>I appreciate
>take a look and tell me if it is correct of it has a flaw.

I do not see a flaw, no.

I can't say whether it is simpler or more elementary. What you do is replace the ratio property of angle bisectors (two of them being used simultaneously) with a property of harmonic ratio which also tackles two ratios simultaneously. Although it all depends on a view point, I would say that your suggestions is as simple as the proof on thepage but less elementary.

For what it is worth...

However, bringing in the harmonic ratio may suggest a generalization to the locus of a fixed cross-ratio. So it is always beneficial to have a second view.

>As the ratio r express the absolute value of ratio of the
>segments AP/BP, there are two point M and N on the straight
>line AB which satisfy the above condition. Indeed r and
>minus r are the barycentric coordinates of M and N with
>respect to to A and B on line AB or in other words the
>simple ratio ABM and ABN are equal an opposed in sign.
>Therefore the cross ratio is harmonic being M and N
>conjugated.
>
>So if we draw a pencil of rays from any arbitrary point in
>the plane to the points A,B,M, and N the harmonic ratio will
>be preserved, however since point P must be selected so ray
>PM be the bisector of angle APB, the ray PN must be the
>other exterior bisector of the angle APB. All the above
>means that the angle NPM must be always 90 degrees and since
>MN is constant P must belong to a circumference of diameter
>MN.