Fractions on a rectangle

$AB = DC = b \\ BC = AD = a \\ BE = x \\ EC = a-x \\ DF = y\\ FC = b-y \\ [\Delta ABE] = \dfrac{1}{2}bx \\ bx = 8 \\ \text{Similarly, } \\ ay = 10 \\ (a-x)(b-y) = 6 \\ ab - ay - bx + xy = 6 \\ ab + xy - (ay+bx) = 6 \\ ab + xy - (10 + 8) = 6 \\ ab + xy = 24 ~~ \boxed{1}\\ \dfrac{BE}{BC} + \dfrac{DF}{DC} = \dfrac{x}{a} + \dfrac{y}{b} \\ = \dfrac{bx + ay}{ab} \\ = \dfrac{18}{ab} \\ bx = 8 \\ x = \dfrac{8}{b} \\ ay = 10 \\ y =\dfrac{10}{a} \\ xy = \dfrac{80}{ab} \\ \text{xy in eqn }\boxed{1} \\ ab + \dfrac{80}{ab} = 24 \\ (ab)^2 + 80 = 24ab \\ (ab)^2 - 24ab + 80 = 0 \\ (ab - 20)(ab - 4) = 0 ab = 20, 4 \\ ab = 20 \\ \boxed{\therefore \dfrac{BE}{BC} + \dfrac{DF}{DC} = \dfrac{18}{20} = \dfrac{9}{10} = \dfrac{a}{b} \\ \implies a +b = 19}$

Let $AB = x, AD = y$ , then the area of rectangle $ABCD$ is $x.y$ and $x.y > 3+4+5=12$ $(1)$

We now can form the equations below according to the triangle area's formulas:

$x(y-CE)=8$ $=>$ $CE=(xy-8)/x$ $(2)$

$y(x-CF)=10$ $=>$ $CF=(xy-10)/y$ $(3)$

$CE.CF=6$ $(4)$

By substituting $(2)$ and $(3)$ into $(4)$ we have $(xy-8)(xy-10)/xy=6$

$=>$ $(xy-20)(xy-4)=0$ $(5)$

$(1)$ and $(5)$ give $xy=20$ , or the area of rectangle $ABCD$ is $20$ , which leads to the area of $CEF$ $(=20-3-4-5=8)$ .

The ratio between $ABE$ and $AECD$ now is equal to $BE/(AD+EC)=4/(8+3+5)=1/4$ , so we have $AD+EC=4.BE => AD+BC=5.BE$ $=>$ $2.BC=5.BE$ $=>$ $BE/BC=2/5$ $(6)$

The ratio between $ADF$ and $ABCF$ now is equal to $DF/(AB+FC)=5/(8+3+4)=1/3$ , so we have $AB+FC=3.DF => AB+CD=4.DF$ $=>$ $2.CD=4.DF$ $=>$ $DF/DC=1/2$ $(7)$

From $(6)$ and $(7)$ we have $BE/BC+DF/DC=2/5+1/2=9/10$ .

Hence $a+b=9+10=19$ .