Dear Alex,

Using this simple lemma and the same proof method we can show also following interesting facts:

Given point P and n points P1, P2,… Pn (n>1) on the circle (O) with diameter d.
s1, s2, … sn are lengths of segments PP1, PP2, … PPn respectively
d1, d2, … dn are distances from P to segments P1P2, P2P3, … PnP1 respectively
The result:
(s1*s2*… *sn)^2 = d1*d2*… *dn*d^n
(When n=2 so we have our lemma.)
Please note that P1P2, P2P3, … PnP1 can be see as one close path passing all n point and each point one time (if I am not wrong it is Hamilton / Euler path in graph theory). For short we name it as "HE path". We can name each point P as a "vertex" and each segment PiPj as a "side" of this path.
Now our result can be formulated as:

Theorem 1:
If P, P1, P2, ... Pn (n>1) are concyclic on one circle with diameter d then
(s1*s2*… *sn)^2 = d1*d2*… *dn*d^n
where
si = segments from P to all vertices of one HE path from Pi (i=1 to n)
di = distances from P to all n sides of this HE path.
or by words
The square of product of all segments from one point to all vertices of one HE path is equal product of all distances from this point to all sides of this HE path multiplied by nth power of diameter of the cycle on which the point and the HE path are concyclic.

Of course, there are a lot of HE paths from n vertices and they have different distances from P to their sides. But they have the same one set of segments PP1, PP2, ... PPn or they have the same product of segments from P to n vertices: s1*s2*s3*...*sn. So we can have:

Theorem 2:
The product of all distances from one point to any HE path from given points are constant if all points are concyclic.
d1*d2*… *dn = (s1*s2*… *sn)^2 / (d^n) = constant

If we choose two HE paths with some common sides then they have also some common distances, therefore we can get some other equal products of distances.

One special example when n = 4
s1 = PP1, s2 = PP2, s3 = PP3, s4 = PP4
From four points P1, P2, P3, P4 we can make three HE paths
1. P1P2P3P4
- sides: P1P2, P2P3, P3P4, P4P1
- distances from P to these sides: d12, d23, d34, d41
2. P1P2P4P3
- sides: P1P2, P2P4, P4P3, P3P1
- distances from P to these sides: d12, d24, d43, d31
3. P2P3P1P4
- sides: P2P3, P3P1, P1P4, P4P2
- distances from P to these sides: d23, d31, d14, d42
(by these dij notations, of course dij = dji)
By our theorems:
d12*d23*d34*d41 = d12*d24*d43*d31 = d23*d31*d14*d42 = (s1*s2*s3*s4)^2/d^4
From these we can have following other equal products of distances:
d12*d34 = d14*d23 = d13*d24

This result can be formulated:
Theorem 3:
The products of two distances from one point on circumcircle of one inscribed quadrilateral to two opposite sides or two diagonals are equal.

Best regards,
Bui Quang Tuan