In Euclidean geometry, Ptolemy's theorem regards the edges of any quadrilateral inscribed within a circle.

Ptolemy's theorem states the following, given the vertices of a quadrilateral are *A*, *B*, *C*, and *D* in that order:

- $ AB \times CD + BC \times AD = AC \times BD $

*If a quadrilateral can be inscribed within a circle, 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.*

## ProofsEdit

**Using Circle Inversion**Edit

We can use properties of circle inversion to help us prove Ptolemy's theorem. We won't go over all of them here, but here are the ones relevant to this proof. For simplicity, we'll let all circles be labelled *O* and all tangent points be labelled *T*. If a point is labelled *A*, its image under inversion is labelled *A _{I}*.

**Abridged Overview of Circle Inversion**Edit

The fundamental process of circle inversion is to invert a point *A* to its image *A _{I}*. The process is as follows for any point inside the circle:

- Draw a circle
*O*. - Draw a point
*A*anywhere inside the circle. - Draw a ray
*OA*. - Draw a line perpendicular to
*OA*, marking a point*T*where the line intersects the circle, thus making the line*AT*. - Join
*OT*. - Draw a line perpendicular to
*OT*at point*T*, marking a point*A*where the line intersects ray_{I}*OA*, thus making the line*TA*. This is a tangent of circle_{I}*O*.

For any point *outside* of the circle:

- Draw a circle
*O*. - Draw a point
*A*anywhere outside the circle. - Draw a ray
*OA*. - Draw a line at
*A*, marking a point*T*where the line intersects the circle, thus making the line*AT*. This is a tangent of circle*O*. - Join
*OT*. This will always be perpendicular to line*AT*. - Draw a line perpendicular to
*OA*at point*T*, marking a point*A*where the line intersects ray_{I}*OA*, thus making the line*TA*._{I}

If a circle *N* is drawn that passes through the center of circle *O*, then the circle inverts to a line that does not pass through the center of that circle. Its distance from the center depends on how close circle *N* is to the edge of circle *O*.

**Property I: Scaling of Inverted Points**Edit

Refer to the above constructions. For simplicity, let point *A* be inside the circle.

We can observe the following:

- △
*OTA*~ △*TA*~ △_{I}A*OA*_{I}T

The three triangles share the same angles, thus they are all similar. We shall refer to △*OTA* and △*OA _{I}T*. Since they're similar, we can deduce the following:

*OT*~*OA*, and_{I}*OA*~*OT*.- $ \therefore \frac{OT}{OA_{I}} = \frac{OA}{OT} $

Multiplying both sides by *OA _{I}* and

*OT*gives us the following:

*OT*^{2}=*OA*×*OA*._{I}

Since *OT* will always be the radius of the circle, let's refer to this line as *r*.
Therefore, *OA* × *OA _{I}* =

*r*

^{2}.

**Property II: Scaling of Inverted Segments**Edit

Now that we have obtained the scaling of points, let's focus on the scaling of inverted segments. For simplicity, we will not draw the circle of inversion here.

In the provided figure, we see two triangles: △*OAB* and △*OA _{I}B_{I}*.

We can observe the following:

- △
*OBA*~ △*OA*_{I}B_{I}

Both triangles share a common angle α from *O*, and both sides scale to give us *r*^{2}, as shown in the previous section. We can therefore deduce the following:

*OA*×*OA*=_{I}*OB*×*OB*_{I}- $ \therefore \frac{OA}{OB} = \frac{OA_{I}}{OB_{I}} $

We can use this similarity to compute *A _{I}B_{I}*.

Since △*OBA* ~ △*OA _{I}B_{I}*:

- $ \frac{A_{I}B_{I}}{OA_{I}} = \frac{AB}{OB} $

- $ A_{I}B_{I} = \frac{AB \times OA_{I}}{OB} $

- $ OA \times OA_{I} = r^{2} \therefore OA_{I} = \frac{r^{2}}{OA} $

- $ \therefore A_{I}B_{I} = \frac{AB}{OB} \times \frac{r^{2}}{OA} = \frac{AB \times r^{2}}{OA \times OB} $

We are now ready to prove Ptolemy's theorem.

**Application**Edit

For simplicity, we will not need to draw the circle of inversion. Let's simply draw a quadrilateral inscribed within a circle and mark its vertices *A*, *B*, *C*, and *D* in that order. We will let *D* be the center of inversion.

This gives us a line with three points *A _{I}*,

*B*, and

_{I}*C*, respectively. We can make a rather simple observation here.

_{I}*A*+_{I}B_{I}*B*=_{I}C_{I}*A*_{I}C_{I}

Since we have a formula for computing how each of these inverted segments are scaled in respect to the original segments, let's apply it to each of these segments. This gives us the following:

- $ \frac{AB \times r^{2}}{AD \times BD} + \frac{BC \times r^{2}}{BD \times CD} = \frac{AC \times r^{2}}{AD \times CD} $

We can factor *r*^{2} out of the whole equation, leaving us momentarily with the conclusion that the radius of inversion has no effect on our result.

- $ \frac{AB}{AD \times BD} + \frac{BC}{BD \times CD} = \frac{AC}{AD \times CD} $

Since we are adding fractions here, we will need to find the greatest common denominator so we can factor it out as well. Observe that we get *AD*, *BD*, and *CD* twice. Therefore, the greatest common denominator is *AD* × *BD* × *CD*. Multiplying by this gives us the following:

- $ \frac{AB \times (AD \times BD \times CD)}{AD \times BD} + \frac{BC \times (AD \times BD \times CD)}{BD \times CD} = \frac{AC \times (AD \times BD \times CD)}{AD \times CD} $

Simplify each fraction.

- $ AB \times CD + BC \times AD = AC \times BD \Box $

## CorollariesEdit

**Pythagorean Theorem**Edit

If the quadrilateral is a rectangle or a square, then Ptolemy's theorem can be used to prove the Pythagorean theorem.

Since the shape is a rectangle or a square, then each side is congruent to its opposite, as are the two diagonals. Given the vertices of a quadrilateral are *A*, *B*, *C*, and *D* in that order:

- $ AB \cong CD; BC \cong AD; AC \cong BD $

Let's designate the letters *a*, *b*, and *c* to each of these pairs.

- $ AB \and CD = a $
- $ BC \and AD = b $
- $ AC \and BD = c $

Let's substitute each of these values in the formula for Ptolemy's theorem. This gives us the following:

- $ a \times a + b \times b = c \times c $
- $ \therefore a^{2} + b^{2} = c^{2} \Box $