Why Altitudes Meet at One Point

An animated proof showing that all three altitudes of a triangle intersect at a single point — the orthocenter — using a parallel construction.

Keywords: orthocenter, altitudes, perpendicular bisector, triangle centers, parallel lines

Prerequisites: circumcenter-proof · Difficulty: intermediate

The altitudes of a triangle all pass through the same point. This special point is called the orthocenter.

An altitude is a line segment from a vertex perpendicular to the opposite side, forming a $90°$ angle.

The Proof Strategy

This proof uses a clever construction:

1. Build an outer triangle $DEF$ using parallel lines
2. Show that $A$, $B$, $C$ are midpoints of $DEF$'s sides
3. Conclude that altitudes of $ABC$ are perpendicular bisectors of $DEF$
4. Apply the circumcenter theorem: perpendicular bisectors meet at one point!

The Proof

Step 1: Construct Outer Triangle $DEF$

We start with triangle $ABC$ and construct an outer triangle $DEF$:

• Through each vertex, draw a line parallel to the opposite side
• These three parallel lines intersect to form triangle $DEF$

This construction creates parallelograms, which will be the key to the proof.

Step 2: $A$, $B$, $C$ are Midpoints

The parallel construction creates parallelograms. In a parallelogram, opposite sides are equal.

This means:

• $A$ is the midpoint of $EF$
• $B$ is the midpoint of $FD$
• $C$ is the midpoint of $DE$

So $ABC$'s sides are actually midsegments of triangle $DEF$!

Step 3: Altitudes are Perpendicular Bisectors

Now draw the altitudes of $ABC$. Each altitude is perpendicular to a side.

Key insight: Each side of $ABC$ is parallel to a side of $DEF$:

• $BC \parallel EF$, and $A$ is the midpoint of $EF$
• So the altitude from $A$ (perpendicular to $BC$) is the perpendicular bisector of $EF$!

Similarly for the other two altitudes. The altitudes of $ABC$ are exactly the perpendicular bisectors of $DEF$!

Step 4: The Orthocenter

We've shown that the altitudes of $ABC$ are the perpendicular bisectors of triangle $DEF$.

From the circumcenter theorem, we know that perpendicular bisectors meet at one point. Therefore, the three altitudes must meet at one point — the orthocenter $H$!

Conclusion

All three altitudes of triangle $ABC$ meet at the orthocenter $H$.

The key insight is that the altitudes of $ABC$ are exactly the perpendicular bisectors of the auxiliary triangle $DEF$. Since perpendicular bisectors always meet at one point (the circumcenter), so do the altitudes!

Fun facts:

• Acute triangle → orthocenter inside
• Right triangle → orthocenter at the right-angle vertex
• Obtuse triangle → orthocenter outside