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To deeply understand the first part of the proof we first generalize it.[br][br]Our cuts will be straight, so let's define them as segments. First consider the area of a circle. A cut from center to edge we call a radial cut, while the radius r is the length of such a cut. We can perform a radial cut in [i]any[/i] direction, due to the circle's supersymmetry all those cuts will be the symmetric and have the same length.[br][br]Now consider a polygon where each edge is tangential to the same circle, an incircle. We call these polygons tangential polygons. In any tangential polygon we can cut from the incircle's center to each edge. Because of symmetry this partitions our polygon into right-angled kites. We can again cut those kites along their axis of symmetry to obtain congruent pairs of right-angled triangles.[br][br]All these right triangles will have one leg of length r. All other legs form the perimeter p of the polygon. We can therefore rearrange the triangles into a simple rectangle where one side has length r while the other has length s := p/2. We call s semi-perimeter. Any tangential polygon's area is thus A = r s.[br][br]For polygons with n > 3 edges tangential polygons are a special case. For n = 3 we get triangles and all three edges are automatically tangential to the same circle: the incircle, just like all three people seated at a round table will automatically be adjacent to each other.[br][br]Thus all triangles contain the patterns shown above, all contain this astounding symmetry our proof has uncovered. And only because of this symmetry we can now describe the area as product of the incircle's radius and the semi-perimeter.

To truly understand how to find r we first need to search for it. This explorative part is usually left out in proofs, since it is not required for a succinct argument.[br][br]To explore different values for x, y, z lets fix two of them to be equal and grow the third.

We can see that for y approaching 0 the radius r also approaches 0.[br]For large y the radius r approaches x = z from below. So r² approaches x z.[br][br]From this we can postulate that r² is equal to x z times y, divided by y plus something positive:[br] [math]r^2=\frac{xzy}{y+c}[/math][br][br]Now since there is nothing special about our choice of y we can repeat the same argument for x and z. We arrive at[br][math]r^2=\frac{xyz}{x+y+z}[/math][br][br]I left out c since the other two variables already provide something positive. If we test the formula with examples it already holds true.[br][br]But how do we prove such a relation?[br][br]Well we have right triangles that fit together perfectly and have the required measures as legs. All we need to do is some multiplication (i.e. scaling) and addition (placing the legs in a line). Lets go:

We have shown that [math]xyz=r^2\left(x+y+z\right)[/math].[br][br]Thus[br][math]A=rs=\sqrt{\frac{xyz}{x+y+z}}\left(x+y+z\right)=\sqrt{xyz\left(x+y+z\right)}=\sqrt{xyzs}=\sqrt{\left(s-a\right)\left(s-b\right)\left(s-c\right)s}[/math][br][br]That concludes our proof, we have found A.

My work is nothing special. The proof's main ideas came from this Mathocube video: https://www.youtube.com/watch?v=6KPSmajeseI[br][br]The second part seemed to have a gap in the reasoning, so I explored it myself, which was a lot of fun. The proof presented here now has a lot of words, but this enabled me to deeply understand it and it is complete. Note that we could also have used the more general geometric mean theorem in the explorative part, however I prefer how xz tangibly forms the square of r in this special case.[br][br]Finally, in the words of the late John Conway I think the proof is most pleasing. https://math.dartmouth.edu/~doyle/docs/heron/heron.txt