Given $\triangle ABC\ :\ A(0,0),B(b,0),C(0,c)$. Point $A'(\frac{b}{2},\frac{c}{2})$. The line $AD\ :\ y=\frac{b}{c}x$ intersects the line $BC\ :\ y=-\frac{c}{b}(x-b)$ in the point $D(\frac{bc^{2}}{b^{2}+c^{2}},\frac{b^{2}c}{b^{2}+c^{2}})$. Point $M(\frac{bc^{2}}{2(b^{2}+c^{2})},\frac{b^{2}c}{2(b^{2}+c^{2})}\ )$. The line $AA'\ :\ y=\frac{c}{b}x$ intersects the line $BM\ :\ y=\frac{bc}{-2b^{2}-c^{2}}(x-b)$ in the point $P(\frac{b^{3}}{c^{2}+3b^{2}},\frac{b^{2}c}{c^{2}+3b^{2}})$. Slope of the line $PC\ :\ m_{PC}=-\frac{c^{3}+2b^{2}c}{b^{3}}$, slope of the line $BC\ :\ m_{BC}=-\frac{c}{b}$. $\tan \angle PCB\ =\ \frac{m_{BC}-m_{PC}}{1+m_{BC} \cdot m_{PC}}=\frac{bc}{b^{2}+c^{2}}=\frac{bc}{BC^{2}}=\frac{b}{BC} \cdot \frac{c}{BC}=\sin C \cdot \cos C$.

Let $PE \perp BC$ Then from triangle $AA'D$ and line $BD$ we can find $\frac{AA'}{PA'}=\frac{2AB^2+BC^2}{BC^2}$ and so $PE=\frac{AB*BC*CA}{2AB^2+BC^2}$ From $\frac{A'E}{A'D}=\frac{A'P}{AA'}$ we have $A'E=\frac{BC|BC^2-2AB^2|}{2(2AB^2+BC^2)}$ And so $CE=\frac{BC^3}{2AB^2+BC^2}$ $\tan\angle PCB = \frac{PE}{CE}=\frac{AB*CA}{BC^2}=\sin C\cdot\cos C$

RagvaloD wrote: Let $PE \perp BC$ Then from triangle $AA'D$ and line $BD$ we can find $\frac{AA'}{PA'}=\frac{2AB^2+BC^2}{BC^2}$ and so $PE=\frac{AB*BC*CA}{2AB^2+BC^2}$ From $\frac{A'E}{A'D}=\frac{A'P}{AA'}$ we have $A'E=\frac{BC|BC^2-2AB^2|}{2(2AB^2+BC^2)}$ And so $CE=\frac{BC^3}{2AB^2+BC^2}$ $\tan\angle PCB = \frac{PE}{CE}=\frac{AB*CA}{BC^2}=\sin C\cdot\cos C$ beautiful ellementarly geometry!