Exagon intersections

Since respective sides of $T_1$ and $T_2$ are parallel, all triangles in the diagram have congruent alternate interior, corresponding, or vertical angles; so they are all similar. Let $DE = a$ , $DL = b$ , and $EL = c$ . Since $AB = 4DE$ , let corresponding sides of $\triangle ABG$ be $4$ times bigger than the sides of $\triangle DEL$ . Also let $\triangle AJF$ be $p$ times bigger than $\triangle DEL$ , $\triangle KBC$ be $q$ times bigger than $\triangle DEL$ , $\triangle DHC$ be $r$ times bigger than $\triangle DEL$ , and $\triangle IEF$ be $s$ times bigger than $\triangle DEL$ .

Let $\theta$ be the angle between $a$ and $b$ , let $b = na$ , and $k = \frac{1}{2}n \sin \theta$ . Then the area of a triangle with sides $ma$ , $mb$ , and $mc$ is $A = \frac{1}{2} \cdot ma \cdot mb \cdot \sin \theta$ $=$ $\frac{1}{2} \cdot ma \cdot m \cdot na \cdot \sin \theta$ $=$ $\frac{1}{2}nm^2a^2 \sin \theta$ $=$ $km^2a^2$ . Therefore, the area of $\triangle GHI$ is $T_1 = k(s + r + 1)^2a^2$ and the area of $\triangle JKL$ is $T_2 = k(p + q + 4)^2a^2$ . Also, the area of $\triangle EFI$ is $A_{EFI} = ks^2a^2$ , the area of $\triangle CDH$ is $A_{CDH} = kr^2a^2$ , and the area of $\triangle ABG$ is $A_{ABG} = 16ka^2$ .

Let $M$ be the area of the exagon. We are given that $M = \frac{1}{2}T_1 = \frac{1}{2}k(s + r + 1)^2a^2$ and $M = \frac{49}{72}T_2 = \frac{49}{72}k(p + q + 4)^2a^2$ . Equating these and simplifying gives us $6(s + r + 1) = 7(p + q + 4)$ .

Since $\triangle GHI \sim \triangle LED$ , we have the proportion $\frac{sa + a + ra}{a} = \frac{sb + pb + 4b}{b} = \frac{rc + qc + 4c}{c}$ . From these we can deduce that $p = r - 3$ and $q = s - 3$ , and substituting these into $6(s + r + 1) = 7(p + q + 4)$ gives us $r + s = 20$ .

We also know that $T_1 = M + A_{EFI} + A_{CDH} + A_{ABG}$ . Substituting known expressions gives us $k(s + r + 1)^2a^2$ $=$ $\frac{1}{2}k(s + r + 1)^2a^2 + ks^2a^2 + kr^2a^2 + 16ka^2$ , which simplifies to $(s + r + 1)^2 = 2(s^2 + r^2 + 16)$ .

This equation and $r + s = 20$ solves to $r = \frac{17}{2}$ and $s = \frac{23}{2}$ , which means $p = \frac{11}{2}$ and $q = \frac{17}{2}$ .

Therefore, $\frac{AB \cdot CD \cdot EF}{BC \cdot DE \cdot FA}$ $=$ $\frac{4a \cdot rb \cdot sc}{qc \cdot a \cdot pb}$ $=$ $\frac{4rs}{qp}$ $=$ $\frac{4 \cdot \frac{17}{2} \cdot \frac{23}{2}}{\frac{17}{2} \cdot \frac{11}{2}}$ $=$ $\frac{92}{11}$ , and $92 + 11 = \boxed{103}$ .

I decided to work with coordinates. The figure can be resized and skewed without changing the given ratios, so I decided to go with $A=(0,0)$ and $D=(10,10)$ . The other points I named and for simplicity, I'll give them the same names as @David Vreken

The line through I called $J,F,E,L$ $y=mx+b$ and gave $G$ coordinates $(0,-c)$ .

Now we can find the coordinates of other points in terms of $b,m,c$ : $B=(\frac{c}{m},0)$ , $C=(10,10m-c)$ , $E=(\frac{10-b}{m},10)$ , $F=(0,b)$ , $H=(\frac{10+c}{m},10)$ , $I=(0,10)$ , $J=(\frac{-b}{m},0)$ , $K=(10,0)$ , $L=(10m+b,10)$ ..

The triangle areas work out to $T_{1}=\frac{(10+c)^{2}}{2m}$ and $T_{2}=\frac{(10m+b)^{2}}{2m}$

We are given that $\frac{1}{2}T_{1}=\frac{49}{72}T_{2}$ , from which we can solve to find $c=\frac{7}{6}(10m+b)-10$ .

We are also given $4\cdot DE=AB$ , or $4\cdot (10-\frac{10-b}{m}) = \frac{c}{m}$ from wich we can solve to find $m=\frac{180-17b}{170}$ and $c=\frac{40}{17}$ .

So now we can give the triangles in terms of $b$ alone: $\frac{1}{2}T_{1}=\frac{110250}{17(180-17b)}$

as well as the area of the exagon, which after much simplifying becomes: $\frac{63500+23800b-2890b^{2}}{17(180-17b)}$

equating these areas gives a quadratic in $b$ whose two solutions are $b=5$ and $b=\frac{55}{17}$

I've run out of time... both values of $b$ end up giving different looking figures but the same final ratio: $\frac{92}{11}$ so $a+b=92+11=\boxed{103}$