Geometry: Finding an angle of a trapezoid Trapezoid $ABCD$ has right angles at $C$ and $D$, and $AD\gt BC$. Let $E$ and $F$ be the points on $AD$ and $AB$, respectively, such that $\angle BED$ and $\angle DFA$ are right angles. Let $G$ be the point of intersection of segments $BE$ and $DF$. If $\angle CED=58°$ and $\angle FDE=41°$, what is $\angle GAB$? I already found that $ABE=41$, $FGB=EGD=BAE=49$, also $BAE=GAB+GAE$. I do not know what to do next. I already gave up. I looked at the answer at the back. The answer is $$17°$$ Can anyone help me how this answer was obtained?

You can prove it without arguments involving trigonometric formulas. Draw diagonal segment $BD$. Look at triangle $ABD$. Segments $DF$ and $BE$ are altitudes in that triangle. Hence $G$ is the orthocenter. Thus $AG$ is the third altitude so $AG$ is orthogonal to $BD$. Therefore $\angle \, GAB = \angle \, GDB$. Now, since the quadrilateral $EBCD$ is a rectangle by assumption and construction, $$\angle \, BDE = \angle \, CED = 58^{\circ}.$$ From here, compute angle $\angle \, GDB$: $$\angle \, GDB = \angle \, BDE - \angle \, FDE = 58^{\circ} - 41^{\circ} = 17^{\circ}.$$ This means that $\angle GAB = \angle GDB = 17^{\circ}$.