Prove equality

∠BAD﹢∠DAC=90，∠FAC+∠DAC=90，

So: ∠BAD=∠FAC.

And AB=AC, AD=AF,

Get △ Abd△ ACF.

So: BD = cf

② Prove verticality

Pass d as DM⊥BC, pass AB and m.

∠ANM=∠ADB-∠BDM=∠ADB-90，

∠EFC =∠ AFC -∠AFE =∠ AFC -90,

It is proved above: △ Abd△ ACF. So: ∠ADB=∠AFC.

So: ∠ANM=∠EFC.

In square ADEF, ADF=∠DFE.

So: ∠ANM+∠ADF=∠EFC+∠DFE.

Namely: ∠MDF=∠DFC

Internal dislocation angles are equal and two straight lines are parallel. So: MD∨FC.

Because: MD⊥BD, so: FC⊥BD.

(2) in the proof ....

After drawing the picture, find out the conditions to prove the congruence of the triangle in the same way as above.

∠DAC is a common angle, so∠∠ FAC =∠ DAB.

And AB=AC, AF=AD.

So: △ Abd△ ACF

So we get FC=DB, ∠ACF=∠ABD.

△ABC is an isosceles right triangle, then: ∠ Abd = ∠ ACB = 45.

So: ∠ ACF = 45.

Then ∠ FCB = ∠ FCA+∠ ACB = 90.

Namely: FC⊥BD.

I hope I can help you!