Given that $\Delta ABC$ is an acute-angled triangle. $H$ is the orthocenter of the triangle. If $R_1$, $R_2$ and $R_3$ are the circumradii of $\Delta AHB$, $\Delta BHC$ and $\Delta CHA$ respectively then $R_1^2+R_2^2+R_3^2-3R^2$ is?
I don't even know how to proceed. Please help me with this question.
I think the answer is $0$ because
$\measuredangle BHC+\measuredangle BAC=180^{\circ}$, $\measuredangle AHC+\measuredangle ABC=180^{\circ}$ and $\measuredangle AHB+\measuredangle ACB=180^{\circ}.$
Let $\alpha$, $\beta$ and $\gamma$ be measures of angles $A$, $B$ and $C$ of our triangle.
Thus, $$\measuredangle AHC=180^{\circ}-\measuredangle HAC-\measuredangle HCA=180^{\circ}-(90^{\circ}-\gamma)-(90^{\circ}-\alpha)=\alpha+\gamma.$$
Thus, $$\measuredangle AHC+\measuredangle ABC=\alpha+\gamma+\beta=180^{\circ},$$
Which says that by law of sines for $\Delta ABC$ and for $\Delta AHC$ we obtain: $$AC=2R\sin\beta$$ and $$AC=2R_3\sin(180^{\circ}-\beta)=2R_3\sin\beta.$$ Thus, $R_3=R$.
Similarly, we see that $R_1=R$ and $R_2=R$.
Thus, $$R_1^2+R_2^2+R_3^2-3R^2=0.$$