Right Triangle and Circle Theorem Let $ABC$ be a triangle such that $\angle BAC$ is a right angle. Suppose $D$ is a point lying on $BC$ such that $BD=1$, $DC =3$ and $\angle ADB=60^{\circ}$, find the length of $AC$.
I was told that there is a neat way to solve the question by considering the circumcircle of $\triangle ABC$. Since $\angle BAC=90^{\circ}$, $BC$ must be a diameter. Let $M$ be the midpoint point $BC$, then it's the centre of the circumcircle. But then I couldn't see anything substantial, please helps,

Let AQ be a chord perpendicular to BC cutting BC at P.

If $PD = x$, then $BP = 1 – x, QP = PA = \sqrt 3 x$ and $AD = 2x$ [$\triangle APD$ is special angled.]
At $P, (1 – x)(x + 3) = (\sqrt 3 x)^2$.
This gives $x = \dfrac {\sqrt {13} - 1}{4}$, after rejecting the negative length
From $\triangle APO$, $(\sqrt 3 x)^2 = 2^2 – (1 + x)^2$ … (*)
From $\triangle APC, AC^2 = (\sqrt 3 x)^2 + (x + 3)^2$ … (#)
Combining (*) and (#), we have $AC^2 = 4x + 12$.
Result follows after putting the value of $x$ back.
