Dividing triangle

From the left figure, the red triangle is similar to $\triangle ABC$ . Therefore, the ratio of their areas $\dfrac {A_{\color{#D61F06}red}}{A_\triangle} = \dfrac {AD^2}{(AD+DB)^2} = \dfrac {5^2}{(5+1)^2} = \dfrac {25}{36}$ . $\implies A_{\color{#D61F06}red} = \dfrac {25}{36}A_\triangle$ , $\implies A_{\color{#3D99F6}blue} = A_\triangle - A_{\color{#D61F06}red} = \dfrac {11}{36}A_\triangle$ .

From the right figure,

$\begin{aligned} \frac {A_{\color{#3D99F6}blue}}{A_\triangle} & = \frac {11}{36} \\ \implies \frac {AE^2}{(AE+EB)^2} & = \frac {11}{36} & \small \color{#3D99F6} \text{Taking square root both sides} \\ \frac {AE}{AE+EB} & = \frac {\sqrt {11}}6 & \small \color{#3D99F6} \text{Divide up and down of LHS by }EB \\ \frac {\frac {AE}{EB}}{\frac {AE}{EB}+1} & = \frac {\sqrt {11}}6 \\ \implies \frac {AE}{EB} & = \frac {\sqrt{11}}{6-\sqrt{11}} \\ & = \frac {\sqrt{11}(6+\sqrt{11})}{(6-\sqrt{11})(6+\sqrt{11})} \\ & = \frac {6\sqrt{11}+11}{25} \end{aligned}$

Therefore, $a+b+c = 6+11+25 = \boxed{42}$ .