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

Ptolemy's Theorem

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 InversionEdit

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 AI.

Abridged Overview of Circle InversionEdit

Circle Inversion

Illustration of circle inversion applied to a point A.

The fundamental process of circle inversion is to invert a point A to its image AI. The process is as follows for any point inside the circle:

  1. Draw a circle O.
  2. Draw a point A anywhere inside the circle.
  3. Draw a ray OA.
  4. Draw a line perpendicular to OA, marking a point T where the line intersects the circle, thus making the line AT.
  5. Join OT.
  6. Draw a line perpendicular to OT at point T, marking a point AI where the line intersects ray OA, thus making the line TAI. This is a tangent of circle O.

For any point outside of the circle:

  1. Draw a circle O.
  2. Draw a point A anywhere outside the circle.
  3. Draw a ray OA.
  4. 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.
  5. Join OT. This will always be perpendicular to line AT.
  6. Draw a line perpendicular to OA at point T, marking a point AI where the line intersects ray OA, thus making the line TAI.
Inversion of a Circle through the Center, Roughly

Inversion of a circle through the center, roughly approximated to illustrate the point.

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 PointsEdit

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

We can observe the following:

OTA ~ △TAIA ~ △OAIT

The three triangles share the same angles, thus they are all similar. We shall refer to △OTA and △OAIT. Since they're similar, we can deduce the following:

OT ~ OAI, and OA ~ OT.
$ \therefore \frac{OT}{OA_{I}} = \frac{OA}{OT} $

Multiplying both sides by OAI and OT gives us the following:

OT2 = OA × OAI.

Since OT will always be the radius of the circle, let's refer to this line as r. Therefore, OA × OAI = r2.

Property II: Scaling of Inverted SegmentsEdit

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.

Inverted Segment

Reference figure for this section.

In the provided figure, we see two triangles: △OAB and △OAIBI.

We can observe the following:

OBA ~ △OAIBI

Both triangles share a common angle α from O, and both sides scale to give us r2, as shown in the previous section. We can therefore deduce the following:

OA × OAI = OB × OBI
$ \therefore \frac{OA}{OB} = \frac{OA_{I}}{OB_{I}} $

We can use this similarity to compute AIBI.

Since △OBA ~ △OAIBI:

$ \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.

ApplicationEdit

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.

Proof of Ptolemy's Theorem

Reference figure for this section.

This gives us a line with three points AI, BI, and CI, respectively. We can make a rather simple observation here.

AIBI + BICI = AICI

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 r2 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 TheoremEdit

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 $
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