Theorem: Ptolemy's Theorem

Let $ABCD$ be a cyclic quadrilateral.
Then:
:$AB \times CD + AD \times BC = AC \times BD$

Proof:
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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.
{{begin-eqn}}
{{eqn | n = a
      | l = AD \cdot A'D
      | r = r^2
      | c = {{Defof|Inversive Transformation}}
}}
{{eqn | n = b
      | l = BD \cdot B'D
      | r = r^2
      | c = {{Defof|Inversive Transformation}}
}}
{{eqn | n = c
      | l = CD \cdot C'D
      | r = r^2
      | c = {{Defof|Inversive Transformation}}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = AD \cdot A'D
      | r = BD \cdot B'D
      | c = {{EuclidCommonNotionLink|1}} from $\text {(a)}$ and $\text {(b)}$
}}
{{eqn | ll= \leadsto
      | l = \dfrac {AD} {BD}
      | r = \dfrac {B'D} {A'D}
      | c = dividing by $A'D \cdot BD$
}}
{{eqn | n = 1
      | ll= \leadsto
      | l = \triangle BAD
      | o = \sim
      | r = \triangle A'B'D
      | c = Triangles with One Equal Angle and Two Sides Proportional are Similar
}}
{{end-eqn}}
From $(1)$ we establish via Triangles with Proportional Sides are Similar that:
{{begin-eqn}}
{{eqn | l = \triangle BAD
      | o = \sim
      | r = \triangle A'B'D
      | c = from $(1)$
}}
{{eqn | l = A'B' : AB
      | r = \dfrac {A'D} {BD}
      | c = Triangles with Proportional Sides are Similar
}}
{{eqn | n = 2
      | ll = \leadsto
      | l = A'B'
      | r = AB \cdot \dfrac {A'D} {BD}
      | c = rearranging
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = \triangle BAD
      | o = \sim
      | r = \triangle A'B'D
      | c = from $(1)$
}}
{{eqn | ll= \leadsto
      | l = \dfrac {AD} {BD}
      | r = \dfrac {B'D} {A'D}
      | c = 
}}
{{eqn | n = 3
      | l = AD
      | r = BD \cdot \paren {\dfrac {B'D} {A'D} }
      | c = rearranging
}}
{{end-eqn}}
Similarly:
{{begin-eqn}}
{{eqn | l = \triangle BCD
      | o = \sim
      | r = \triangle C'B'D
      | c = {{mutatis}} from $\text {(b)}$ and $\text {(c)}$
}}
{{eqn | l = \dfrac {B'C'} {BC}
      | r = \dfrac {B'D} {CD}
      | c = Triangles with Proportional Sides are Similar
}}
{{eqn | n = 4
      | l = B'C'
      | r = BC \cdot \paren {\dfrac {B'D} {CD} }
      | c = 
}}
{{end-eqn}}
and noting the collinearity of $A'$, $B'$ and $C'$:
{{begin-eqn}}
{{eqn | l = \triangle ACD
      | o = \sim
      | r = \triangle C'A'D
      | c = {{mutatis}} from $\text {(a)}$ and $\text {(c)}$
}}
{{eqn | l = \dfrac {A'C'} {AC}
      | r = \dfrac {A'D} {CD}
      | c = 
}}
{{eqn | n = 5
      | l = A'C'
      | r = AC \cdot \paren {\dfrac {A'D} {CD} }
      | c = 
}}
{{end-eqn}}
Finally: 
{{begin-eqn}}
{{eqn | l = A'B' + B'C'
      | r = A'C'
      | c = 
}}
{{eqn | l = AB \cdot \paren {\dfrac {A'D} {BD} } + BC \cdot \paren {\dfrac {B'D} {CD} }
      | r = AC \cdot \paren {\dfrac {A'D} {CD} }
      | c = substituting from $(2)$, $(4)$ and $(5)$
}}
{{eqn | l = AB \cdot CD \cdot A'D + BC \cdot BD \cdot B'D
      | r = AC \cdot BD \cdot A'D
      | c = multiplying through by $CD \cdot BD$
}}
{{eqn | l = AB \cdot CD + BC \cdot BD \cdot \paren {\dfrac {B'D} {A'D} }
      | r = AC \cdot BD
      | c = dividing everything by $A'D$
}}
{{eqn | l = AB \cdot CD + BC \cdot AD
      | r = AC \cdot BD
      | c = substituting for $AD$ from $(3)$
}}
{{end-eqn}}
The result follows.
{{qed}}