Qn: Prove that in a triangle ABC 
the altitude through C, the median through B
and the internal angle bisector through A are concurrent iff $sin C = cos B\times tan A$.
Attempt1: By cevas theorem we can say that $\frac{CE}{AE}\times \frac{AF}{FB}\times \frac{CD}{DB} = 1$ . Going backwards from given equation, I get $ \ \ \ tanA = \frac{sinC}{cosB}$. Then by using right triangles I obtain $\ tanA = \frac{CF}{FA} \ \ , \ \ cosB= \frac{BF}{CB}$. $\\ \ $ Using this information how can I prove equation above to hold if and only if lines are concurrent?
We will calculate the following expresion: $$\frac{\sin {\angle{EBA}}}{\sin {\angle{EBC}}}\cdot \frac{\sin {\angle{DAC}}}{\sin {\angle{FCA}}}\cdot\frac{\sin {\angle{FCB}}}{\sin {\angle{DAB}}}$$ Since $\angle{DAB}=\angle{DAC}$ expression becomes $$\frac{\sin {\angle{EBA}}}{\sin {\angle{EBC}}}\cdot\frac{\sin {\angle{FCB}}}{\sin {\angle{FCA}}}$$ Now $\angle{FCA}=90-A,\angle{FCB}=90-B$
Another hand if we use Sin Law for triangles $BEA$ and $ECB$ we get that $$\frac{EA}{\sin \angle{EBA}}=\frac{BE}{\sin \angle{A}},\quad \frac{BE}{\sin \angle{C}}=\frac{CE}{\sin \angle{EBC}}$$ If we multiply those two and since $EC=EA$ we get that $$\frac{\sin \angle{EBA}}{\sin \angle{EBC}}=\frac{\sin \angle{A}}{\sin \angle{C}}$$ So combine those two equations we get that $$\frac{\sin {\angle{EBA}}}{\sin {\angle{EBC}}}\cdot\frac{\sin {\angle{FCB}}}{\sin {\angle{FCA}}}=\frac{\sin \angle{A}}{\sin \angle{C}}\cdot \frac{\sin \angle{90-B}}{\sin \angle{90-A}}=\frac{\sin \angle{A}}{\sin \angle{C}}\cdot \frac{\cos \angle{B}}{\cos \angle{A}}=\frac{\tan \angle{A}\cdot \cos \angle{b}}{\sin \angle{C}}$$ So from Ceva's theorem about trigometry we get that the altitude through C, the median through B and the internal angle bisector through A are concurrent if $$\frac{\sin {\angle{EBA}}}{\sin {\angle{EBC}}}\cdot \frac{\sin {\angle{DAC}}}{\sin {\angle{FCA}}}\cdot\frac{\sin {\angle{FCB}}}{\sin {\angle{DAB}}}=1$$ So from our calculating $$\frac{\sin {\angle{EBA}}}{\sin {\angle{EBC}}}\cdot \frac{\sin {\angle{DAC}}}{\sin {\angle{FCA}}}\cdot\frac{\sin {\angle{FCB}}}{\sin {\angle{DAB}}}=1$$ iff $\tan \angle{A}\cdot \cos \angle{b}=\sin \angle{C}$