192 In the square $A B C D$, points $E$ and $F$ are on sides $A B$ and $C D$ respectively, and $A E=\frac{1}{m} A B, D F=\frac{1}{n} D C, D E$ intersects $A F$ at point $G, G H \perp A B$, with the foot of the perpendicular being $H$. Try to express $\frac{A H}{A B}$ in terms of $m$ and $n$.
Solution: As shown in Figure 4, let the line $DE$ intersect the line $BC$ at point $L$, and the line $AF$ intersect the line $BC$ at point $K$. Draw a line through point $D$ parallel to $AK$ intersecting the line $BC$ at point $P$.
Given $AE=\frac{1}{m} AB, DF=\frac{1}{n} DC$, we know
$EB=\frac{m-1}{m} AB, FC=\frac{n-1}{n} DC$.
It is easy to see that $\frac{LB}{AD}=\frac{BE}{AE}=\frac{\frac{m-1}{m} AB}{\frac{1}{m} AB}=m-1$.
Similarly, $\frac{CK}{AD}=n-1$.
Since $\frac{LK}{AD}=\frac{LB}{AD}+\frac{BC}{AD}+\frac{CK}{AD}=m+n-1$, we have
$\frac{AB}{AH}=\frac{LD}{GD}=\frac{LP}{KP}=\frac{LP}{AD}=\frac{LK}{AD}+1=m+n$.
Therefore, $\frac{AH}{AB}=\frac{1}{m+n}$.
(Tian Yonghai, Longjiang Province Lianghua City High School, 152054)