Given that triangles \( \triangle AYB \) and \( \triangle XYZ \) are similar (denoted by \( \triangle AYB \sim \triangle XYZ \)), we can use the properties of similar triangles to set up a proportion based on the lengths of their corresponding sides.
From the problem, we know:
- \( AB = 5 \)
- \( AY = 12 \)
- \( AY = AX \)
- \( BY = BZ \)
Since \( \triangle AYB \sim \triangle XYZ \), the ratios of the corresponding sides are equal:
\[ \frac{AB}{XZ} = \frac{AY}{XY} = \frac{BY}{YZ} \]
Let's denote \( BY \) as \( BZ = x \).
From the similarity of triangles, we also know that:
- Since \( AB \) corresponds to \( XZ \), we express \( XZ \) in terms of \( AB \).
- Since \( AY \) corresponds to \( XY \), we know that \( AY = 12 \).
- Hence, we can find the length ratios.
Now, since we have \( AB = 5 \) and \( AY = 12 \), we need to find the values for corresponding sides and ultimately \( BZ \).
Using the properties of similar triangles: \[ \frac{AB}{XZ} = \frac{AY}{XY} \] We can re-arrange this proportion:
Let \( XY \) correlate to \( BZ \), as \( BY = BZ \).
Setting up the proportion directly leads to: \[ \frac{5}{XZ} = \frac{12}{x} \]
To make further progress, let's consider the ratio itself using analogous side lengths:
Assuming \( AY \) corresponds to \( XY \): Let:
- \( XZ \) be the same relationship because \( AB \) and \( AY \) are known.
Next step, we need to find a common ratio. We know:
- The triangles' similarity will maintain the ratios \( 5/x = 12/x \)
Thus evaluating gives:
- \( x = BY = BZ = x \)
To find \( x \): Cross-multiplying gives: \[ 5x = 12XZ \implies x = \frac{12}{5}XZ \implies BZ = BY = \frac{12}{5} \times 5 \]
However, we realize that we need to denote \( BZ = k \), thus review the base relating \( X/Z \):
- Similar triangles yield: \[ y = 12k \implies k = 1 \]
So if \( k \) leads back to direct ratios:
Solving for direct lengths: \[ \frac{x}{5} = \frac{AY}{12}, thus here use common \( BY \).
Solving gives: [ X = \left(5 \cdot \frac{12}{5}\right) \]
So ultimately: \[ BZ = 12/5 = 2\frac{2}{5} \] and preserving: The final value of \( BZ\): \[ \boxed{2 \frac{2}{5}} \]
Thus \( BZ\) resolves as \( 2\frac{2}{5} \) in simplified degrees of one.