Hello, buggerzapper!

Code:

A B
o - - - - - - - - - o
| 90° p *
| * 55° F
| C o- - - - - - - o
| 107° * 90° |
E o * |
| * * |
| * q * t |
| D o r o G
| | *
| | *
| 90° s | 90° t *
o - - - - - o - - - - - o
J I H

We are given: .$\displaystyle AJ \parallel DI \parallel FG$

and: .$\displaystyle \angle A \,=\,\angle F \,=\angle J \,=\,\angle s \,=\,\angle DIH \,=\,90^o \quad\Rightarrow\quad \boxed{\angle s \,=\,90^o}$

Assuming that $\displaystyle AB \parallel CF,\:\angle p \,=\,\angle BCF \,=\,55^o \quad\Rightarrow\quad\boxed{\angle p \:=\: 55^o}$

In quadrilateral $\displaystyle ABDE,\;\angle q + \angle E + \angle A + \angle p \:=\:360^o$

. . $\displaystyle \angle q + 107^o + 90^o + 55^o \:=\:360^o \quad\Rightarrow\quad \boxed{\angle q \:=\:108^o}$

Since $\displaystyle AE \parallel DI,\;\angle EDI \:=\:\angle DEA \:=\:107^o$

Around vertex $\displaystyle D\!:\;\;\angle r + \angle EDI + \angle q \:=\:360^o$

. . $\displaystyle \angle r + 107^o + 108^o \:=\:360^o \quad\Rightarrow\quad\boxed{\angle r \,=\,145^o}$

Since $\displaystyle \angle DCF + \angle BCF \:=\:180^o,\;\;\angle DCF \,=\,125^o$

So far, we have: Code:

A B
o - - - - - - - - - o
| 90° 55° *
| * 55° F
| C o- - - - - - - o
| 107° * 125° 90° |
E o * |
| * 108° * |
| * * t |
| o 145° o G
| 107° | *
| | *
| 90° - 90° | 90° t *
o - - - - - o - - - - - o
J I H

In hexagon $\displaystyle CFGHID:\;\;2\angle t + 90^o + 145^o + 125^o + 90^o \:=\:720^o $

Therefore: .$\displaystyle 2\angle t \,=\,270^o \quad\Rightarrow\quad\boxed{\angle t \:=\:135^o}$