A geometry problem by Walid Mustafa

Draw h = 1 = CG $\bot$ AB, and CF || DE, F on AB.
$Solving ~right~ \Delta s ~ with ~ height ~~h=1. \\$
$In ~\Delta ~CAG,~~\color{#3D99F6}{AG = 1}, ~~~\therefore ~ ~ A_1=\frac 1 2 *h*AG= \dfrac 1 2.\\ In ~\Delta ~CGF,~~\color{#3D99F6}{GF =\dfrac 1 {\sqrt3}}, ~~~\therefore ~ ~ A_2=\frac 1 2 *h*GF=\dfrac 1 {2*\sqrt3}.\\ Area~\Delta ~CAF=A_1+A_2=\dfrac 1 2 +\dfrac 1 {2* \sqrt3}=\frac 1 2*\dfrac {1 +\sqrt3} { \sqrt3}.\\ In ~\Delta ~CGB,~~GB = \sqrt3, ~~~\therefore ~ ~\color{#EC7300}{AB=1+\sqrt3}, ~~ A_1+A_2+A_3=\frac 1 2 *h*AB=\dfrac {1+\sqrt3} 2. \\ \therefore~Area~\Delta ~EAD=\frac 1 2 * (A_1+A_2+A_3)=\dfrac {1+\sqrt3} {4}. ~~Given ~ condition.\\ \Delta~EAD\text{~} \Delta ~CAF.~~~~~~\\ Let~K^2=\dfrac{\Delta~EAD}{ \Delta ~CAF}=\dfrac{\dfrac {1+\sqrt3}4}{ \frac 1 2*\dfrac {1 +\sqrt3} { \sqrt3}}=\dfrac {\sqrt3} 2. \\$ $\dfrac{AD}{AB}=\dfrac{AF*K}{AB}=(AG+GF)*\dfrac K {AB}$ $=(1+\frac 1 {\sqrt3} ) * \dfrac { \sqrt { \dfrac {\sqrt3} 2 } } {1+\sqrt3}= \dfrac 1 {\sqrt{2*\sqrt3} } = \color{#D61F06}{0.53728 }$

$Divide\quad the\quad triangle\quad ABC\quad by\quad drawing\quad the\quad altitude\quad from\quad C\quad \\ to\quad side\quad AB.\quad Call\quad the\quad foot\quad of\quad the\quad altitude\quad F.\quad This\quad creates\\ two\quad right\quad angled\quad triangles\quad \triangle ACF\quad and\quad \triangle BCF.\quad Let\quad the\quad \\ length\quad of\quad the\quad altitude\quad CF\quad be\quad 1.\quad Using\quad our\quad trigonometric\\ functions\quad we\quad find\quad AB=1+\sqrt { 3 } ,\quad BC=\quad 2,\quad and\quad AC=\sqrt { 2 } .\quad \\ Therefore\quad the\quad Area\quad of\quad \triangle ABC=\frac { 1+\sqrt { 3 } }{ 2 } .\\ This\quad means\quad that\quad DE\quad divides\quad the\quad triangle\quad into\quad equal\quad areas\quad \\ of\quad \frac { 1+\sqrt { 3 } }{ 4 } .\quad Using\quad this\quad information\quad it\quad is\quad clear\quad that\quad E\quad \\ must\quad lie\quad on\quad AC.\\ \\ Now\quad we\quad have\quad \triangle ADE.\quad We\quad find\quad the\quad lengths\quad of\quad the\quad sides\quad \\ of\quad \triangle ADE\quad \quad the\quad same\quad way\quad we\quad found\quad it\quad for\quad \triangle ABC,\quad by\quad \\ dividing\quad it\quad using\quad an\quad altitude\quad from\quad E\quad to\quad AD.\quad Let\quad the\quad \\ length\quad of\quad this\quad altitue\quad be\quad p.\\ \\ We\quad obtain\quad AD=\frac { (3+\sqrt { 3 } )p }{ 3 } ,\quad DE=\frac { 2p\sqrt { 3 } }{ 3 } ,\quad AE=p\sqrt { 2 } .\\ \\ \therefore \quad \frac { 1+\sqrt { 3 } }{ 4 } =\frac { 1 }{ 2 } \times p\sqrt { 2 } \times \frac { 2p\sqrt { 3 } }{ 3 } \times \sin { 75 } \\ =>\quad p=\frac { \sqrt [ 4 ]{ 3 } }{ \sqrt { 2 } } \\ \therefore \quad \quad \frac { AD }{ AB } =\frac { \frac { 3+\sqrt { 3 } }{ 3 } \times \frac { \sqrt [ 4 ]{ 3 } }{ \sqrt { 2 } } }{ 1+\sqrt { 3 } } =\frac { 1 }{ \sqrt [ 4 ]{ 12 } } \approx \boxed { 0.537 } .$