Proving Orthocenter Property of Triangle ABC

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SUMMARY

The orthocenter property of triangle ABC states that for the orthocenter H, the equations HA² + BC² = HB² + AC² = HC² + AB² hold true. This relationship can be derived using the properties of the parallelogram PAHB, where PB is perpendicular to BC. The derived equation HA² + BC² = 4R² demonstrates the connection between the orthocenter and the circumradius R of triangle ABC.

PREREQUISITES
  • Understanding of triangle properties, specifically orthocenters and circumcircles.
  • Familiarity with the concept of parallelograms in geometry.
  • Knowledge of the Pythagorean theorem as it applies to triangle geometry.
  • Basic understanding of trigonometric relationships in triangles.
NEXT STEPS
  • Study the properties of the circumcircle and its radius in relation to triangle geometry.
  • Explore the derivation of the orthocenter and its significance in triangle properties.
  • Learn about the relationships between triangle sides and angles using trigonometric identities.
  • Investigate other triangle centers, such as the centroid and circumcenter, and their properties.
USEFUL FOR

Mathematicians, geometry enthusiasts, and students studying advanced triangle properties will benefit from this discussion, particularly those interested in the relationships between triangle centers and their geometric implications.

Albert1
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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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