Talk:Ptolemy's Theorem/Proof 3

Hope you don't mind, I worked through it carefully and I think I have made everything clear. As it stood, it was not immediately obvious which triangle similarity was being exploited at any one time.

I have clarified exactly what is being used at each stage. Now I believe I am convinced of the proof of this, although the diagram does not make it obvious. Have not been able to make a better one though. --prime mover (talk) 23:48, 13 November 2023 (UTC)

Very happy to have your help. --Telliott99 (talk) 01:00, 14 November 2023 (UTC)

FWIW, here is how I think about the logic:

$\triangle BAD \sim \triangle A'B'D$

which gives these ratios of sides:

A'D BD

A'B' AB

B'D AD

Take any four that form "four corners". We want A'B' and few other primes so:

A'B' AB

A'D BD

This gives an an expression for

(1) A'B' = AB x A'D / BD

All 3 pairs of triangles have the same relationships.

Sub B for A and C for B in (1)

(2) B'C' = BC x B'D / CD

Sub C for B in (1)

(3) A'C' = AC x A'D / CD

Clear the denominators.

Finally, we need something with A'D and B'D and no other primes.

From the first set of ratios:

A'D BD

B'D AD

And we're there.

If we could write the paired triples clearly it might be easier to understand.

But I think what you have is a big improvement and thank you.--Telliott99 (talk) 01:49, 14 November 2023 (UTC)