So triangles bcd and aec are similar. so this makes lines ac and bc similar and lines db and ae similar.
to solve this problem, we just need to do some algebra
[tex] \frac{ae}{ac} = \frac{bd}{bc} [/tex]
now we substitute those values in
[tex] \frac{7x + 2}{9 + 3} = \frac{4}{3} [/tex]
which results in
[tex] \frac{7x + 2}{12} = \frac{4}{3} [/tex]
then multiply both sides by 12
[tex] 7x + 2 = 16 [/tex]
soon you should get x=2 with the length of ae being 16
