Note: I had posted the proof for verification on math stackexchange a few days ago, but I did not reach any conclusion. The discussion is dormant. I am just curious if my approach is wrong. I'm posting my proof here.

Prove: $ frac { mathrm { overline {B}} { mathrm { overline {AB}}} = frac { mathrm { overline {B & # 39; C & # 39; C & # 39; ;}}} { mathrm { overline {BC}}} = frac { mathrm { overline {A}} { mathrm { overline {AC}}} = mathrm {K} $, or $ mathrm {K} $ is a constant.

Proof: in the figure above, $ Delta { mathrm {A & B} C & # 39;} $ is registered in the largest radius circle $ mathrm {R & # 39;} and $ Delta { mathrm {ABC}} $ is registered in the smallest radius circle $ mathrm {R} $. Right here, $ Delta { mathrm {A & B} C & # 39;} $ is the enlarged version of $ Delta { mathrm {ABC}} $.

A well-known result of elemental geometry is the length of the arc $ l $ from a circle, the angle $ mathrm { theta} $ underpinned by the bow and rays $ mathrm {R} $ join the ends of the arc using the equation, $ l = mathrm {R cdot theta} $.

From the figure and the result above, we see that $ mathrm {arc A <B> = mathrm {R & # 39; theta} $ and $ mathrm {arc AB} = mathrm {R} theta $, or $ mathrm { theta} $ is $ angle { mathrm {A & # 39;}} $. By taking the ratios we get, $$ frac { mathrm {arcA} { mathrm {arcAB}} = frac { mathrm {R} # { mathrm {R}} $$

Similarly, $$ frac { mathrm {arcB C} { mathrm {arcBC}} = frac { mathrm {R} # { mathrm {R}} $$ and $$ frac { mathrm {arcA} C {} { mathrm {arcAC}} = frac { mathrm {R} # { mathrm {R}} $$

The diameter whose length is $ mathrm {D} $ is the longest agreement of a circle. The other chords are reduced versions of the diameter. Mathematically, $$ mathrm {P} = mathrm {B} cdot mathrm {D} $$

Or $ mathrm {B} $ is a scaling factor, whose value is between $ 0 and $ 1 $-and or $ mathrm {P} $ is the length of the rope.

Figure 2

Let $ mathrm {P} $ in Figure 2, the length diameter. The length of the diameter changes from $ mathrm {D} $ at $ mathrm {D & # 39;}. It is obvious that the scale factor is the same ($ mathrm {B} = $ 1) in the largest and the smallest circle, the ratio of the lengths of diameter in each circle is equal to $ 1 $.

figure 3

Now let $ mathrm {P} $ is the length of an agreement that underlies an angle of $ frac { pi} {3} $ radians in the center of a circle (see Figure 3). The length of the string is equal to the length of the radius of the circle because the string is part of an equilateral triangle. As the length of the agreement goes from $ mathrm {P} $ at $ mathrm {$ P $, it is again obvious that the scale factor is the same ($ mathrm {B} = frac {1} {2} $) in the largest circle and the smallest circle, the length of the string in each circle is equal to half the length of the diameter.

We see a motive. The model tells us that for a fixed $ theta $, the scale factor is the same in all concentric circles. In other words, the scale factor is independent of the radius length and is a constant for a constant. $ theta $.

Following the pattern we have, $$ mathrm {P} propto mathrm {D} $$

That implies, $$ mathrm {P} propto mathrm {R} $$

This result and the result $ l = mathrm {R cdot theta} $(for constant $ mathrm { theta} $) together involve, $$ mathrm {P} propto l $$

Returning to

figure 1 we see, $$ overline { mathrm {B} = mathrm {C} cdot mathrm {arcA & # 39; B}

In the same sector, $$ overline { mathrm {AB}} = mathrm {C} cdot mathrm {arcAB} $$

Or $ mathrm {C} $ is a constant. By taking the ratios we get, $$ frac { mathrm {arcA} B {AB}} { mathrm {arc AB}} = frac { overline { mathrm {A} B}} { overline { mathrm {AB}}} $$

Similarly, $$ frac { mathrm {arcB C} { mathrm {arc BC}} = frac { overline { mathrm {B & # 39; C}} { overline { mathrm {BC}}} $$

and

$$ frac { mathrm {arcA C} { mathrm {arc AC}} = frac { overline { mathrm {C }} { overline { mathrm {AC}}} $$

The results above reflect, $$ frac { mathrm { overline {B }} { mathrm { overline {AB}}} = frac { mathrm { overline {B & # 39; C & # 39;}}} { mathrm { overline {BC}}} = frac { mathrm { overline {C} {{mathrm { overline {AC}}} = mathrm { frac {R & # 39;} {R}}

Or $ frac { mathrm {R} # { mathrm {R}} = mathrm {K} $, a constant. This concludes the proof.