Ptolemy's Theorem

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Theorem

Let $ABCD$ be a cyclic quadrilateral.

Then:

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


Proof 1

PtolemysTheorem.png

Let $ABCD$ be a cyclic quadrilateral.

By Angles in Same Segment of Circle are Equal:

$\angle BAC = \angle BDC$

and:

$\angle ADB = \angle ACB$

By Construction of Equal Angle, construct $E$ on $AC$ such that:

$\angle ABE = \angle CBD$

Since:

$\angle ABE + \angle CBE = \angle ABC = \angle CBD + \angle ABD$

it follows that:

$\angle CBE = \angle ABD$


By Equiangular Triangles are Similar:

$\triangle ABE$ is similar to $\triangle DBC$

and:

$\triangle ABD$ is similar to $\triangle EBC$

Thus:

$\dfrac {AE} {AB} = \dfrac {CD} {BD}$

and:

$\dfrac {CE} {BC} = \dfrac {DA} {BD}$


Equivalently:

$AE \times BD = AB \times CD$

and:

$CE \times BD = BC \times DA$


Adding:

$AE \times BD + CE \times BD = AB \times CD + BC \times DA$


Factorizing:

$\paren {AE + CE} \times BD = AB \times CD + BC \times DA$


But:

$AE + CE = AC$

so:

$AC \times BD = AB \times CD + BC \times DA$

$\blacksquare$


Proof 2

Ptproof7.png

Let $\Box ABCD$ be a cyclic quadrilateral, with diagonals $AC$ and $BD$.

By Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles:

$\angle ABC$ is supplementary to $\angle ADC$

As well:

$\angle BAD$ is supplementary to $\angle BCD$

Construct two triangles $\triangle A'B'C'$ and $\triangle C'D'E'$ congruent to $\triangle ABC$ and $\triangle CDE$ respectively, with $B'C'D'$ collinear.

Ptproof8.png


Although this article appears correct, it's inelegant. There has to be a better way of doing it.
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$E$ is the same point that was $A$ in Figure $1$.

Hence:

$ED = AD$

$\angle ABC$ and $\angle CDE$ are supplementary.

By Equal Corresponding Angles or Supplementary Interior Angles implies Parallel Lines:

$AB \parallel ED$

By construction:

$AC = EC$

Now, scale the sides of $\triangle CDE$ by the length of $AB$.

Also scale the sides of $\triangle ABC$ by the length of $DE$.

Ptproof9.png

Let the scaled figure be $A'B'F'G'$.

$A'B' = AB \cdot DE = F'G'$

By Quadrilateral is Parallelogram iff One Pair of Opposite Sides is Equal and Parallel:

$\Box A'B'F'G'$ is a parallelogram

By construction:

$\angle A'C'B' + \angle A'C'F'$ has the same measure of angle as the original $\angle BCD$.

Therefore the supplementary angle $\angle A'C'G'$ has the same measure as the supplementary angle $\angle BAD$.

By construction:

$A'C' = AC \times DE$
$B'C' = BC \times DE$

Also by construction:

$C'G' = CE \times AC = AD \cdot AC$
$C'F' = CD \times AC$

Therefore $A'C'$ and $C'G'$ are in proportion with scale factor $AC$.

By Triangles with One Equal Angle and Two Sides Proportional are Similar:

$\triangle A'C'H' \sim \triangle ABD$

Thus, the length of $A'G'$ of the new construct is equal to $BD$ times the scale factor $AC$ Euclid:Proposition/VI/4:

$A'G' = AC \times BD$

And:

$B'F' = B'C' + C'F'$

Substituting:

$B'F' = BC \times DE + CD \times AC$

But by Opposite Sides and Angles of Parallelogram are Equal:

$A'G' = B'C'F' = B'F'$

Therefore

$AC \times BD = BC \times DE + CD \times AC$

$\blacksquare$


Proof 3

Ptproof16.png

Let an arbitrary circle $K$ be drawn in the plane.

Let $A$, $B$, $C$, and $D$ be arbitrary points on $K$.

By definition, $\Box ABCD$ is a cyclic quadrilateral.


We are to show that $AB \cdot CD + BC \cdot AD = AC \cdot BD$.


Let $T$ be an inversive transformation such that:

the inversion center of $T$ is $D$
the inversion circle $O$ for $T$ is chosen such that $K$ is completely inside $O$
the radius of $O$ is $r$.

