give it a try..

$\angle$ AGH = 45 (DIAGONALS BISECT THE ANGLES = 90/2).....

tan $\angle$ AFH= 1/2.....

tan $\angle$ AEH=1/3......

tan[ $\angle$ AFH+ $\angle$ AEH]= tan $\angle$ AFH+tan $\angle$ AEH/1-tan $\angle$ AFH tan $\angle$ AEH} = $\frac{1/2+1/3}{1-1/2*1/3}$ = $\frac{5/6}{5/6}$ = 1........

So $\angle$ AFH+ $\angle$ AEH=45........

SO $\angle$ AGH+ $\angle$ AFH+ $\angle$ AEH = 45 + 45 = $\boxed{90}$

let angle AFH = x1 and angle AEH = x2 tan (x1 + x2) =[ {tan(x1) + tan(x2)} / {1 - tan(x1).tan(x2)} ] = [ { (1/2) + (1/3) } / { 1 - (1/2).(1/3) } ] = [ { (5/6) } / { (5/6 ) } ] = 1 ; now taking tan inverse on both sides
x1 + x2 = tan ^(-1) { 1} x1 + x2 = 45 and angle AGH is already 45 so AGH + AFH + AEH = 45+45 = 90