- #1
Albert1
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Point $H$ is the orthocenter of $\triangle ABC$
prove :$HA^2+BC^2=HB^2+AC^2=HC^2+AB^2$
prove :$HA^2+BC^2=HB^2+AC^2=HC^2+AB^2$
Very nice solution.Albert said:hint:
A solutiopn of the diagrm of this problem is given,now it is obvious ,hope someone can solve it
yes, you got it !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.
The orthocenter property of a triangle states that the three altitudes of a triangle intersect at a single point, known as the orthocenter.
The orthocenter property of a triangle can be proved using several methods, such as coordinate geometry, vector algebra, or trigonometry. One common method is to use the perpendicular bisectors of the sides of the triangle to show that they intersect at a single point, which is the orthocenter.
Yes, the orthocenter property of a triangle can be proved using only geometry. One method is to construct the altitudes of the triangle and show that they intersect at a single point. Another method is to use the properties of perpendicular bisectors to prove the orthocenter property.
Yes, the orthocenter property of a triangle is always true. It is a fundamental property of triangles and is not dependent on the shape or size of the triangle.
The orthocenter property of a triangle is important in geometry as it helps to define and identify different types of triangles. It also has practical applications in fields such as architecture and engineering, where the orthocenter is used to determine the height of a building or the placement of supports for bridges.