How to solve this congruence of triangles geometrically?

Question

Find x.

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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 $$x = \frac{9 \sqrt{2} \cdot sen(135) \cdot sen(13)}{sen(13) \cdot sen(143)}=\frac{9 \sqrt{2} \cdot sen(135)}{sen(143)}$$

Answer 1

In the figure $ x = 3 $. However, the lenght that you seek is $ 5x = 15 $

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

Answer 2

Maybe there is a shorter way:

Here, you can use what you called breast theorem on $\Delta ABC$, you will find the same answer.