Ptolemy's Incredible Theorem - Part 2

3 min read
Ptolemy's Incredible Theorem - Part 2

In part-1 of the series, we proved Ptolemy's theorem, and saw that its was really useful in proving various other well known mathematical facts. It turns out that the theorem is a special case of yet another theorem (actually, an inequality), which is much more powerful.

Ptolemy's Inequality

The statement of Ptolemy's inequality is

For any four points $A,B,C$ and $D$ in a plane, we have

$$AB \cdot CD + AD \cdot BC \ge AC \cdot BD$$
The equality holds if and only if $ABCD$ is a cyclic quadrilateral.

You might have seen a similar version for triangles, which is called the triangle inequality.

Inequalities like these are incredibly efficient at solving some really hard problems in mathematics and phyiscs. We are going to use this inequality in future articles.

The proof

Although, the proof here is easy to follow and self contained, the methods used are inspired from another article, where we have discussed the case of equality, instead of inequality. So, I recommend you to go through that one as well.

Let us start by drawing a diagram which contains 4 points in the plane as the vertices of a quadrilateral $ABCD$.

Now, we make a construction by marking two points $X$ and $Y$ on each of the diagonals respectively, such that $\angle BAX = \angle CAD$ and $\angle ABY = \angle ACD$ (diagram below).

Next, we consider the triangles $\Delta ABE$ and $\Delta ACD$ (diagram below),

and observe that -

  • $\angle ABE = \angle ACD$ (By construction)
  • $\angle EAB = \angle DAC$ (By construction)

We know that if two corresponding angles of two triangles are equal, then such triangles are called similar triangles. If two triangles are similar then the ratios of the corresponding sides are also equal . That means,

$$\frac{AB}{AC} = \frac{BE}{CD} \Longrightarrow AB \cdot CD = AC \cdot BE \tag{1}$$

We could also take another pair of sides -

$$\frac{AB}{AC} = \frac{AE}{AD} \tag{2}$$

Similarly, we consider another pair of triangles, $\Delta DAE$ and $\Delta CAB$ (diagram below).

We already know that $\angle DAC = \angle BAE$, by construction. If we add another angle to both sides of the equation, it won't change anything . So, we add $\angle CAE$ on both sides to get -

$$\angle DAC + \angle CAE = \angle BAE + \angle CAE. \tag{3}$$

But, we can see from the diagram that,

  • $\angle DAC + \angle CAE = \angle DAE$
  • $\angle BAE + \angle CAE = \angle CAB$

Therefore, we can rewrite equation $(3)$ as

$$\angle DAE = \angle CAB. \tag{4}$$

Now, looking at equations $(2)$ and $(4)$, we can see that we have satisfied the requirements of similar triangles - this time we have established the equality of one angle and ratio of a pair of sides containing this angle. This is called a Side - Angle - Side criterion of similarity.

If these triangles ($\Delta DAE$ and $\Delta CAB$) are similar, we can obtain the equality,

$$\frac{DE}{BC} = \frac{AD}{AC} \Longrightarrow BC \cdot AD = AC \cdot DE \tag{5}$$

Finally, we add equations $(1)$ and $(5)$, to get

$$\begin{aligned}AB \cdot CD + BC \cdot AD =& AC \cdot BE + AC \cdot DE \\ =& AC\cdot \left( BE + DE\right) \end{aligned} ~~~~~~ \tag{6}$$

In the beginning of the section, we talked about the existence of a triangle inequality. Now is the time to use it!

It states that -

For any triangle, the sum of the lengths of any two sides must be greater than or equal to the length of the third side.

Using this fact, for $\Delta DEB$ in the diagram above, we can write -

$$BE + DE \ge BD \tag{7}$$

Therefore, using equation $(7)$ we can write equation $(6)$ as

$$AB \cdot CD + BC \cdot AD \ge AC \cdot BD $$

which is what we wanted to prove.

Next time, we will see this result being used in real world problems.

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