In square ABCD, be = 3, EF = 5, DF = 4, how many degrees is the angle BAE + angle DCF?

In square ABCD, be = 3, EF = 5, DF = 4, how many degrees is the angle BAE + angle DCF?

The value of BAE + DCF is 45 degrees
Proof: if AF is connected, then ∠ DAF = ∠ DCF
Make DM vertical dB, make DM = be = 3, connect am and FM
Then, ADM = Abe = 45 degrees and ad = ab
So ⊿ ADM ≌ Abe, am = AE
In addition, DM ^ 2 + DF ^ 2 = 9 + 16 = 25 = FM ^ 2, FM = 5 = EF; AF = AF. we get ⊿ EAF ≌ ⊿ MAF, ∠ EAF = MAF
The result is: dam + DAE = BAE + DAE = 90 degrees
So ∠ EAF = ∠ MAF = 45 degrees, then ∠ BAE + ∠ DAF = ∠ BAE + ∠ DCF = 45 degrees