See here

I believe this picture is enough. Just prove $H$ is also the orthoheart of $KMC$

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Throw the configuration onto the complex plane,unit circle, vertices on the unit circles, such that $c=\frac{1}{b}$. Since $OF \parallel BC$ we must have that $\frac{f}{\overline{f}}=\frac{b-c}{\overline{b-c}}=-bc=-1$, which implies that $f=-\overline{f}$. But since we have that $F \in AB$, we must have that $\overline{f}=\frac{a+b-f}{ab}$. Combining the two together we get that $f=-\frac{a+b}{ab-1}$. Since $H$ is the orthocenter we have that $h=a+b+c=a+\frac{b^2+1}{b}$, implying that $m=\frac{a+h}{2}=a+\frac{b^2+1}{2b}$. Plugging in what we have we get that: $$\frac{c-m}{\overline{c-m}}=-\frac{m-f}{\overline{m-f}}$$clearly implying that $CM \perp MF$

We use complex numbers with $(ABC)$ as the unit circle and $OF$ as the real line. Observe that $B$ and $C$ are reflections across the imaginary line so $bc=-1$. Clearly we have $M=\frac{A+H}2=\frac{a+a+b+c}2=\frac{2a+b+c}2$. By Lemma 8, we have $F=\frac{a+b}{ab+1}$ and clearly $C=c$. Their conjugates are $\overline{M}=\frac{\frac2a+\frac1b+\frac1b}{2},\overline{F}=F=\frac{a+b}{ab+1},\overline{C}=\frac1c$, respectively. It remains to check that $\frac{m-f}{m-c}+\overline{\left(\frac{m-c}{m-f}\right)}=0$ which can be rewritten as $\frac{m-f}{\overline{m}-\overline{f}}+\frac{m-c}{\overline{m}-\overline{c}}=0$. Upon plugging in the values and substituting $c=-\frac1b$, we now wish to show $$\frac{\frac{2a+b-1/b}2-\frac{a+b}{ab+1}}{\frac{2/a+1/b-b}2-\frac{a+b}{ab+1}}+\frac{\frac{2a+b-1/b}2-\left(-\frac1b\right)}{\frac{2/a+1/b-b}2-(-b)}=0$$which happens to be true upon expansion. $\blacksquare$

franzliszt wrote: We use complex numbers with $(ABC)$ as the unit circle and $OF$ as the real line. Observe that $B$ and $C$ are reflections across the imaginary line so $bc=-1$. Clearly we have $M=\frac{A+H}2=\frac{a+a+b+c}2=\frac{2a+b+c}2$. By Lemma 8, we have $F=\frac{a+b}{ab+1}$ and clearly $C=c$. Their conjugates are $\overline{M}=\frac{\frac2a+\frac1b+\frac1b}{2},\overline{F}=F=\frac{a+b}{ab+1},\overline{C}=\frac1c$, respectively. It remains to check that $\frac{m-f}{m-c}+\overline{\left(\frac{m-c}{m-f}\right)}=0$ which can be rewritten as $\frac{m-f}{\overline{m}-\overline{f}}+\frac{m-c}{\overline{m}-\overline{c}}=0$. Upon plugging in the values and substituting $c=-\frac1b$, we now wish to show $$\frac{\frac{2a+b-1/b}2-\frac{a+b}{ab+1}}{\frac{2/a+1/b-b}2-\frac{a+b}{ab+1}}+\frac{\frac{2a+b-1/b}2-\left(-\frac1b\right)}{\frac{2/a+1/b-b}2-(-b)}=0$$which happens to be true upon expansion. $\blacksquare$ Wow didn't know Wolframalpha could do this. But guess you lost more time writing than eventually calculating yourself

Well yeah I didn't feel like typing out all my calculations this time lol Also check with WA that your bash works in the end or else rip