### Given a $\Delta ABC$ in which $\angle B = 90^\circ$ and $AB = \sqrt{3} BC$. Prove that $\angle C = 60^\circ$. B C A

1. Let $D$ be the midpoint of the hypotenuse $AC$.
Join $BD$.
2. Now, we have \begin{aligned} &AC^2 = AB^2 + BC^2 &&[\text{ By pythagoras' theorem }] \\ \implies &AC^2 = (\sqrt{ 3 } BC)^2 + BC^2 &&[\because \text{ AB = } \sqrt{3 } \text{ BC (given) } ] \\ \implies &AC^2 = 4BC^2 \\ \implies& AC = 2BC \\ \implies & 2CD = 2BC &&[\because \text{ D is the midpoint of AC}] \\ \implies & CD = BC &&\ldots\text{ (i) } \end{aligned}
3. We know that the midpoint of the hypotenuse of a right triangle is equidistant from the vertices. \begin{aligned} \therefore \space BD = CD &&\ldots\text{ (ii) } \\ \end{aligned} From $\text{(i)}$ and $\text{(ii)}$, we get $$BC = BD = CD$$ Therefore, $\Delta BCD$ is equilateral and hence $\angle C = 60^\circ$.