Irish Mathematical Olympiad

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