Altitudes of Triangle Meet at Point

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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 $\triangle ABC$.


AltitudesIntersectAtOrthocenter.png


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

Newton altitude.png

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


Sources

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