \frac{EC}{ED} &= \frac{BC}{DA}\\[5pt] EC &= \frac{ED\cdot BC}{DA} \end{align*} Finding $EB$ is a little harder: \begin{align*} (1)\div (3):\quad\frac{EA}{ED}\div\frac{EA}{EB} &= \frac{AB}{CD}\div\frac{DA}{BC}\\[5pt] \frac{EB}{ED} &= \frac{AB\cdot BC}{CD\cdot DA}\\[5pt] EB &= \frac{ED\cdot AB\cdot BC}{CD\cdot DA} \end{align*} Now we can sub these formulae into our expressions for $AC$ and $BD$, and simplify: \begin{align*} \frac{AC}{BD} &= \frac{EA + EC}{EB + ED}\\[5pt] &= \frac{\cancel{ED}\cdot AB\div CD + \cancel{ED}\cdot BC\div DA}{\cancel{ED}\cdot AB\cdot BC\div(CD\cdot DA) + \cancel{ED}}\\[5pt] &= \frac{AB\div CD + BC\div DA}{AB\cdot BC\div(CD\cdot DA) + 1} \cdot \frac{CD\cdot DA}{CD\cdot DA}\\[5pt] &= \frac{AB\cdot DA + BC\cdot CD}{AB\cdot BC + CD\cdot DA} \end{align*}

   

4. Maybe we’re not quite satisfied with that, and we’d like some more similar triangles to see what else we can figure out. We might decide to create one. Noting that $\angle ADB = \angle ACB = \delta$, we could create a triangle similar to $\Delta ABC$, by placing $F$ on $BD$ such that $\angle DAF = \angle CAB = \alpha$: This ensures that $\Delta ABC \sim \Delta AFD$. Have we created any other similar triangles? Show hint

   When we placed $F$, we also created $\Delta ABF$. Is it similar to any other triangles in the diagram?

   Show answer

   \begin{align*} \angle FAB &= \alpha + \angle FAC = \angle DAC \\ \angle FBA &= \angle DCA \quad (\text{angles standing on arc }AD)\\ \therefore \Delta ABF &\sim \Delta ACB. \end{align*}

   

5. Use these two similar triangles to find another equation relating the four sides $AB$, $BC$, $CD$, $DA$, and the two diagonals $AC$ and $BD$. This is Ptolemy’s theorem, which will prove very useful in Section 11.4.2: Angle Formulae. Show hint

   $BD = BF + FD$.

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   The first triangle gives the equation \[\frac{DF}{BC} = \frac{AD}{AC}\] and the second gives \[\frac{BF}{CD} = \frac{AB}{AC}\] allowing us to find expressions for $BF$ and $DF$: \begin{align*} DF &= \frac{BC\cdot AD}{AC}\\[5pt] BF &= \frac{AB\cdot CD}{AC} \end{align*} \begin{align*} BD &= DF + BF\\[5pt] &= \frac{BC\cdot AD}{AC} + \frac{AB\cdot CD}{AC}\\[5pt] AC\cdot BD &= BC\cdot AD + AB\cdot CD \end{align*} Or, in words, the product of the diagonals is the sum of the products of opposite sides of a cyclic quadrilateral.