Proving Orthocenter Property of Triangle ABC

[Albert1]

Point $H$ is the orthocenter of $\triangle ABC$

prove :$HA^2+BC^2=HB^2+AC^2=HC^2+AB^2$

 

[Albert1]

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

 

[caffeinemachine]

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.

 

[Albert1]

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 !