Orthocentre - concurrency of the altitudes

The orthocentre is the intersection of the three altitudes. The proof of their concurrency uses a brilliant idea: build a larger triangle for which the altitudes become the perpendicular bisectors of the sides.

Theorem — Concurrency of the altitudes

The three altitudes of a triangle meet at a single point (orthocentre).

The altitudes of $ABC$ (green) are the perpendicular bisectors of the sides of the outer triangle $DEF$.

Proof — The altitudes are concurrent

1. From each vertex of $ABC$ we draw the parallel to the opposite side, forming the larger triangle $DEF$.
2. By construction, $ABDC$ is a parallelogram ($AB\parallel CD$, $AC\parallel BD$), hence $AB\cong CD$. Likewise, $ABCE$ is a parallelogram, hence $AB\cong CE$.
3. It follows that $CD\cong CE$, that is, $C$ is the midpoint of $DE$.
4. Now comes the key idea: the altitude $CH$ is perpendicular to $AB$, and $AB\parallel DE$ (by construction), hence $CH\perp DE$ at the midpoint $C$. But then $CH$ is the perpendicular bisector of the side $DE$!
5. Likewise, the other altitudes of $ABC$ are perpendicular bisectors of the sides of $DEF$. Since the perpendicular bisectors of a triangle are concurrent, the altitudes of $ABC$ are concurrent.

∎

Links

Topics: Euclidean circle
Concepts: Altitude · Orthocentre
Skills: Proving · Synthetic geometry