If $\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma }$ are lengths of the altitudes of a triangle ABC with area $∆$, then $\frac{{\mathrm{\Delta }}^2}{{\mathrm{R}}^2}\left(\frac{1}{{\mathrm{\alpha }}^2}+\frac{1}{{\mathrm{\beta }}^2}+\frac{1}{{\mathrm{\gamma }}^2}\right)=$

$\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma }$ are the lengths of the altitudes $\Rightarrow \mathrm{\Delta }=\frac{1}{2}\mathrm{a\alpha }=\frac{1}{2}\mathrm{b\beta }=\frac{1}{2}\mathrm{c\gamma }$
$$\begin{gathered} \Rightarrow \mathrm{\alpha }=\frac{2\mathrm{\Delta }}{\mathrm{a}},\mathrm{\beta }=\frac{2\mathrm{\Delta }}{\mathrm{b}},\mathrm{\gamma }=\frac{2\mathrm{\Delta }}{\mathrm{c}} \\ \therefore \frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+\frac{1}{{\gamma }^2}=\frac{a^2+b^2+c^2}{4{∆}^2}=\frac{4R^2\left({\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C\right)}{4{∆}^2} \\ \Rightarrow \frac{{∆}^2}{R^2}\left(\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+\frac{1}{{\gamma }^2}\right)={\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C \end{gathered}$$