parmenides51 wrote: Restore the acute triangle $ABC$ given the vertex $A$, the base of the altitude drawn from the vertex $B$ and the center of the circle circumscribed around triangle $BHC$ (point $H$ is the orthocenter of triangle $ABC$). FTFY The question looks WAY better after rephrasing: rephrased problem wrote: You are given the orthocenter and circumcenter of triangle $BHC$, alongwith the foot of $C$-altitude of $BHC$. Construct point $B$ and $C$. Let $O$ be the circumcenter of $BHC$. Then the circle with center as midpoint of $AO$ and passing through the foot of $C$-altitude is the nine-point circle of $BHC$. Take a homothety of ratio $2$ from $A$ and you get the circumcircle of $BHC$. Consider the line joining $A$ and the $C$-altitude of $BHC$. It hits the circle at $C$!....... .........ugh, it meets at TWO points? What do we do? All efforts wasted, dump this solution One fact that we never used till now was that triangle $ABC$ was acute! (Frankly speaking, I forgot that in the midst of solving this ). This causes the intersection farther from $A$ to be $C$. Now it's simple. Draw a line perpendicular to $AC$ at the foot of perpendicular from $B$. The point where it meets the circumcircle is $B$, done! Sadly this line can meet the circle at two points as well. But we do know that one of them is $B$ and the other point is $H$. (Friendly reminder: $C$-altitude of $BHC$ and $B$-altitude of $ABC$). As we have seen in the solution earlier, such an ambiguity is resolved with the acute angle condition. And that shall happen again! $\angle BAC$ is acute which implies $\angle BHC$ is obtuse. So why is that important? It is because a triangle might have upto $3$ acute angles, but not more than one obtuse angle! So in triangle $BHC$ (which we have constructed, and are just distinguishing between $B$ and $H$) the only obtuse angle is contained by $H$. The other point is $B$. Note: Rephrasing problems is one trick which is used properly, can prove itself to be magical. And, geometry is quite prone to configuration issues. Stuff like 'triangle $ABC$ is acute' should be put to proper use to resolve them.