Ptolemy’s Theorem And Circle Inversion

In Euclidean geometry, “Ptolemy’s theorem” (2024) is a relation between the four sides and two diagonals of a cyclic quadrilateral (a quadrilateral whose vertices lie on a common circle). The theorem is named after the Greek astronomer and mathematician Ptolemy (Claudius Ptolemaeus). Ptolemy used the theorem as an aid to creating his table of chords, a trigonometric table that he applied to astronomy.

If the vertices of the cyclic quadrilateral shown in Figure 1 are $A$, $B$, $C$, and $D$ in order, then the theorem states that:

\begin{align}
AC \cdot BD \; = \; AB \cdot CD \; + \; BC \cdot AD \\
\end{align}

This relation can be verbally expressed as follows:

If a quadrilateral is cyclic then the product of the lengths of its diagonals is equal to the sum of the products of the lengths of the pairs of opposite sides.

Moreover, the converse of Ptolemy’s theorem is also true:

In a quadrilateral, if the sum of the products of the lengths of its two pairs of opposite sides is equal to the product of the lengths of its diagonals, then the quadrilateral can be inscribed in a circle i.e. it is a cyclic quadrilateral.

There are several very important corollaries that naturally fall out from this theorem, and we’re discussing what’s probably the most significant one below in section 2. But first, it’s always nice to see exactly why a mathematical statement like this is inarguably true in the first place. So we’re showing a very straight forward way to prove Ptolemy’s theorem in section 1 below.

1

Ptolemy’s Theorem Proof

Before we explore a proof with circle inversion, we should look at a nice way to prove Ptolemy’s Theorem with geometric construction as shown in Figure 3.

We can start by placing a point $Z$ on $DB$ such that angle $\angle DAZ$ is equal to angle $\angle CAB$. From this we’ll be able to construct two similar triangles that will help us to determine the equality of Ptolemy’s Theorem shown in Figure 2.

In our Basics of Circle Theorems article, the first theorem states that angles in the same segment are equal. Since $AB$ is a chord of the circle and it divides the circle into two segments, any angles subtended from it are going to be equal. This means angle $\angle ADB$ is equal to $\angle ACB$.

Remember that the three angles in a triangle will always add up to 180 degrees. So if two different triangles each have two of the same angles, the third angle obtained by subtracting their sum from 180 will also be the same for both triangles. From this we can see how the following two triangles are similar, since all three angles in each will now be equal.

\begin{align}
\bigtriangleup ADZ \sim \bigtriangleup ABC \\
\end{align}

With these two triangles established, we can work with their ratios to see if we can get something relating to the Ptolemy’s Theorem formula. From the similarity above, we can see our first set of equal ratios,

\begin{align}
\frac{DZ}{AD} = \frac{BC}{AC} \\
\end{align}

Now let’s see if we can create another relationship with two more similar triangles. If we look at chord $DA$, we can see that angles $\angle ACD$ and $\angle ABD$ will also be equal. In addition, since we stated earlier that angles $\angle DAZ$ and $\angle CAB$ are equal, they share a common angle $\angle CAZ$ which can be subtracted from both. So we can let $\alpha$ be each of these identical new angles shown in Figure 3.

\begin{align}
\alpha = \angle DAZ \; – \; \angle CAZ = \angle CAB \; – \; \angle CAZ \\
\end{align}

This gives us enough information to construct two more similar triangles

\begin{align}
\bigtriangleup ABZ \sim \bigtriangleup ACD \\
\end{align}

And from this similarity we get two more of the ratios we’re after,

\begin{align}
\frac{BZ}{AB} = \frac{CD}{AC} \\
\end{align}

Now we can work our way from the left side of Ptolemy’s Theorem in Figure 2, by observing that $BD = (BZ + DZ)$,

\begin{align}
AC \cdot BD = AC \cdot (BZ + DZ) \\
\end{align}

By cross multiplying our two ratio equalities from above, we can solve for the values of $AC \cdot BZ \;$ and $AC \cdot DZ$

\begin{align}
AC \cdot BZ = AB \cdot CD \\
AC \cdot DZ = BC \cdot AD \\
\end{align}

Substituting these values into the right side of our new equation, we have our final proof of Ptolemy’s Theorem,

\begin{align}
AC \cdot BD \; = \; AB \cdot CD \; + \; BC \cdot AD \\
\end{align}

2

Proving The Pythagorean Theorem

Now that we’ve seen a standard geometric proof of Ptolemy’s Theorem, we can see how easily it would be able to prove the Pythagorean Theorem as shown in Figure 4.

Any square or rectangle will meet the definition of a cyclic quadrilateral, since you’ll always be able to find a circle that intersects with all four corners of these shapes. We can also quickly verify that the following equalities would hold true,

\begin{align}
AC = BD = c \\
AB = CD = a \\
BC = AD = b \\
\end{align}

Plugging in the values $c$, $a$, and $b$ into the formula for Ptolomey’s Theorem will reveal a wonderful proof of the Pythagorean Theorem!

\begin{align}
c^2 = a^2 + b^2 \\
\end{align}

According to many mathematicians, the Pythagorean Theorem is considered the most important theorem in mathematics. It’s a fundamental concept in Euclidean geometry with wide applications across various fields, and it’s considered to be one of the oldest non-trivial theorems in existence. But if we’ve found a different formula that can actually provide a proof of this other very important theorem, what does that say about theorem used for the proof?

