S A N D A , văd că nu vrei să socializam (!hic) // A Problem with the TRANSVERSAL Line // O Problema cu linia TRANSVERSALA

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          Let us consider the problem taken from page 384 :

                        "17948*.  An arbitrary point  $D$  is chosen on side  $BC$  of triangle 

                        $ABC$ , and an arbitrary point  $P$  is considered on segment  $AD.$ 

                  An arbitrary line passing through point  $P$  intersects sides  $AB\; and\; AC$

                  at points  $M \;and\; N.$  If we denote  $\frac{BD}{DC}=m\;\; and\;\;\frac{AP}{PD}=n$ ,

                  then the relation 

$\frac{MB}{MA}+m\cdot \frac{NC}{NA}=\frac{m+1}{n} \tag{E}$

                    holds.

I. Șiclovan, University Lecturer, Petroșani."

SOLUTION  CiP

          Let's use the notation :       $\frac{\overline{MB}}{\overline{MA}}=-k\;\;,\;\;\frac{\overline{NC}}{\overline{NA}}=-l\;\;\;\;\;\;(1)$

.

The relationship  (E)  to be proven is written as follows:

$\fbox{$k+m\cdot l=\frac{m+1}{n}$}$                                                       (2)

Using the formulas established here or here, it follows from  (1)  that :

$\overrightarrow{PM}=\frac{k}{1+k}\cdot \overrightarrow{PA}+\frac{1}{1+k}\cdot \overrightarrow{PB} \tag{3}$

$\overrightarrow{PN}=\frac{l}{1+l}\cdot \overrightarrow{PA}+\frac{1}{1+l}\cdot \overrightarrow{PC}\tag{4}$

     We will take the result  (1)  established here , writing

  $\overrightarrow{PC}=-\frac{1+m}{mn}\cdot \overrightarrow{PA}-\frac{n}{mn}\cdot \overrightarrow{PB}$

and then we can continue in  (4) :

 $\overrightarrow{PN}=\frac{l}{1+l}\cdot \overrightarrow{PA}+\frac{1}{1+l}\cdot \left (-\frac{1+m}{mn}\cdot \overrightarrow{PA}-\frac{1}{m}\cdot \overrightarrow{PB} \right)=\frac{lmn-1-m}{(1+l)mn}\cdot \overrightarrow{PA}-\frac{1}{(1+l)m}\cdot \overrightarrow{PB}$

     Since the points  $M,\; P\; and \;N$  are collinear, we have, for a certain scalar  $\lambda$

 $\overrightarrow{PN}=\lambda \cdot \overrightarrow{PM}$

And since the vectors  $\overrightarrow{PA}\; and \;\overrightarrow{PB}$  are linearly independent, the equality of the scalars follows from the  (3)  and the final form of  (4) :

$\begin{cases}\frac{lmn-1-m}{(1+l)mn}=\frac{\lambda\cdot k}{1+k}\\-\frac{1}{(1+l)m}=\frac{\lambda}{1+k}\end{cases}$

From the above, eliminating  $\lambda$  yields

$\frac{lmn-1-m}{(1+l)mn}=k\cdot \left (-\frac{1}{(1+l)m}\right )\Rightarrow lmn-1-m=-kn\Rightarrow nk+lmn=m+1\Rightarrow$

$\Rightarrow k+lm=\frac{m+1}{n}.$

We obtained  (2).

$\blacksquare\;\;\;\;\;\blacksquare$