Proof of Ptolemy's Theorem

Question

I am trying to prove using similar triangles without considering an arbitrary point on the diagonals which is the most common proof what we see in text books.

Let $HI=a$, $IJ=b$, $JC=c$ ,$CH=d$, $HL=x$, $LJ=e-x$, $CL=f-y$ and $LI=y$

we have $$\Delta HLC \sim \Delta ILJ $$ so we have

$$\frac{d}{b}=\frac{f-y}{e-x}=\frac{x}{y} \tag{1}$$ Also

$$\Delta CLJ \sim \Delta HLI$$ so

$$\frac{a}{c}=\frac{e-x}{y}=\frac{f-y}{x} \tag{2}$$ from $(1)$ and $(2)$ is it possible to prove that $$ac+bd=ef$$

Answer 1

from $(1)$ and $(2)$ is it possible to prove that $ac + bd = ef$

Not directly, as far as I can see. For one thing, Ptolemy's theorem "decays" nicely to $a c = a c$ in the degenerate case where $I \equiv J, b = 0, e = a, f = c$, while similarity-based proofs would not directly translate to the trivial case.

That said, the similarity of both pairs of triangles is equivalent to calculating the Power of the Point $L$ with respect to the given circle $LH\;LJ = LI\; LC$. The power of the point is related to the Inversion transformation, and it is indeed possible to prove Ptolemy by Inversion. As said, however, that's not exactly an immediate consequence of the given similarity.

Answer 2

As a practice to me learning geometry, I write an answer to this question :^) Note: $[ab]$ denotes the distance between point $a$ and point $b$

Introduction to geometric inversion:

For an origin centered circle C of radius R, geometric inversion map is defined in the following way:

$$ \tilde{z} = I_C (z) = \frac{R^2}{\overline{z}}$$

From this we get that a relation between conjugate and inverted point,

$$ \tilde{z} \cdot \overline{z} = R^2$$

Note that:

$$ | \tilde{z} | | \overline{z} | = | \tilde{z}| |z| = R^2$$ Also note that the transformation is involutive (self inverse)

Now let's see this transformation's effect on geometrical objects such as triangles:

Under inversion $ a \to \tilde{a}$ and $ b \to \tilde{b}$, note that:

$$ [qa][q \tilde{a} ] = R^2$$

Also,

$$ [qb][ q \tilde{b} ] =R^2$$

Hence,

$$ \frac{ [qa]}{[qb] } = \frac{ [q \tilde{b} ] }{ [ q \tilde{a} ]}$$

Hence with the common angle q, this implies

$$ \triangle qab \sim \triangle q\tilde{b} \tilde{a}$$

Hence,

$$ \frac{ [ab]}{ [\tilde{a} \tilde{b}] } = \frac{[qa]}{[q \tilde{b}]}$$

Now,

$$ [ q \tilde{b} ] = \frac{R^2}{ [qb] }$$

This and rearrangement gives:

$$ [ \tilde{a} \tilde{b} ] = \frac{R^2}{ [ qa] [qb]}[ab] \tag{1}$$

How are circles passing through center of inverting circle mapped under inversion?

Consider a Line outside the circle, take two points $a$ and $b$ and consider the inverses $ \tilde{a}$ and $ \tilde{b}$, from the previous triangle arguements we find $ \angle q \tilde{b} \tilde{a} = \frac{\pi}{2}$, now notice that as we vary $b$ this angle condition remains same. From inscribed angle theorem, we can say that locus of $ \tilde{b}$ is a circle.(2)

Proof

Under inversion of a circle K centered at $a$, the circle containing the cyclic quadrilateral gets mapped abcd gets mapped to a line L (2)with the $ b \to \tilde{b}$ , $ c \to \tilde{c}$ , $ d \to \tilde{d}$

It is obvious that:

$$[ \tilde{b} \tilde{c} ] + [\tilde{c} \tilde{d}] = [\tilde{b} \tilde{d}]$$

Now,

$$ \frac{[bc]}{[ab][ac]} + \frac{[cd]}{[ac][ad]} = \frac{[bd]}{[ab][ad]}$$ (1)

$$ [bc][ad] + [cd][ab] = [bd][ac]$$

QED

All pictures from Tristan Needham's Visual Complex Analsysis