$\definecolor{A}{RGB}{0,148,79}\color{A}\fbox{Solution}$ Let $b,c,e,f,p$ be complex numbers lying on unit circle centered at $0$. Then $$b+c=0,\ a=\frac{bf(c+e)-ce(b+f)}{bf-ce}=\frac{f(c+e)+e(b+f)}{e+f},\ 2c'=b+p+f-\frac{bf}{p}.$$Because $C'H\perp CE$ $$\frac{c'-h}{c-e}=-\overline{\left(\frac{c'-h}{c-e}\right)}\iff \frac{b+p+f-\frac{bf}{p}-2h}{c-e}=\frac{\frac{1}{b}+\frac{1}{p}+\frac{1}{f}-\frac{p}{bf}-2\overline{h}}{\frac{c-e}{ce}}.$$Analogously $B'H\perp BF$ implies $$ \frac{c+p+e-\frac{ce}{p}-2h}{b-f}=\frac{\frac{1}{c}+\frac{1}{p}+\frac{1}{e}-\frac{p}{ce}-2\overline{h}}{\frac{b-f}{bf}}.$$Eliminating $\overline{h}$ from both equations and using $b+c=0$ gives something very nice: $$h=\frac{p(e+f)+b(e-f)}{2p}.$$Therefore $$h-e=\frac{(b-p)(e-f)}{2p},\ h-f=\frac{(b+p)(e-f)}{2p}\implies \frac{h-e}{h-f}=\frac{b-p}{b+p}=-\overline{\left(\frac{h-e}{h-f}\right)}\iff \definecolor{A}{RGB}{0,255,79}\color{A}EH\perp HF.\blacksquare$$#1756