Ptolemy's Theorem/Proof 3
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Theorem
Let $ABCD$ be a cyclic quadrilateral.
Then:
- $AB \times CD + AD \times BC = AC \times BD$
Proof
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.
$\blacksquare$