Isosceles trapezoid and rhombus from angle bisectors in a square

Proof:

(a) Since $\angle DAZ = \angle DCE = 90°$, $AD = CD$ (sides of square), and $\angle ADZ = \angle CDE = 22.5°$, we have $\triangle AZD \cong \triangle CED$ (ASA).

Therefore $AZ = CE \quad (1)$ and $DZ = DE \quad (2)$.

We have $ZB = AB - AZ = BC - CE = BE$ from $(1)$, so $ZB = BE$.

Since $\angle ZBE = 90°$, we get $\angle BZE = 45°$ and $\angle AZH = 45°$ (vertical angles).

With $\angle HAZ = 90°$, we have $\angle AZH = \angle AHZ = 45°$, so $AH = AZ \quad (3)$.

From $(1)$ and $(3)$: $AH = CE \quad (4)$.

Since $\angle BAH = \angle DCE = 90°$, $AB = CD$, and from $(4)$ $AH = CE$, we have $\triangle AHB \cong \triangle DCE$ (SAS).

Therefore $HB = DE \quad (5)$.

Also, $BE \parallel AD$ and $BE \parallel HD \quad (6)$.

From $(5)$, $(6)$, and the fact that $BH$ and $DE$ intersect (since $\angle EDH + \angle BHD < 180°$), we conclude that $HBED$ is an isosceles trapezoid. $\square$

(b) Since $AH = EC$ from $(4)$ and $AH \parallel EC$, quadrilateral $HECA$ is a parallelogram. Therefore $HE \parallel AC \quad (7)$.

Since $AC \perp BD$ and $HE \parallel AC$, we have $HE \perp BD$, thus $ZT \perp BD$.

Point $Z$ lies on the angle bisector $DZ$ of $\angle ADB$. Since $ZA \perp DA$ and $ZT \perp DB$, we have $ZA = ZT \quad (9)$.

Triangles $ADZ$ and $TDZ$ are congruent (SAS): $\angle ZAD = \angle ZTD = 90°$, $ZA = ZT$ from $(9)$, and $ZD$ common. Therefore $DA = DT \quad (10)$.

Triangles $ADF$ and $TDF$ are congruent (SAS): $DF$ common, $\angle ADF = \angle TDF = 22.5°$, and $DA = DT$ from $(10)$. Therefore $AF = TF$ and $\angle DAF = \angle DTF \quad (11)$.

From $(9)$: $AZ = ZT$ and $\angle ZAT = \angle ZTA$.

From $(11)$: $AF = TF$ and $\angle FAT = \angle FTA$.

Since $\angle ZAT = \angle FAT$ (same angle), we have $\angle ZAT = \angle ZTA = \angle FAT = \angle FTA$.

In triangle $ZAF$, since $\angle ZAF = \angle ZFA$, we have $AZ = AF$.

Similarly, in triangle $ZTF$, since $\angle ZTF = \angle ZFT$, we have $ZT = TF$.

Therefore $AZ = ZT = TF = FA$, so $AZTF$ is a rhombus. $\square$

Your solution good but it can be simpler:

1- By symmetry we have:

$CG=AH\Rightarrow DH= BK$

$\angle HBK=\angle EDG= 22.5^o$

$\angle HKB=\angle EGE=45^o$

So for ASA we have:

$\Rightarrow \triangle HKB \cong\triangle EDG$

$\Rightarrow BH= DE$

Since $BE||DH$, then quadrilateral BHDE is isosceles trapezoid.

2-$\triangle ZDE\cong \triangle ZBD\Rightarrow ZA =ZT \space\space\space\space\space\space(1)$

as they are the attitudes of these triangles.DZ is coincident on the altitude of isosceles triangle DHB , together with (1) results in:

$AF=FT$

In triangle DHG we have:

$AD= DC$

$DH= DG$

$\Rightarrow AC||GH$

So in quadrilateral AZTF we have:

$AF||TZ$ and by symmetry we have $FT||AZ$

which together with (1) results in the quadrilateral AZTF is a rhombus.

Here's a trigonometric approach.

By reflection in $BD$, $|EC|=|AZ|$, and because $\triangle AHZ$ is isosceles, $|HA|=|AZ|$.

$\triangle HBD$ is isosceles, therefore $|HD|=|BD|=\sqrt{2}$.

Therefore $|AH|=\sqrt{2}-1$ and from $\triangle DAZ, \tan \frac\pi8=\sqrt{2}-1$.

The x-coordinate of $F$ is then $x$ where $\frac{x}{1-x}=\sqrt{2}-1$, so $x=1-\frac1{\sqrt{2}}$, which gives it the same coordinate as the line $EZ$ intersection with the line $BD$ with reflection.

H3nce $FT||AB$ and $AZTF$ is a rhombus.