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.

Ia
|
|
|ra          Io
|            |ro
----X----C-------Y------A
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)}.