Ptolemy's Incredible Theorem - Part 1


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Ptolemy's Incredible Theorem - Part 1

Ptolemy was an ancient astronomer, geographer, and mathematician who lived from (c. AD 100 – c. 170). He is most famous for proposing the model of the "Ptolemaic system", where the Earth was considered the center of the universe, and the stars revolve around it. This was the generally accepted view for many centuries, until Copernicus (19 February 1473 – 24 May 1543) proposed the theory where the sun was the center with earth an other planets revolving around it.

Although, Ptolemy's most famous idea turned out to be wrong, he did a lot of really useful work in geometry and we are going to discuss one of those very nice ideas.

Ptolemy's Theorem

The statement of the theorem is

If a quadrilateral $ABCD$ is inscribed in a circle, then

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

The proof of this theorem is quite straightforward. We only require basic understanding of the properties of circles and triangles. Let us see how it is done.

The Proof

We first draw a quadrilateral $ABCD$, which is inscribed in a circle. Such quadrilaterals are called cyclic quadrilaterals.

The labels on the line segments denote their respective lengths.

Now, we make a construction - choose a point $X$ on $BD$, such that $\angle AXD =\angle ABC$ (diagram below)

In order to prove the theorem, we need to concentrate on the triangles $\Delta ABX$ and $\Delta ACD$ (diagram below) -

Using the basic geometry of circles and quadrilaterals, we can observe that

  • $\angle AXD = \angle ABC$ (By construction)
  • $\angle AXD + \angle AXB = 180^{\circ}$ (Supplementary angles)

Using these two equations, we get

$$\angle ABC + \angle AXB = 180^{\circ}$$

Similarly, we also have,

  • $\angle ABC + \angle ADC = 180^{\circ}$ (Opposite angles of a cyclic quadrilateral)

Since the right hand side of the last two equations are equal, we obtain

$$\bcancel{\angle ABC} + \angle ADC = \bcancel{\angle ABC} + \angle AXB \Longrightarrow \angle ADC = \angle AXB$$

We also have,

  • $\angle ABX = \angle ACD$ (Angles in the same segment of the circle are equal). In this case, the segment is $AD$

So, what we have established is that - Two out of three angles in $\Delta ABX$ and $\Delta ACD$ are equal, which implies that these are similar triangles.

Similarity of triangles is one of the central ideas in geometry. What makes this idea so useful is the fact that - For any pair of similar triangles, the ratios of the corresponding sides are equal.

Therefore, we can write

$$\frac{AB}{AC} = \frac{BX}{CD} \Longrightarrow BX = \frac{ab}{f} \tag{1}$$

Repeating the steps discussed above, we can also establish that the triangles $\Delta ABC$ and $\Delta AXD$ are similar (diagram below).

Therefore, we can write

$$\frac{AD}{AC} = \frac{DX}{CB} \Longrightarrow DX = \frac{cd}{f} \tag{2}$$

Using equations $(1)$ and $(2)$, we can write

$$BX + DX = \frac{ab}{f} + \frac{cd}{f} = \frac{ab+cd}{f}$$

But, from the diagram it is clear that

$$BX+DX = BD = e,$$

and we can combine the two equations to obtain the final result

$$e = \frac{ab+cd}{f} \Longrightarrow ab+cd = ef$$

Therefore, we have proved that

$$AB \cdot CD + AD \cdot BC = AC \cdot BD$$,

which is the Ptolemy's theorem.

It all looks fine, but why do we care about this theorem?

The Pythagoras' Theorem

In order to understand the importance of Ptolemy's theorem, we are going to see that a simple derivation of the famous Pythagoras' theorem is possible as a special case. In the following diagram, we have chosen the sides of the quadrilateral such that

$AB=CD=a$, $BC=AD=b$, $BD=AC=c$.

This means that the quadrilateral is a rectangle with $\angle BCD = 90^{\circ}$. If we apply the Ptolemy's theorem, we will get

$$AB \cdot CD + AD \cdot BC = AC \cdot BD \Longrightarrow a^2 + b^2 = c^2,$$

which is the Pythagoras' theorem for $\Delta BDC.$ This means, Ptolemy's theorem is more powerful than Pythagoras' theorem!

Let's take a look at one more application.

Trigonometric Identities

Ptolemy's theorem also provides easy derivations for trigonometric relations involving sum of angles. Before we see how it is done, we have to understand a basic geometrical fact. In the following diagram,

we have a circle of radius $r$. A line segment $BC$ subtends an angle of $\theta$ on the circle. Our job is to find the length of $BC$ in terms of $r$ and $\theta$.

We make use of another fundamental theorem called, inscribed angle theorem. It states that

An angle $\theta$ inscribed in a circle is half of the central angle $2\theta$ that subtends the same arc on the circle.

Next, we draw a perpendicular $OD$ on the segment $BC$. Since line segments $OC$ and $OB$ are both the radii (plural of 'radius') of the circle, the $\Delta BOC$ is and isosceles triangle, with $OD$ as the angle bisector as well as side bisector. This is just due to symmetry (angle bisector theorem). So, we have

  • $\angle COD = \frac{1}{2} \angle COB = \theta$
  • $CD = \frac{1}{2} \cdot BC$

Using the basic trigonometric formula, we get

$$\sin{\theta} = \frac{Perpendicular}{Hypoteneuse} = \frac{CD}{OC} \Longrightarrow CD = r \sin{\theta}$$

Therefore, the length of the line segment $BC$ is $2r\sin{\theta}$.

Now we are ready to use Ptolemy's theorem to obtain a very useful trigonometric identity. Let us look at the following diagram

Here, we have a cyclic quadrilateral such that one of its diagonals, AC is along the diameter of the circle. If we use the inscribed angle theorem as discussed previously, it is clear that $\angle ABC$ and $\angle ADC$ are both $90^{\circ}$, because these are subtended by the segment $AC$, which is the diameter of the circle, making an angle of $180^{\circ}$ (flat line) at the center.

Therefore, we conclude that $\Delta ABC$ and $\Delta ADC$ are right angled, and we can use basic trigonometry as discussed earlier, to get

  • $AB = AC \cos{\alpha}$
  • $BC = AC \sin{\alpha}$
  • $CD = AC \sin{\beta}$
  • $DA = AC \cos{\alpha}$
  • $BD = 2r \sin{\left(\alpha+\beta\right)}$ (using inscribed angle theorem)

Finally, we can use Ptolemy's theorem to get

$$\left(AC \cos{\alpha} \times AC \sin{\beta}\right) + \left(AC \sin{\alpha} \times AC \cos{\beta}\right) = AC \times 2r \sin{\left(\alpha+\beta\right)}$$

Since, $AC$ is the diameter of the circle, its value is $2r$. Therefore, after simplification, we get the expression,

$$\left( \cos{\alpha} \times \sin{\beta}\right) + \left( \sin{\alpha} \times \cos{\beta}\right) = \sin{\left(\alpha+\beta\right)}$$

This is a well known trigonometric identity. In a similar manner, a lot more identities can be derived.

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