So we should move on now to showing another interesting way to prove what might be the king of all theorems! There are several ways to prove Ptolemy’s Theorem, one by using similar triangles as we demonstrated above in section 1, but also with trigonometry, or even with complex numbers. Then in addition to that, we think what might be the most interesting and elegant proof we’ve seen is using circle inversion. Before we get into that kind of proof however, we need to discuss some of the details about what circle inversion is and how it works.

3

Circle Inversion or Plane Inversion

“Inversive geometry” (2024) is the study of inversion, a transformation of the Euclidean plane that maps circles or lines to other circles or lines and that preserves the angles between crossing curves. Many difficult problems in geometry become easier to solve when an inversion is applied.

To invert a number in arithmetic usually means to take its reciprocal. A closely related idea in geometry is that of “inverting” a point. In the plane, the inverse of a point P with respect to a reference circle shown in Figure 5 with center $O$ and radius $r$ is a point $P’$, lying on the ray from $O$ through $P$ such that,

\begin{align}
OP \cdot OP’ \; = \; r^2 \\
\end{align}

This is called circle inversion or plane inversion. The inversion taking any point $P$ (other than $O$) to its image $P’$ also takes $P’$ back to $P$, so the result of applying the same inversion twice is the identity transformation. This makes it a self-inversion (i.e. an involution).

For a point inside the circle, the nearer the point to the center, the further away its transformation. While for any point (inside or outside the circle), the nearer the point to the circle, the closer its transformation.

4

Compass And Straightedge Construction

Figure 6 shows how to go about mapping the points for circle inversion using a compass and a straight edge. Once you have the construction technique, this is one of the methods used to derive proofs for various types of circle inversions.

Point Outside The Circle
To construct the inverse $P’$ of a point $P$ outside a circle Ø:

• Draw the segment from $O$ (center of circle Ø) to $P$.
• Let $M$ be the midpoint of $OP$.
• Draw the circle ${\mathbb{C}}$ with center $M$ going through $P$.
• Let $A$ and $A’$ be the points where Ø and ${\mathbb{C}}$ intersect.
• Draw segment $AA’$.
• $P’$ is where $OP$ and $AA’$ intersect.

Point Inside The Circle
To construct the inverse $P$ of a point $P’$ inside a circle Ø:

• Draw ray $OP’$ from $O$ (center of circle Ø) through $P’$.
• Draw a vertical line through $P’$ perpendicular to $OP’$.
• Let $A$ be one of the points where circle Ø and the vertical line intersect.
• Draw the segment $r = OA$.
• Draw a tangent line through $A$ perpendicular to $r$.
• $P$ is where the extended ray $OP’$ and the tangent line intersect.

Finding a Ratio of Segments
We can also see from Figure 6 that the following two triangles are similar,

\begin{align}
\bigtriangleup OAP’ \sim \bigtriangleup OAP
\end{align}

From this we can get the ratio of segments and show that these are related to $r^2$,

\begin{align}
\frac{OA}{OP’} = \frac{OP}{OA} \; \equiv \; OP’ \cdot OP = r^2 \\\\
\end{align}

In other words, the old distance from the center times the new distance from the center is always fixed, and it is equal to $r^2$.

5

Circle Inversion Properties

Figure 7 shows examples with inversions of circles $A$ to $J$ with respect to the red circle at $O$. Circles $A$ to $F$, which pass through $O$, map to straight lines. Circles $G$ to $J$, which do not, map to other circles. The reference circle and line $L$ map to themselves. Circles intersect their inverses, if any, on the reference circle.

The inversion of a set of points in the plane with respect to a circle is the set of inverses of these points. The following properties make circle inversion useful.

• A circle that passes through the center $O$ of the reference circle inverts to a line not passing through $O$, but parallel to the tangent to the original circle at $O$, and vice versa; whereas a line passing through $O$ is inverted into itself (but not pointwise invariant).
• A circle not passing through $O$ inverts to a circle not passing through $O$. If the circle meets the reference circle, these invariant points of intersection are also on the inverse circle. A circle (or line) is unchanged by inversion if and only if it is orthogonal to the reference circle at the points of intersection.

