The triangles are similar because the corresponding base angles are congruent and they share one vertex.
From here, we can have a proportion in their sides.
[tex]\frac{AD}{DE}=\frac{AB}{BC}[/tex]
From the problem, AD = 6, DE = 9 and BC = 15
AB will be :
[tex]\begin{gathered} \frac{6}{9}=\frac{AB}{15} \\ \text{Cross multiply :} \\ 6(15)=9AB \\ 90=9AB \\ AB=\frac{90}{9}=10 \end{gathered}[/tex]
The answer is 10
