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An altitude of a triangle is the whole line through one vertex that is orthogonal to the line through the opposite side. For vertex $A$ and opposite vertices $B,C$, its equation in position vectors is $(\mathbf{x}-\mathbf{a})\cdot (\mathbf{c}-\mathbf{b})=0$. The three such lines meet at the orthocenter; they need not meet inside the triangle.

The medial triangle joins the three midpoints of a triangle's sides. Each side is parallel to the corresponding original side and has half its length: $((\mathbf{x}+\mathbf{z})/2)-((\mathbf{x}+\mathbf{y})/2)=(\mathbf{z}-\mathbf{y})/2$. Each altitude of a triangle in the medial triangle is a perpendicular bisector of an original side, so its orthocenter is the original triangle's circumcenter.

For positive coordinate intercepts $A=(a,0,0)$, $B=(0,b,0)$ and $C=(0,0,c)$, their plane equation is $\mathbf{n}\cdot \mathbf{r}=1$. The closest point on a plane to the origin is $\mathbf{d}=\mathbf{n}/\Vert \mathbf{n}{\Vert }^2$, with distance $1/\Vert \mathbf{n}\Vert$. Its barycentric coordinate system coefficients are proportional to $a^{-2},b^{-2},c^{-2}$ and sum to one.

The projection is also the orthocenter of triangle $ABC$. Indeed $\mathbf{a}\cdot (\mathbf{b}-\mathbf{c})=0$, and $\mathbf{d}\cdot \mathbf{b}=\mathbf{d}\cdot \mathbf{c}$ because each equals $1/\Vert \mathbf{n}{\Vert }^2$. Hence $(\mathbf{a}-\mathbf{d})\cdot (\mathbf{b}-\mathbf{c})=0$. Cycling the vertices gives all three altitude directions. This connects an ambient-space orthogonal projection with intrinsic triangle geometry.

The altitude of a triangle through $A$ has equation $(\mathbf{a}-\mathbf{d})\cdot (\mathbf{b}-\mathbf{c})=0$, and the altitude of a triangle through $B$ has equation $(\mathbf{b}-\mathbf{d})\cdot (\mathbf{a}-\mathbf{c})=0$. Their normal vectors $\mathbf{b}-\mathbf{c}$ and $\mathbf{a}-\mathbf{c}$ are linearly independent because the triangle is nondegenerate. Consequently these two lines meet in exactly one point $D$.

Substitute its position vector $\mathbf{d}$ in the supplied identity. Its first term vanishes by the first altitude of a triangle equation, and its second term vanishes by the second altitude of a triangle equation. HenceThis is exactly the condition that $D$ lies on the altitude of a triangle through $C$. The three lines therefore meet in a unique point, the orthocenter. This argument includes right triangles and triangles whose orthocenter lies outside them.

Let $A=(Y+Z)/2$, $B=(Z+X)/2$, and $C=(X+Y)/2$, interpreting these expressions as position vectors. The resulting medial triangle is nondegenerate: for example,Thus $BC$ is parallel to $YZ$, and the same calculation holds cyclically. The altitude of a triangle through $A$ is consequently the line through the midpoint of $YZ$ that is orthogonal to $YZ$, namely its perpendicular bisector. Each remaining altitude of a triangle similarly gives one of the other two perpendicular bisectors.

By iii, the medial triangle has a unique orthocenter, so the three perpendicular bisectors meet in one point. Every point on the perpendicular bisector of $YZ$ is equally far from $Y$ and $Z$: expanding the squares of the two Euclidean distances gives $2\mathbf{q}\cdot (\mathbf{z}-\mathbf{y})=\Vert \mathbf{z}{\Vert }^2-\Vert \mathbf{y}{\Vert }^2$. Applying this to the three sides shows that the common point is equally far from $X,Y,Z$. It is the circumcenter.

The vector $\mathbf{a}$ is orthogonal to both $\mathbf{b}$ and $\mathbf{c}$, so $\mathbf{a}\cdot (\mathbf{b}-\mathbf{c})=0$. Also $\mathbf{d}\cdot \mathbf{b}=\mathbf{d}\cdot \mathbf{c}=1/S$, hence $\mathbf{d}\cdot (\mathbf{b}-\mathbf{c})=0$. Subtraction gives $(\mathbf{a}-\mathbf{d})\cdot (\mathbf{b}-\mathbf{c})=0$. Both vectors are nonzero, soThis is the altitude property of the orthogonal projection onto an intercept plane: the projection point is the triangle's orthocenter.