Because the quadrilateral ABCD is a square

When ∠DCB=90 degrees, AC score ∠BCD (square)

So ∠ACB= 1/2∠DCB=45 degrees.

And because AC=EC (known)

So ∠CAE=∠CEA (the two base angles of an isosceles triangle are equal)

And because ∠CAE+∠CAE=∠BCA (sum of triangle outer angles)

So ∠CAE=45÷2=22.5 degrees.

2.

The passing point d is DG⊥AB.

So ∠DGA=∠DGB=90 degrees

And because DF⊥AC and DE⊥BC (known)

So ∠DFA=∠DFC=90 degrees ∠DEB=∠DEC=90 degrees.

Because ∠ACB=90 degrees

So the quadrilateral CEDF is a rectangle.

And because DA points ∠CAB, DB points ∠CBA.

So ∠CAD=∠BAD, ∠CBD=∠ABD.

At △ Debbie and △DGB.

∠DGB=∠DGB，∠CBD=∠ABD，DB=DB

So congruence, DE=DG(AAS)

Similarly, △DFA is equal to △DGA, and DF=DG.

So DE=DF

So it is a square.

3.

Because quadrilateral ABCD is a square and quadrilateral ECGF is a square (known)

So CE=CG, CD=CB, ∠ECG=∠ECB

So △BCE is equal to △ △DCG(SAS).

So BE=DG

4.

Because the quadrilateral ABCD is a square

So BC=BA, ∠CBA=90 degrees.

Because AE⊥l, refer to ⊥l.

So ∠AEB=∠CFB=90 degrees

And because ∠FBC+∠FCB=∠FBC+∠EBA=90 degrees

So ∠FCB=∠EBA

In Delta △CFB and Delta △ Bia.

∠FCB=∠EBA，∠AEB=∠CFB，CB=AB

So △CFB is equal to △BEA(AAS).

So BE=CF=3.

At Rt△ABE

AB= root sign 10 (Pythagorean theorem)

Exactly. Hope to see