AD=X ESTWART in triangle ABD:1+y^2(x-1)-BK^2x=x(x-1) BK=(x(x-2))^.5 and y=(4x^2-3)^.5/2-1/2 Substituing this values in ESTWART we obtain:4x^4-8x^3-12=0 and solve x=2^(1/3)+1

Well, it took me some time to decipher the above solution! Let $AK=x, \; BD=y, \; BK=z$ and $E\equiv CD \cap BK$. Due to $\angle DBE=30^\circ$ and $BE\perp CD$ we get $DE=\frac{y}{2}\;$ $(1)$, next, from $\Delta DEK \sim \Delta ABK$ we have $\frac{DE}{AB}=\frac{DK}{AK}$, with (1) getting $\frac{\frac{y}{2}}{1}=\frac{1}{x}$ , or $y=\frac{2}{x}$ , $(2)$. Now, applying Stewart to $\Delta ABD$ and cevian $BK$ we get $AB^2\cdot DK+BD^2\cdot AK=BK^2\cdot AD+AD\cdot AK\cdot BD$ or, after substituting (1) and (2) and performing some small calculations: $2(x+2)=x^3(x+2)$, i.e. $x=\sqrt[3]{2}$, as $x=-2$ seems not convenient, hence $AB=1+\sqrt[3]{2}$ done. Best regards, sunken rock