The square of 1.PF =PD×PA is obtained by tangent theorem.
∠PED=∠BCE is obtained by PE parallel to BC.
And because ∠BCE=∠BAD (the circumferential angle of the same arc pair)
So ∠PED=∠PAE, so the two triangles are similar.
So PD×PA = PE square of pe.
So PE=PF
(Anyone who doesn't understand the theorem can ask me again.)
3.
Even OD. Then OD=OB, so ∠ABC=∠ODB.
Again ∠ABC=∠ACB
So ... .
So OD is parallel to AC. Because DE is perpendicular to AC, DE is perpendicular to OD.
So it is tangent (OD is radius).
Even AD. It is found that ABD is a right triangle (diameter pairs are called right angles). Then in the delta ADC. Seeing AE×EC=DE×DE from projective theorem
But DE×DE=EF×EB is obtained from the tangent theorem.
So we get AE×EC=EF×BE.