Altitudes of Triangle Meet at Point
Theorem
Let $\triangle ABC$ be a triangle.
The altitudes of $\triangle ABC$ all intersect at the same point.
Proof 1
From Triangle is Medial Triangle of Larger Triangle, it is possible to construct $\triangle DEF$ such that $\triangle ABC$ is the medial triangle of $\triangle DEF$.
Let $AJ$, $BG$ and $CH$ be the altitudes of $\triangle ABC$.
By definition of medial triangle, $A$, $B$ and $C$ are all midpoints of the sides $DF$, $DE$ and $EF$ of $\triangle DEF$ respectively.
As $DF$, $DE$ and $EF$ are parallel to $BC$, $AC$ and $AB$ respectively, $AJ$, $BG$ and $CH$ are perpendicular to $DF$, $DE$ and $EF$ respectively.
Thus, by definition, $AJ$, $BG$ and $CH$ are the perpendicular bisectors of $DF$, $DE$ and $EF$ respectively.
The result follows from Perpendicular Bisectors of Triangle Meet at Point.
$\blacksquare$
Proof 2
In $\triangle ABC$ construct the altitude from vertex $A$ to side $BC$ at $P$.
Also draw the altitude from vertex $B$ to side $AC$ at $Q$.
By Two Straight Lines make Equal Opposite Angles:
- $\angle AOQ = \angle BOP$
Given:
- $\angle AQO = \angle BPO = $ one right angle.
By Triangles with Two Equal Angles are Similar:
- $\triangle AOQ \sim \triangle BOP$
and:
- $\triangle AOQ \sim \triangle APC \sim \triangle BOP$
By the definition of similarity:
- $\dfrac {OP} {BP} = \dfrac {CP} {AP}$
Rearrange:
- $OP = \dfrac {BP \cdot CP} {AP}$
Note there is no term related to side $AC$.
If we had used side $AB$ instead of $AC$, we would have the same result.
Therefore, the altitude from vertex $C$ to side $AB$ meets both $AP$ and $BQ$ at $O$.
$\blacksquare$
Also see
- Definition:Orthocenter: the name given to that point where the altitudes intersect
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {IV}$. Pure Geometry: Plane Geometry: The orthocentre
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): altitude
- For a video presentation of the contents of this page, visit the Khan Academy.