$30^\circ$ Is Always Good

Relevant wiki: Similar Triangles - Problem Solving - Medium

We extend the straight line $AB$ to the point $E$ such that the $\triangle BAC \sim \triangle BED$ . Hence, like $\triangle BAC$ , the $\triangle BED$ is also a right triangle with $\angle BED = 90^\circ$ .

Let $m$ denote the distance $AE$ , then $\cos(\angle EAD) = \dfrac{AE}{AD} \Rightarrow \cos(180^\circ - 90^\circ - 30^\circ) = \dfrac m{AD} \Rightarrow AD = 2m$ .

By Pythagorean theorem , $(ED)^2 = (AD)^2 - (AE)^2 \Rightarrow ED = \sqrt 3 m$ .

Let $AC = h, BC = x\Rightarrow 1^2 + h^2 = x^2$ .

Since $\triangle BAC \sim \triangle BED$ , then $\dfrac{AB}{BE} = \dfrac{BC}{BD} \Rightarrow \dfrac 1{1+m} = \dfrac x{1+x} = \dfrac1{1 + 1/x} \Rightarrow m = \dfrac1x$ .

Similarly, $\dfrac{AC}{ED} = \dfrac{BC}{BD} \Rightarrow \dfrac h{\sqrt3 m} = \dfrac x{1+x}$ . With $m = \dfrac1x$ , then $h = \dfrac{\sqrt3}{1+x}$ .

Because $1^2 + h^2 = x^2$ , then $1 + \dfrac3{(1+x)^2} = x^2 \Rightarrow x^4 + 2x^3 = 2x + 4 \Rightarrow (x+2)x^3 = 2(x+2)\Rightarrow x^3 = 2 \Rightarrow \boxed{ x = 2^{1/3} }.$

Let $AC=x$ , $BD=z$ and $\angle ACB=\theta$ . Apply law of cosines on $\triangle ABD$ :

$(x+1)^2=1+z^2-2(1)(z)\cos 120^\circ \implies x^2+2x=z^2+z$

Apply law of sines on $\triangle ABC$ :

$\sin \theta=\dfrac{1}{x}$

Then on $\triangle BCD$ :

$\dfrac{\sin(\pi-\theta)}{\sin 30^\circ}=\dfrac{z}{1} \implies \sin\theta=\dfrac{z}{2}$

Then, $z=\dfrac{2}{x}$ .

Combining both equations we get:

$x^2+2x=\left(\dfrac{2}{x}\right)^2+\dfrac{2}{x}$

$x^4+2x^3-2x-4=0 \implies (x+2)(x^3-2)=0$ .

Finally, $\boxed{x=\sqrt[3]{2}}$ .