How to solve this congruence of triangles geometrically? Find x.
  
  




Solution with trigonometry:


  First there are the missing angles ...
  
  Superior = $180 - 13 - 32 = 135$
  
  Bottom = $180 - 13 - 24 = 143$
  
  By the breast theorem we have to
  $$\frac{9 \sqrt{2}}{sen(13)}=\frac{Diagonal}{sen(135)} \to Diagonal=\frac{9 \sqrt{2} \cdot sen(135)}{sen(13)}$$
  Also in other triangle...
  $$\frac{x}{sen(13)}=\frac{Diagonal}{sen(143)} \to x = \frac{Diagonal \cdot sen(13)}{sen(143)}$$
  Using what we know diagonally
  $$x = \frac{9 \sqrt{2} \cdot sen(135) \cdot sen(13)}{sen(13) \cdot sen(143)}=\frac{9 \sqrt{2} \cdot sen(135)}{sen(143)}$$
Maybe there is a shorter way:



Here, you can use what you called breast theorem on $\Delta ABC$, you will find the same answer.
In the figure $ x = 3 $. However, the lenght that you seek is $ 5x = 15 $ 

In your case $$x = \frac{9 \sqrt{2} \cdot sen(135)}{sen(143)} = \frac{9 \sqrt{2} \cdot \frac{1}{\sqrt 2}}{\frac{3}{5}} = 15 $$