Altitudes of Triangle Meet at Point/Proof 1

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.

$\blacksquare$