# A Strictly Geometrical Proof of the Altitude Theorem

The **Altitude Theorem** or **Geometric Mean Theorem** is a result from
high-school geometry. In a right triangle, the *altitude* $h$ on the
hypotenuse divides the hypotenuse into two segments, $p$ and $q$. The
theorem now states that $h^2 = pq$ or, equivalently, $h = \sqrt{pq}$:
the altitude equals the geometric mean of the segments of the hypotenuse.

The content of this theorem is a bit surprising, because the altitude and the hypotenuse segments seem geometrically somewhat “unrelated”: it’s not clear how one could (geometrically) be transformed into the other. And although the theorem can be proven in many different ways, many of the proofs are at least partially algebraic, and therefore do not provide an intuitive, geometric sense why it is true. But it turns out that a very elegant, strictly geometric proof of this proposition can be constructed.

The first hint is the realization that the two triangles formed by the
altitude are *similar*: they all have the same angles. To see why this is
true, focus on the left, “blue”, triangle. The two angles $\alpha$ and $\beta$
add up to 90 degrees (the right angle contributes another 90 degrees to
make it a total 180 degrees in the triangle). But the top angles from both
the left “blue” and the right “red” triangle *also* add up to 90 degrees, by
construction of the overall right triangle! Hence, the top angle of the
red triangle equals $\beta$, so that the left, right, and overall triangles
are all similar.

(By the way, this observation already leads to a first, though algebraic, proof of the proposition: if the triangles are similar, then the ratio of their legs must be equal, hence: $\frac{p}{h} = \frac{h}{q}$, which leads to $h^2 = pq$.)

Now construct the following rectangle, composed of the two partial triangles created by the altitude (two copies of each), and the two areas in question, $h^2$ and $pq$.

The essential observation now is that the diagonal *divides this
rectangle in half*, which means that the areas above and below the diagonal
are equal. But the triangles above and below the diagonal are
equal and hence cancel each other out, which ultimately leads to the
conclusion that the rectangle $pq$ and the square $h^2$ must *also* be
equal - which concludes our proof.

I certainly make no claims of this proof being new! But I like its strictly geometrical nature: all the lengths and areas involved are directly visible in the previous figure. The proof is also refreshing because it does not employ the Pythagorean Theorem. And finally, it provides the intuition behind the Altitude Theorem: both $h^2$ and $pq$ are constructed above the respective sides of the two partial triangles - all that was necessary was to arrange the red and the blue triangle appropriately!