Well since no one has tackled even the first proof I will give the full derivation of 1) AP/AB = phi

Please follow the figure as I have described with the same lettering.

PT^2 = AP.BP (theorem)

i.e. AB^2 = AP ( AP -AB )

Whence AP^2 - AP.AB - AB^2 = 0

or (AP/AB)^2 - (AP/AB) -1 = 0

Thus AP/ AB = (1 + sq.rt.5 ) /2 = phi QED

Now, if C is a point in PA such that PC = PT, find CA/CB

HINT : It is also equal to phi. Please try to prove it.

Mally