A problem with "Crossed Ladders Theorem" In the following diagram, $AY \parallel BZ$, $AB$ is base. $M$ is $5$ above $N$ and $N$ is $4$ above $O$. What is the height of the triangle $\Delta AOB$.

My Work
There is a theorem named Crossed Ladder Theorem.
In the following Diagram $\dfrac{1}{h} = \dfrac{1}{H_1} + \dfrac{1}{H_2}$

I think this problem can be transformed into this. But I don't know how to. Any hint will be helpful.
Note that there is no statement about the angle between $AY$ and $AB$. The only constraint is that $AY$ and $BZ$ are parallel. It is therefore permissible to move $Y$ to give (amongst many others) the different arrangements shown below:



In all of these $|MC|=5$ and $|EO|=4$ and you will see that $|OF|=h$ is independent of the angle between $YA$ and $AB$.

Of particular interest is the diagram on the right, where $YA$ is perpendicular to $AB$.

This is essentially the same as your diagram for the crossed ladders:



$AM=H_2=h+4+5$

$BN=H_1=h+4$ 

Using the relationship you described, you now have:

$$\frac 1 h = \frac 1 {h+9} + \frac 1 {h+4}$$
Note that there is no statement about the angle between $AY$ and $AB$. The only constraint is that $AY$ and $BZ$ are parallel. It is therefore permissible to move $Y$ to give (amongst many others) the different arrangements shown below:



In all of these $|MC|=5$ and $|EO|=4$ and you will see that $|OF|=h$ is independent of the angle between $YA$ and $AB$.

Of particular interest is the diagram on the right, where $YA$ is perpendicular to $AB$.

This is essentially the same as your diagram for the crossed ladders:



$AM=H_2=h+4+5$

$BN=H_1=h+4$ 

Using the relationship you described, you now have:

$$\frac 1 h = \frac 1 {h+9} + \frac 1 {h+4}$$