So we use barycentric coordinates.

It's obvious that un-normalized, , so we get a normaized . Similarly, .

Okay, so then we get the point by intersecting the lines and .


Let be such that . It's obvious that , so we find that $Q'_y = \frac{s-b}{a} - \frac{s-b}s} = \frac{(s-a)(s-b)}{sa}$. Also, since it lies on the line , we get that . Hence,

Let . Then the displacement vector of is which we want to show has length . Let and . Using the inversion distance formula, as well as , this becomes equivalent to:

\[ \begin{align*} -\frac{1}{4} \cdot \frac{16K^2}{4s^2} &= a^2 \cdot \frac{jk}{(2as)^2} + b^2 \cdot \frac{ak}{4as^2} + c^2 \cdot \frac{aj}{4as^2} \\ &= \frac{jk + b^2k + c^2j}{4s^2} \\ \Leftrightarrow -16K^2 &= 4(jk+b^2k+c^2j) \end{align*} \]

But we also know that by Heron, and

\[ \begin{align*} j &= 2(s-a)(s-b)-ab \\ &= 2s^2-2(a+b)s+ab \\ &= \frac{1}{2} (a+b+c)^2 - (a+b)(a+b+c) + ab \\ &= \frac{1}{2}(a^2+b^2+c^2) + (ab+bc+ca)-(a^2+2ab+b^2+ca+ab)+ab \\ &= \frac{1}{2}(ac^2-a^2-b^2} \end{align*} \]

Similarly,

So now we just compute

\[ \begin{align*} 4jk + 4b^2k+4c^2j &= (c^2-a^2-b^2)(b^2-a^2-c^2) + 2b^2(b^2-a^2-c^2) + 2c^2(c^2-a^2-b^2) \\ &= (a^2+(b^2-c^2))(a^2+(c^2-b^2)) + 2b^4 + 2c^4 -2a^2b^2 - 2a^2c^2 - 4b^2c^2 \\ &= a^4-(b^2-c^2)^2  + 2b^4 + 2c^4 -2a^2b^2 - 2a^2c^2 - 4b^2c^2 \\ &= a^4+b^4+c^4-2a^2b^2 - 2b^2c^2-2c^2a^2 \\ &= -16K^2 \end{align*} \]

as desired.