Let $A'$, $B'$, and $C'$ be the images of $A$, $B$, and $C$ under $T$ respectively.

By Inverse of Circle Through Inversion Center is Straight Line Not Through Inversion Center:

$A'$, $B'$ and $C'$ are collinear.


\(\text {(a)}: \quad\) \(\ds AD \cdot A'D\) \(=\) \(\ds r^2\) Definition of Inversive Transformation
\(\text {(b)}: \quad\) \(\ds BD \cdot B'D\) \(=\) \(\ds r^2\) Definition of Inversive Transformation
\(\text {(c)}: \quad\) \(\ds CD \cdot C'D\) \(=\) \(\ds r^2\) Definition of Inversive Transformation


\(\ds AD \cdot A'D\) \(=\) \(\ds BD \cdot B'D\) Common Notion $1$ from $\text {(a)}$ and $\text {(b)}$
\(\ds \leadsto \ \ \) \(\ds \dfrac {AD} {BD}\) \(=\) \(\ds \dfrac {B'D} {A'D}\) dividing by $A'D \cdot BD$
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \triangle BAD\) \(\sim\) \(\ds \triangle A'B'D\) Triangles with One Equal Angle and Two Sides Proportional are Similar


From $(1)$ we establish via Triangles with Proportional Sides are Similar that:

\(\ds \triangle BAD\) \(\sim\) \(\ds \triangle A'B'D\) from $(1)$
\(\ds A'B' : AB\) \(=\) \(\ds \dfrac {A'D} {BD}\) Triangles with Proportional Sides are Similar
\(\text {(2)}: \quad\) \(\ds \leadsto \ \ \) \(\ds A'B'\) \(=\) \(\ds AB \cdot \dfrac {A'D} {BD}\) rearranging


and:

\(\ds \triangle BAD\) \(\sim\) \(\ds \triangle A'B'D\) from $(1)$
\(\ds \leadsto \ \ \) \(\ds \dfrac {AD} {BD}\) \(=\) \(\ds \dfrac {B'D} {A'D}\)
\(\text {(3)}: \quad\) \(\ds AD\) \(=\) \(\ds BD \cdot \paren {\dfrac {B'D} {A'D} }\) rearranging


Similarly:

\(\ds \triangle BCD\) \(\sim\) \(\ds \triangle C'B'D\) mutatis mutandis from $\text {(b)}$ and $\text {(c)}$
\(\ds \dfrac {B'C'} {BC}\) \(=\) \(\ds \dfrac {B'D} {CD}\) Triangles with Proportional Sides are Similar
\(\text {(4)}: \quad\) \(\ds B'C'\) \(=\) \(\ds BC \cdot \paren {\dfrac {B'D} {CD} }\)


and noting the collinearity of $A'$, $B'$ and $C'$:

\(\ds \triangle ACD\) \(\sim\) \(\ds \triangle C'A'D\) mutatis mutandis from $\text {(a)}$ and $\text {(c)}$
\(\ds \dfrac {A'C'} {AC}\) \(=\) \(\ds \dfrac {A'D} {CD}\)
\(\text {(5)}: \quad\) \(\ds A'C'\) \(=\) \(\ds AC \cdot \paren {\dfrac {A'D} {CD} }\)


Finally:

\(\ds A'B' + B'C'\) \(=\) \(\ds A'C'\)
\(\ds AB \cdot \paren {\dfrac {A'D} {BD} } + BC \cdot \paren {\dfrac {B'D} {CD} }\) \(=\) \(\ds AC \cdot \paren {\dfrac {A'D} {CD} }\) substituting from $(2)$, $(4)$ and $(5)$
\(\ds AB \cdot CD \cdot A'D + BC \cdot BD \cdot B'D\) \(=\) \(\ds AC \cdot BD \cdot A'D\) multiplying through by $CD \cdot BD$
\(\ds AB \cdot CD + BC \cdot BD \cdot \paren {\dfrac {B'D} {A'D} }\) \(=\) \(\ds AC \cdot BD\) dividing everything by $A'D$
\(\ds AB \cdot CD + BC \cdot AD\) \(=\) \(\ds AC \cdot BD\) substituting for $AD$ from $(3)$

The result follows.

$\blacksquare$


Source of Name

This entry was named for Claudius Ptolemy.


Sources

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