Credits to PMO

$\text{Join FE, and FC .} \\ \text{Since BCDE, is a rectangle, so BCDE is a circle. }\\ \text{ In quadrilateral BCDF sum of opposite angles}\\ \angle DFB+\angle BCD=180. ~~~ \therefore ~ it ~ is ~cyclic. \\ \implies ~~ \color{#3D99F6}{BCDEF ~is ~a ~ circle.}\\ \angle CEB=\angle DEB - \angle DEC = 32.\\ Angles ~ on ~ chord ~ FC,~~\angle CDF =\angle CEF,\\ \implies ~ ~ \angle CDE - \angle FDE= \angle CEB+\angle BEF \\ \implies ~ ~90 - 41= 32+\angle BEF.~~~~\therefore ~\angle BEF=17.\\ In ~~ ~\color{#3D99F6} { quadrilateral ~ AFGE} ,\\ \angle s ~ at ~ F, ~ and ~ E ~are ~ 90^o, \therefore ~ it ~ is ~cyclic.\\ \therefore ~ \angle GAB= \angle BEF= \Large ~~~~~\color{#D61F06} {17}$

Observe that $BFED$ , $AFGE$ , $BCDE$ are all cyclic quadrilaterals.

In $BFED$ , we have $\angle FDE = \angle FBE = { 41 }^{ 0 }$ .

Also, in $BCDE$ , we have $\angle DCE = { 32 }^{ 0 } = \angle DBE$ .

Now observe that in $\triangle ABD$ $DF$ and $BE$ are altitudes intersecting at $G$ .

Hence, $G$ is the orthocenter of $\triangle ABD$ which implies $\angle AGD = { 180 }^{ 0 } - \angle ABD = { 180 }^{ 0 } - ({ 41 }^{ 0 } + { 32 }^{ 0 }) = { 107 }^{ 0 }.$

As $FGD$ is a straight line, we have $\angle AGD + \angle AGF = { 180 }^{ 0 }$ $\Rightarrow \angle FGA = { 73 }^{ 0 }$ .

Applying angle sum rule in triangle $FAG$ , we get $\angle GAF = \boxed { { 17 }^{ 0 } } .$