3 Squares inside triangles

${In}\:\bigtriangleup{ABC}\:\:\:\frac{{OA}}{{OC}}=\frac{{AM}}{{BM}}\:\:\:\Rightarrow\frac{{AC}}{{OC}}=\frac{{AB}}{{MB}}\Rightarrow\frac{{AC}}{{AB}}=\frac{{OC}}{{MB}}$

${In}\:\bigtriangleup{LHC}\:\left({large}\:{hadron}\:{Collider}\:!:\right){and}\:\bigtriangleup{OLX}_{\mathrm{4}}$ ${Since}\:\bigtriangleup{LHC}\sim\bigtriangleup{OLX}_{\mathrm{4}}$

$\frac{{LH}}{{CH}}=\frac{{OX}_{\mathrm{4}} }{{X}_{\mathrm{4}} {L}}\:\Rightarrow\frac{\mathrm{5}}{{HC}}=\frac{\mathrm{13}-\mathrm{5}}{\mathrm{5}}\Rightarrow{HC}=\frac{\mathrm{25}}{\mathrm{8}}$

Similarly in $\bigtriangleup{GJB}\:{and}\:{MX}_{\mathrm{4}} {G}\:,\frac{{MX}_{\mathrm{3}} }{{GX}_{\mathrm{3}} }=\frac{{GJ}} {{JB}}\Rightarrow\frac{\mathrm{13}-\mathrm{12}}{\mathrm{12}}=\frac{\mathrm{12}}{{BJ}}$ $\implies{BJ}=\mathrm{144}$

${OC}={OL}+{LC}=\sqrt{\left(\mathrm{13}-\mathrm{5}\right)^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} }+\sqrt{\frac{\mathrm{25}^{\mathrm{2}} }{\mathrm{64}}+\mathrm{25}}=\frac{\mathrm{13}\sqrt{\mathrm{89}}}{\mathrm{8}}$

${BM}={GM}+{BM}=\sqrt{\mathrm{12}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }+\sqrt{\mathrm{12}^{\mathrm{2}} +\mathrm{12}^{\mathrm{4}} }=\mathrm{13}\sqrt{\mathrm{145}}$

$\frac{{AC}}{{AB}}=\frac{{OC}}{{MB}}=\frac{\sqrt{\mathrm{89}}}{\mathrm{8}\sqrt{\mathrm{145}}}$

So answer is $\boxed{89+145}=234$

Since the quest is to find a ratio, we can ignore the real triangular side lengths and just use a similar ones. CA's side inclination is √25 = 5 units to the right and √169 - √25 = 13 - 5 = 8 units up, for a slope of 8/5, while AB's side inclination is √144 = 12 units to the right and √169 - √144 = 13 - 12 = 1 units down, for a slope of -1/12 = -8/96 just to make the nominators, or the equal heights, a match. The ratio is equal to √(8² + 5²) : √(8² + 96²) = √(8² + 5²) : 8√(1² + 12²) = √(64 + 25) : 8√(1 + 144) = √89 : 8√145.

Answer
= 89 + 145
= 234