Additional properties include:

• If a circle q passes through two distinct points $A$ and $A$’ which are inverses with respect to a circle $k$, then the circles k and q are orthogonal.
• If the circles $k$ and $q$ are orthogonal, then a straight line passing through the center $O$ of $k$ and intersecting $q$, does so at inverse points with respect to $k$.
• Given a triangle $OAB$ in which $O$ is the center of a circle $k$, and points $A’$ and $B’$ inverses of $A$ and $B$ with respect to $k$, then
  $\angle OAB = \angle OB’A’$ and $\angle OBA = \angle OA’B’$
• The points of intersection of two circles $p$ and $q$ orthogonal to a circle $k$, are inverses with respect to $k$.
• If $M$ and $M’$ are inverse points with respect to a circle $k$ on two curves $m$ and $m’$, also inverses with respect to $k$, then the tangents to $m$ and $m’$ at the points M and M’ are either perpendicular to the straight line MM’ or form with this line an isosceles triangle with base MM’.
• Inversion leaves the measure of angles unaltered, but reverses the orientation of oriented angles.

Examples in two dimensions

• Inversion of a line is a circle containing the center of inversion; or it is the line itself if it contains the center
• Inversion of a circle is another circle; or it is a line if the original circle contains the center
• Inversion of a parabola is a cardioid
• Inversion of hyperbola is a lemniscate of Bernoulli

6

Circle Inversion Proof of Ptolemy’s Theorem

In Figure 8, if we choose an auxiliary circle Ø of radius $r$ centered at $D$, we can invert the circumcircle of a cyclic quadrilateral $ABCD$ into a line. Since we’re choosing the origin point of the circle Ø to be the center of inversion, this means point $D$ is not mapped to the line. So we’ve simplified the inversion from 4 points to 3 points, making the problem even easier to solve. When we look at the segments on the resulting inversion line, we can write a formula for $A’C’$.

\begin{align}
A’C’ = A’B’ + B’C’ \\\\
\end{align}

We can also see from Figure 8 that the following triangles are similar,

\begin{align}
\bigtriangleup DAB \sim \bigtriangleup DA’B’ \\\\
\bigtriangleup DBC \sim \bigtriangleup DB’C’ \\\\
\end{align}

So by the law of similar triangles, we can show how our old and new segments are related in terms of ratios, and then solve for the inverted segments on the line,

\begin{align}
\frac{AB}{DA} = \frac{A’B’}{DB’} \\\\
A’B’ = \frac{AB \cdot DB’}{DA} \\\\
\end{align}

We can get $B’C’$ and $A’C’$ in a similar fashion, and then rewrite the formulas for our inverted segments,

\begin{align}
A’B’ = \frac{AB \cdot DB’}{DA} \\\\
B’C’ = \frac{BC \cdot BD’}{DC} \\\\
A’C’ = \frac{AC \cdot DC’}{DA} \\\\
\end{align}

Next, we can multiply each expression by $(DA \cdot DC)/ DB’$ to remove the denominators and repeating terms,

\begin{align}
\frac{AB \cdot DB’}{DA} \cdot \frac{DA \cdot DC}{DB’} \; = \; AB \cdot DC \; = \; A’B’ \\\\
\frac{BC \cdot BD’}{DC} \cdot \frac{DA \cdot DC}{DB’} \; = \; BC \cdot DA \; = \; B’C’ \\\\
\frac{AC \cdot DC’}{DA} \cdot \frac{DA \cdot DC}{DB’} \; = \; \frac{AC \cdot DC’ \cdot DC}{DB’} \; = \; A’C’ \\\\
\end{align}

Since we established a formula earlier which shows that,

\begin{align}
DC \cdot DC’ \; = \; DB \cdot DB’ \; = \; r^2 \\\\
\end{align}

we can also make the following substitution for the $A’C’$ segment equation above,

\begin{align}
\frac{DC’}{DB’} = \frac{DB}{DC} \\\\
\end{align}

Now our equations for $A’B’$, $ \; B’C’$ and $A’C’$ can be rewritten in the following way,

\begin{align}
A’B’ = AB \cdot DC \\\\
B’C’ = BC \cdot DA \\\\
A’C’ = AC \cdot DB \\\\
\end{align}

And this gives us our final requirement for proving Ptolemy’s Theorem! You can see how both expressions below are identical,

\begin{align}
A’C’ = A’B’ + B’C’ \\\\
AC \cdot DB \; = \; AB \cdot DC \; + \; BC \cdot DA \\\\
\end{align}

Conclusion

To see a really cool example of Ptolemy’s Theorem in action, check out our Heptagon Seven Pointed Star Geometry article, where we use it to show how the “optics equations” can be derived from a heptagon.

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Cite This Page As:

The Math Zone. “Ptolemy’s Theorem And Circle Inversion.” From MathZone.io — A Modern Exploration of Ancient Mathematics. https://mathzone.io/ptolemys-theorem-and-circle-inversion/

Work Sited:

Wikipedia contributors. (2024, October 5). Ptolemy’s theorem. In Wikipedia, The Free Encyclopedia. Retrieved 19:09, October 18, 2024, from https://en.wikipedia.org/w/index.php?title=Ptolemy%27s_theorem&oldid=1249586504

Wikipedia contributors. (2024, September 11). Inversive geometry. In Wikipedia, The Free Encyclopedia. Retrieved 13:36, October 21, 2024, from https://en.wikipedia.org/w/index.php?title=Inversive_geometry&oldid=1245232439