MHB Proving Orthocenter Property of Triangle ABC

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The discussion focuses on proving the orthocenter property of triangle ABC, specifically the equation HA² + BC² = HB² + AC² = HC² + AB². Participants note that the configuration of points forms a parallelogram, with PB being perpendicular to BC. This leads to the conclusion that HA² + BC² equals 4R², where R is the circumradius. The solution is acknowledged as clear and effective. The proof highlights the relationship between the orthocenter and the triangle's circumcircle.
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Point $H$ is the orthocenter of $\triangle ABC$

prove :$HA^2+BC^2=HB^2+AC^2=HC^2+AB^2$
 
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Albert said:
Point $H$ is the orthocenter of $\triangle ABC$

prove :$HA^2+BC^2=HB^2+AC^2=HC^2+AB^2$
hint:
A solutiopn of the diagrm of this problem is given,now it is obvious ,hope someone can solve it
 

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Albert said:
hint:
A solutiopn of the diagrm of this problem is given,now it is obvious ,hope someone can solve it
Very nice solution.

It is clear that $PAHB$ is a parallelogram and that $PB$ is perpendicular to $BC$.

Thus $HA^2+BC^2=4R^2$, where $R$ is the radius of the circumcircle.
 
caffeinemachine said:
Very nice solution.

It is clear that $PAHB$ is a parallelogram and that $PB$ is perpendicular to $BC$.

Thus $HA^2+BC^2=4R^2$, where $R$ is the radius of the circumcircle.
yes, you got it !
 

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