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The Circle in the Corner

Traditional · Antiquity · OtherDifficultyPen & paper

A right triangle's incircle ties its area to its semiperimeter in one formula and to its legs in another — setting the two equal, and simplifying, is enough to force $c^{2} = a^{2} + b^{2}$ .

Let triangle $A B C$ have its right angle at $A$ , legs $A B = a$ and $A C = b$ , hypotenuse $B C = c$ . Taking the legs as base and height, its area is $\frac{1}{2} ab$ .
Inscribe its incircle — the circle tangent to all three sides — centered at $I$ with radius $r$ .
The incircle touches both legs, and the angle between them at $A$ is square, so the two tangent points sit exactly $r$ from $A$ along each leg — the same short distance either way.
For any triangle, the distance from a vertex out to a tangent point equals the semiperimeter $s = \frac{1}{2} (a + b + c)$ minus the side directly opposite it. From $A$ , the opposite side is $B C = c$ , so that distance is $s - c$ .
Both descriptions name the same length: $r = s - c = \frac{1}{2} (a + b - c)$ .
Now join $I$ to each vertex. This cuts the triangle into three smaller triangles, one on each side, every one of them standing exactly $r$ tall — the incircle's radius, measured straight out to that side.
Their areas add up to the whole: $\frac{1}{2} a r + \frac{1}{2} b r + \frac{1}{2} cr = r \cdot \frac{1}{2} (a + b + c) = r \cdot s$ .
Two expressions for the same area: $\frac{1}{2} ab = r \cdot s = \frac{1}{2} (a + b - c) \cdot \frac{1}{2} (a + b + c) = \frac{1}{4} [(a + b)^{2} - c^{2}]$ .
Multiply through by $4$ : $2 ab = (a + b)^{2} - c^{2} = a^{2} + 2 ab + b^{2} - c^{2}$ . The $2 ab$ on each side cancels, leaving $c^{2} = a^{2} + b^{2}$ .

Therefore c² = a² + b². ∎

The right angle is doing quiet work here: it’s the reason the two tangent points nearest $A$ land the same distance out along each leg, which is what lets a general fact about any triangle’s incircle — tangent length equals semiperimeter minus the opposite side — collapse into the clean formula $r=\tfrac12(a+b-c)$ . Drop the right angle and the incircle still exists, but that shortcut disappears with it.