# A Perspective on Altitudes

**Figure 1.**Pushing the triangle into a corner.

*B*is the point at which the line

_{i}*A*intersects with the opposite side.

_{i}PAs it turns out, embedding the problem in three dimensions yields an additional insight. To begin, we shove an arbitrary acute** ^{1}** triangle \(A_1A_2A_3\) into the corner of a rectangular quadrant, as shown in Figure 1; each vertex now lies on a coordinate axis.

I claim that the concurrency point of the altitudes is precisely the foot \(P\) of the perpendicular from the origin onto the plane of the triangle.

**Figure 2.**The tangent cone.

### Proof

With \(P\) defined as in the previous sentence, let \(B_1\) be the point at which the line \(A_1P\) intersects with side \(A_2A_3\).

I claim that \(A_1B_1\) is an altitude of the triangle. Indeed, \(A_2A_3 \perp OP\) (since \(A_2A_3\in {\rm plane}(A_1A_2A_3) \perp OP\)) and \(A_2A_3\perp \ OA_1\) (since \(A_2A_3\in{\rm plane}(O A_2A_3)\perp OA_1\)). In summary, because \(A_2A_3\) is normal to two lines (\(OP\) and \(OA_1\)) in the plane \(POA_1\), it is normal to every line in that plane and thus to \(A_1B_1\). So \(A_1B_1\) is indeed an altitude. The same argument applies to \(A_i B_i\) for \(i=2, 3\), meaning that all altitudes pass through \(P\). Q.E.D.

### Proof 2

Here is a slightly different way to express essentially the same idea. Referring to Figure 2, construct the circular cone tangent to the plane of the triangle, with vertex \(A_1\) and axis \(A_1O\). Define \(B_1\) as the point at which the line of tangency intersects with side \(A_2A_3\). Now \(A_1B_1 \perp A_2A_3\), according to Figure 3a, and \(A_1B_1\) passes through \(P\), according to Figure 3b (with \(P\) defined as above). This shows that the altitude from an arbitrarily chosen vertex passes through \(P\). Q.E.D.

**Figure 3.**In

**(a)**, the generator is orthogonal to the base for a right circular cone. In

**(b)**, the foot of the perpendicular to a tangent plane from a point on the axis lies on the line of tangency.

*The figures in this article were provided by the author.*

** ^{1}** Unfortunately, this approach does not seem to extend to obtuse triangles; or perhaps I am not acute enough to find an extension.