Triangle Problem from Brazilian Math Olympics

[Wellingson]

In the acutangle triangle ABC, the heights BE and CF intersect and H, with E on the AC side and F on the AB side. Suppose that the circumcenter of ABC belongs to segment EF. Demonstrate that HA² = HB² + HC²

 

[Deleted member 4993]

In the acutangle triangle ABC, the heights BE and CF intersect and H, with E on the AC side and F on the AB side. Suppose that the circumference of ABC belongs to segment EF. Demonstrate that HA² = HB² + HC²

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Please explain:

"Suppose that the circumference of ABC belongs to segment EF."

 

[Wellingson]

I gave the best possible improvement in the statement

 

[Deleted member 4993]

In the acutangle triangle ABC, the heights BE and CF intersect and H, with E on the AC side and F on the AB side. Suppose that the circumcenter of ABC belongs to segment EF. Demonstrate that HA² = HB² + HC²

What is the definition of circumcenter?

 

[Wellingson]

The circumncent is the intersection of the satellite media ones a triangle. This point is the center of the circumscribed circumscribed in the triangle, which will go through its vertices. An important property of the circumncent is the fact that it is equidistant of its vertices.