Theorem of the three perpendiculars

From the perpendicularity criterion follows one of the most useful results of the geometry of space.

Theorem — Theorem of the three perpendiculars

Let $r$ be a line $\perp$ to a plane $\pi$ at a point $H$, and let $s$ be a line of $\pi$ not passing through $H$. Let $K$ be the foot of the perpendicular from $H$ to $s$ (in the plane $\pi$). Then, for any point $V$ on the line $r$ (with $V\ne H$), the line $VK$ is perpendicular to $s$.

Proof

Since $r\perp\pi$ at $H$ and $s\subset\pi$, we have $r\perp s$. Moreover $HK\perp s$ by construction (it is the perpendicular dropped from $H$ to $s$ in the plane). Consider the plane $\alpha$ that contains the line $r$ and the points $H$ and $K$: both $r$ and $HK$ lie in $\alpha$. The line $s$ is therefore perpendicular to two distinct lines of $\alpha$, hence $s\perp\alpha$. Since $VK$ lies in the same plane $\alpha$, it follows that $s\perp VK$. $\blacksquare$

The line $r$ is perpendicular to the plane at $H$; $K$ is the foot of the perpendicular from $H$ to $s$. Then $VK\perp s$.

Observation — Application: right pyramid

The theorem of the three perpendiculars explains why, in a right pyramid, the height $VH$ falls at the centre of the circle inscribed in the base. If $VH\perp$ to the plane of the base, then $VH$ is perpendicular to every line of the base, in particular to the base edges. Hence the slant height of each lateral face is $\perp$ to the base edge, and the feet of these perpendiculars turn out to be all equidistant from $H$: that is, $H$ is the centre of the circle inscribed in the base.

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Topics: Synthetic geometry in space
Concepts: Line-plane perpendicularity · Pyramid · Theorem of the three perpendiculars
Skills: Proving · Synthetic geometry