Geometric Mean in Right Triangles

Drop a perpendicular from the right angle to the hypotenuse. You now have three triangles — the original and two small ones — and all three are similar to each other.

That similarity produces two relationships worth memorising.

The altitude is the geometric mean of the two pieces it creates:
altitude² = (left piece)(right piece)

Each leg is the geometric mean of the whole hypotenuse and the piece next to it:
leg² = (whole hypotenuse)(adjacent piece)

The geometric mean of a and b is √(ab). So if the hypotenuse splits into 4 and 9, the altitude is √36 = 6.

The trap is pairing a leg with the *far* piece. Each leg goes with the piece it touches.

The altitude from the right angle to the hypotenuse creates two triangles each similar to the original and to each other, by AA — each shares an acute angle with the original and has a right angle.

The proportions that follow are the geometric mean relations. With hypotenuse split into segments p and q adjacent to legs a and b respectively, and altitude h:

h² = pq a² = cp b² = cq

where c = p + q. Each is a statement that one length is the geometric mean √(xy) of two others.

Adding the last two recovers the Pythagorean theorem: a² + b² = c(p + q) = c², which is one of the standard proofs of it.

The geometric mean of a and b is √(ab), and it always lies between them, satisfying √(ab) ≤ (a+b)/2 with equality only when a = b — the arithmetic-geometric mean inequality.

Matching each leg to its adjacent segment is what the similarity statement dictates; pairing a leg with the far segment is the standard error.