Those who have learned secant theorem (anyway, I didn't learn it in junior high school, and the high school teacher said that I had learned it in junior high school! If not, go and have a look. Not explained in detail here)

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.