Irish Mathematical Olympiad

Let $x = |AD|$ and $y = |BC|$. From $\triangle ADB$ we obtain $\sin(\angle ADB) = \frac{1}{x}$ and from $\triangle BDC$ we get $\frac{\sin(\angle BDC)}{y} = \frac{\sin(30^\circ)}{1} = \frac{1}{2}$. Since $\sin(\angle ADB) = \sin(\angle BDC)$, this gives $\frac{1}{x} = \frac{y}{2}$, hence $y = \frac{2}{x}$.

The cosine theorem for $\triangle ABC$ gives
$$-\frac{1}{2} = \cos(\angle ABC) = \frac{1 + y^2 - (1+x)^2}{2y} = \frac{1 + \left(\frac{2}{x}\right)^2 - (1+x)^2}{\frac{4}{x}}$$
Simplifying, we obtain $x^4 + 2x^3 - 2x - 4 = 0$. Rewriting this polynomial as $x(x^3 - 2) + 2(x^3 - 2) = (x^3 - 2)(x + 2)$, we see that $x = \sqrt{2}$.