Given a triangle $ABC$ with $D,E,F$ respectively on $BC,CA,BA$. If $P$ is the intersection point of $AD$ and $EF$, is $(\frac{AB}{AF} \times DC) + (\frac{AC}{AE} \times DB) - (\frac{AD}{AP} \times BC)=0$?
If triangle $ABC$ is equilateral and $D,E,F$ respectively divide $BC,CA,BA$ into halves, I think it's easy to prove that statement is true. But what if the triangle is scalene and we don't know whether $D,E,F$ respectively divide $BC,CA,BA$ into halves?
Proving the given statement is equivalent to proving that $(\frac {FB}{AF}\times DC)+(\frac {CE}{AE}\times BD)=(\frac {PD}{AP}\times BC)$
Extend $EF$ to meet $BC$ at point $Q$. Applying Menelaus' Theorem on $\triangle ABD$ and $\triangle ADC$ considering $FEQ$ as the transversal gives,
$\frac{FB}{AF}\cdot \frac{AP}{PD}\cdot \frac{DQ}{BQ}=1$ and $\frac{CE}{AE}\cdot \frac{AP}{PD}\cdot \frac{DQ}{CQ}=1$
$(\frac {FB}{AF}\times DC)+(\frac {CE}{AE}\times BD)$
$=\frac{PD}{AP}\{\left(\frac{FB}{AF}\times \frac{AP}{PD}\times DC\right)+\left(\frac{CE}{AE}\times \frac{AP}{PD}\times BD\right)\}$
$=\frac{PD}{AP}\{\left(\frac{BQ\times DC}{DQ}\right)+\left(\frac{CQ\times BD}{DQ}\right)\}$
Now, $\left(BQ\times DC\right)+\left(CQ\times BD\right)$ can be written as,
$\left(BQ\times DC\right)+\left(BQ-BC\right)\left(BC-DC\right)$
This simplifies to $BC\cdot DQ$.
Hence, $\left(\frac {FB}{AF}\times DC\right)+\left(\frac {CE}{AE}\times BD\right)$
$=\frac{PD}{AP}\left\{\left(\frac{BQ\times DC}{DQ}\right)+\left(\frac{CQ\times BD}{DQ}\right)\right\}=\frac{PD}{AP}\times BC$