Altitudes of Triangle Meet at Point

Theorem

Let $\triangle ABC$ be a triangle.

The altitudes of $\triangle ABC$ all intersect at the same point.

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.

$\blacksquare$

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 $Q$.

$\angle AOQ = \angle BOP$

Given:

$\angle AQO = \angle BPO = $ one right angle.

$\triangle AOQ \sim \triangle BOP$

and:

$\triangle AOQ \sim \triangle ACP \sim \triangle BOP$

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 $O$.

$\blacksquare$

Definition:Orthocenter: the name given to that point where the altitudes intersect

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
2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): altitude (geometry)