Corona virus quarantine. Home schooling. My teenage daughter asks: “Daddy, how do you compute the area of this triangle?” Well, that is something that I haven't done since I was a teenager. And then only as “base x height / 2”. This wasn't a right triangle, and my daughter didn't yet have trigonometry, so how do you figure the height? She googled and found Heron's formula.

Heron's remarkable formula states that the area of a triangle is `\Delta = \sqrt{s(s - a)(s - b)(s - c)}` where `a`, `b`, `c`, are the side lengths and `s` is the semiperimeter `(a + b + c) / 2`. Why would this be true? Where do these values come from? I found messy proofs using algebra, geometry, trigonometry, polynomials, complex numbers, determinants, and four dimensions. Noe of them appeared particularly enlightening.

Apparently, I am not the only one looking for a more insightful proof. The following is from an email dialog between Peter Doyle and John Horton Conway (yes, the John Horton Conway, who sadly just fell victim to a Corona virus infection):

 You were once keen to know the best proof of Heron.  The best
I've found is this:

    DELTA^2 = ro.so.ra.sa = so.sa.sb.sc

since from the diagram below we have  ra/sb = sc/ro, the triangles
Ia X C  and  C Y Io  being similar in view of the angles  Io C V = C/2
and  Ia C X  the complement of this.

          |ra          Io
          |            |ro
            sb     sc

   I write  so,sa,sb,sc  for the usual  s,s-a,s-b,s-c, and remark
that the formulae  DELTA = rx.sx  (x=0,a,b,c)  have well-known very
simple proofs.

    It would be nice to find a simple symmetric 4-dimensional proof,
but I've not yet managed to do so.

   Regards,  JHC

That is in fact a nice proof (even though it is not symmetric) because it shows `s - a`, `s - b`, `s - c`, and `s`, and it teaches us some interesting properties of the incircle and excircles. The following description is a bit longer than Conway's ASCII art, but hopefully clearer.

Let's start with the incircle. Its center is the intersection `I` of the angle bisectors (in light gray).

If we knew its radius `r`, we would be in good shape. Then we could carve up the area `\Delta` of the triangles into the areas of the three triangles below.

Tip: In all these diagrams, try dragging the blue points. It gives you a good feel how the remainder of the diagram is constructed.

`\Delta = 1/2 ar + 1/2 br + 1/2 cr = sr`

But we don't know `r`. To get there, let's measure the distances from the triangle vertices to the tangential points `X`, `Y`, `Z` of the incircle. This is where `s - a`, `s - b`, `s - c` come in:

To see why, first observe that the two yellow segments must have the same length since the triangles `AYI` and `AZI` are congruent—the gray line `AI` bisects the angle `ZAY`. Similarly, the red and the green segments are the same. Let's call them `sa`, `sb`, `sc` for a moment. Then `2 sa + 2 sb + 2 sc = a + b + c = 2s`, and `sa + sb = c`. Therefore, `sc = s - c`, and similar for `sa` and `sb`.

Now, let's draw one of the excircles. The excircle corresponding to the vertex `A` has center `O` that lies on the extension of the line segment `AI`. It touches the opposing side `a` and the extensions of the adjacent sides `b` and `c`. Let's call the touch points `T`, `U` and `V`.

Remarkably, the line from `A` to `V` has three segments of length `s - a`, `s - b`, and `s - c` (as shown below), for a total length of `s - a + s - b + s - c = 3s - (a + b + c) = s`.

Let's now see why `BV` has length `s - c`. Note that `BV` and `BT` have the same length, as do `CT` and `CU`. Moreover, `AU` and `AV` also have the same length, and the sum of their lengths is `a + b + c = 2s`, as you can see by “folding” `CU` to `CT` and `BV` to `BT`. Therefore, each of them has length `s`. Then the length of `BV` is `s - (s - a) - (s - b) = s - c`.

On to our final step to determine the incircle radius `r`. Note that the two right triangles marked below are similar. As Conway wrote, “in view of angles”. ∡ZBI = β/2, so ∡ZIB = π/2 - β/2. And ∡OBV = ∡TBV / 2 = (π - β) / 2.

Therefore, `(s - b)/r = t / (s - c)`, where `t` is the radius of the outcircle. A dilation with center `A` maps the incircle to the outcircle. Its scale factor must be `s / (s - a)` since it also maps `Z` to `V`. That yields the radius of the outcircle as `t = r s / (s - a)`. Plugging into the first equation yields `(s - b)/r = r s / ((s - a)(s - c))`, or `r^2 = ((s - a)(s - b)(s - c))/s`.

At the very beginning, we saw that the area `\Delta = sr`. Now that we know `r`, we see that `Delta = s \sqrt{((s - a)(s - b)(s - c))/s} = \sqrt{s(s-a)(s-b)(s-c)}`.

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