Problem 1609

Let $|BC| = a$, $|AD| = b$, $|AB| = |CD| = c$. We are given $b^2 = a c$ and need to prove $\angle CBD = 30^\circ$.
$$\frac{a^2 + c^2 - (b + c)^2}{2 a c} = \cos(90^\circ + \angle CBD) = -\sin(\angle CBD). \text{ Hence } \sin(\angle BCD) = \frac{b^2 + 2 b c - a^2}{2 a c}.$$
$$\text{Also } \sin(\angle CBD) = \frac{c}{a} \sin(\angle BDC) = \frac{c}{a} \sin(\angle BDA) = \frac{c}{a} \cdot \frac{c}{b}.$$
Hence
$$\frac{b^2 + 2 b c - a^2}{2 a c} = \frac{c^2}{a b}$$
Thus $b^3 + 2 b^2 c - a^2 b - 2 c^3 = 0$. Using $b^2 = a c$ we obtain $(a b - 2 c^2)(c - a) = 0$.
If $a = c$ then $|AC| = b + c > c + c$ which contradicts the triangle inequality. Hence $a b = 2 c^2$ and from previous calculations $\sin(\angle CBD) = \frac{c^2}{a b} = \frac{1}{2}$. Thus $\angle CBD = 30^\circ$.
$$\text{Conversely, } \sin(\angle CBD) = \frac{b^2 + 2 b c - a^2}{2 a c} = \frac{1}{2} \text{ implies } b^2 + 2 b c - a^2 - a c = 0.$$
Also $\sin(\angle CBD) = \frac{c^2}{a b} = \frac{1}{2}$ implies $a b = 2 c^2$. Substituting for $a$ we obtain $b^2 + 2 b c - \frac{4 c^4}{b^2} - \frac{2 c^3}{b} = 0$. Factorising we obtain $(b^3 - 2 c^3)(b + 2 c) = 0$. As $b + 2 c > 0$, $b^3 = 2 c^3 = 2 a b c$. Hence $b^2 = a c$.