Related to concurrence of lines in triangles

Question

In triangle $ABC$, the straight lines $AD, BE, CF$ are drawn through a point $P$ to meet $BC, CA, AB$ at $D, E, F$ respectively. Prove that $PD/AD + PE/BE + PF/CF = 1$ and $AP/AD + BP/BE + CP/CF= 2.$

Answer 1

If a parallel to $BC$ is drawn from $P$, we can state that it cuts the perpendicular to $BC$ in ratio of $PD/AD$. This implies
$$\frac{\frac12BC\cdot PD}{\frac12 BC\cdot AD}=\frac{PD}{AD}=\frac{[BPC]}{[ABC]} \tag{1}$$

Similarly, $$ \frac{PF}{CF} =\frac{[APB]}{[ABC]} \qquad\qquad \frac{PE}{BE} =\frac{[APC]}{[ABC]} \tag{2} $$

Therefore, $$\frac{PD}{AD}+\frac{PE}{BE}+\frac{PF}{CF} = \frac{[BPC]+[APC]+[APB]}{[ABC]} =\frac{[ABC]}{[ABC]}= 1 \tag{3}$$

Also, $$\frac{AP}{AD}=\frac{[ABP]}{[ABD]}=\frac{[APC]}{[APD]}= \frac{[ABP]+[APC]}{[ABD]+[APD]}=\frac{[ABP]+[APC]}{[ABC]}\tag{4}$$ (because perpendicular from $AD$ and $AP$ to $B$ will be same and that to $C$ will be same)

Similarly, $$\frac{BP}{BE}=\frac{[BPC]+[BPA]}{[ABC]} \qquad\qquad \frac{CP}{CF}=\frac{[CPA]+[CPB]}{[ABC]} \tag{5} $$

Therefore, $$\begin{align} \frac{AP}{AD}+\frac{BP}{BE}+\frac{CP}{CF} &=\frac{[APB]+[APC]+[BPC]+[BPA]+[CPA]+[CPB]}{[ABC]} \\[4pt] &=\frac{2\cdot[ABC]}{[ABC]} \\[4pt] &=2 \tag{6} \end{align}$$