Because AB=AD

So ∠ Abd = (180-2x)/2 = 90-x.

And ∠BFC =∠ Baff +∠ABF.

∠BAD=∠BAF+∠FAD

So ∠DBC =∠ fashion.

And ∠DBC and ∠FAD are the circumferential angles of circular arc CD.

therefore

∠DBC =∠ Fashion

So there is ∠DBC=∠DBC.

So ∠ FBC = ∠ Abd = 90-X.

In triangle ABD and FBC, there are two angles corresponding to each other, so they are similar.

So the triangular FBC is also isosceles.

So ∠ FCB = ∠ FBC = 90-X.

The quadrilateral ABCD is inscribed in a circle.

So the sum of diagonal angles is 180.

So ∠BAD+∠BCD= 180.

That is 2x+90-x+∞FCD = 180.

So ∠FCD=90-x

In the triangle FDC, ∠ FDC =180-(∠ FCD+∠ DFC) =180-(90-x+x) = 90.

So CD⊥DF

(2) Go to class first and write to you when I come back.