[tex]\qquad \qquad\huge \underline{\boxed{\sf Answer}}[/tex]
In the given question ~
Line segments BC and DE are parallel to one another therefore,
The Angles :
[tex] \longrightarrow \sf \angle ABC = \angle ADE[/tex]
[tex] \longrightarrow \sf \angle ACB = \angle AED [/tex]
[ By corresponding Angle property ]
Now, in the given figure the Triangles ABC and ADE are similar to one another. by AA criteria of similarity.
since we know, The sides of similar triangles are in equal proportion, let's use the property ~
[tex]\qquad \sf \dashrightarrow \: \dfrac{x}{5} = \dfrac{9}{9 + 6} [/tex]
[tex]\qquad \sf \dashrightarrow \: x = \dfrac{9}{15} \times 5[/tex]
[tex]\qquad \sf \dashrightarrow \: x = \dfrac{9}{3} [/tex]
[tex]\qquad \sf \dashrightarrow \: x = 3 \: units[/tex]
