Using vectors, prove that all three altitudes of a triangle…

Using vectors, prove that all three altitudes of a triangle meet at a single point (the orthocentre).

Step 1: Setup

Let $\vec a,\vec b,\vec c$ be the position vectors of the vertices $A,B,C$ with respect to any origin $O$: \[\overrightarrow{BC}=\vec c-\vec b,\qquad \overrightarrow{CA}=\vec a-\vec c,\qquad \overrightarrow{AB}=\vec b-\vec a.\]

Step 2: Intersection of two altitudes

The altitudes from $A$ and $B$ are perpendicular to the non-parallel sides $BC$ and $CA$, so they are not parallel and meet at one point $H$ with position vector $\vec h$:

\[ (\vec h-\vec a)\cdot(\vec c-\vec b)=0 \qquad (1) \]

\[ (\vec h-\vec b)\cdot(\vec a-\vec c)=0 \qquad (2) \]

Step 3: $H$ lies on the third altitude

Expand both conditions: $\vec h\cdot\vec c-\vec h\cdot\vec b-\vec a\cdot\vec c+\vec a\cdot\vec b=0 \qquad (1')$.

\[ \vec h\cdot\vec a-\vec h\cdot\vec c-\vec a\cdot\vec b+\vec b\cdot\vec c=0 \qquad (2') \]

Adding them, the terms $\vec h\cdot\vec c$ and $\vec a\cdot\vec b$ cancel: $\vec h\cdot\vec a-\vec h\cdot\vec b-\vec c\cdot\vec a+\vec c\cdot\vec b=0$.

\[ (\vec h-\vec c)\cdot(\vec a-\vec b)=0\quad\Longleftrightarrow\quad \overrightarrow{CH}\cdot\overrightarrow{AB}=0. \qquad (3) \]

So $CH\perp AB$: the point $H$ also lies on the altitude from $C$. (If $H=C$, the statement is trivial.) Therefore all three altitudes pass through $H$. $\blacksquare$

\[(1)+(2)\ \Longrightarrow\ \overrightarrow{CH}\cdot\overrightarrow{AB}=0\]

Step 4: Special case: right triangle

If $\angle A=90^\circ$, the altitude from $B$ (perpendicular to $CA$) is the side $BA$, and the altitude from $C$ (perpendicular to $AB$) is the side $CA$. They meet at $A$, so the orthocentre is the right-angle vertex. The altitude from $A$ passes through $A$ as well.