Dear Alex,

I continue study these distances from one points to segments of other points on one circle.
Using the same lemma and Ptolemy's theorem I can prove following fact:

Given one inscribed convex polygon X1X2X3...Xn-1Xn (n>2) and one point P on its circumcircle
We use following notations (i, j can be interchanged)
aij = length of side XiXj
dij = distance from P to sidelines XiXj
Rij = ratio aij/dij
then the set of ratios R12, R23, R34,...R(n-1)n, Rn1 hold a property: one is sum of others

First, we prove for n = 3, the triangle case
Given triangle ABC and one point P on circumcircle of ABC
a, b, c = sides of ABC
da, db, dc = distances from P to sides BC, CA, AB
The result:
Three ratios a/da, b/db, c/dc hold a property: one is sum of other two.
Proof:
We use following lemma: in the triangle ABC h = b*c/(2*R)
where h = altitude from A, b = AC, c = AB, R = circumradius of ABC
With triangle PBC
da = PB*PC/(2*R)
Similarly we have:
a/da = a/(PB*PC)*(2*R) = a*PA*(2*R)/(PA*PB*PC)
b/db = b/(PC*PA)*(2*R) = b*PB*(2*R)/(PA*PB*PC)
c/dc = c/(PA*PB)*(2*R) = c*PC*(2*R)/(PA*PB*PC)
Triangle ABC divides its circumcircle into three arc, say (BC), (CA), (AB)
If P on arc (BC) then by Ptolemy's theorem:
a*PA = b*PB + c*PC therefore a/da = b/db + c/dc. It is what should be proved.

For n>3, we use mathematical induction method.

Suppose now our theorem is true for n-1 and convex polygon X1X2...Xn divides circumcircle into n arcs and P is on arc X1X2. Consider triangle X1X2X3. This triangle devides circumcircle into three arcs and P is on arc X1X2 (It is true because polygon is convex).

X1X3X4...Xn is n-1 side convex polygon and P is on arc X1X3 (It is true because polygon is convex) so:
R13 = R34 + R45 + ... + Rn1 (our theorem for n-1)
In the triangle X1X2X3
R12 = R23 + R13 = R23 + R34 + R45 + ... + Rn1
So the theorem is true for any n>3
We can also know which ratio is sum of other: the ratio of side respective with the arc where P is.

Please correct for me if I have made any mistake!

Best regards,
Bui Quang Tuan