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