Geometry question on triangles involving ratio of lines drawn from vertices to any arbitrary point P in the interior of the triangle.

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 $$\frac{PD}{AD} + \frac{PE}{BE} + \frac{PF}{CF} = 1$$ and $$\frac{AP}{AD} + \frac{BP}{BE} + \frac{CP}{CF} = 2$$

I tried using the following lemma - "If the diagonals of a quadrilateral ABCD meet at O. then $\frac{\triangle ABC}{\triangle ADC} = \frac{BO}{OD}$" (To clarify, $\triangle ABC$ means area of the triangle)

I applied this to quadrilaterals ACDF, ABDE, and BCEF. Thus I got $\frac{PD}{AD} = \frac{\triangle FDC}{ACDF}, \frac{PE}{BE} = \frac{\triangle AED}{ABDE}, \frac{PF}{CF} = \frac{\triangle BEF}{BCEF}$. After this, I am unable to proceed further. Also tried something similar to the second part of the question, but to no avail. Any help is appreciated.

Answer 1

premise: It is a good idea to start from the ratio of areas to arrive at a ratio of lengths. Now because appears the lines $AD,BE,CF$ which goes through all the triangle $ABC$, I think it's reasonable to use its area throughout the proof.

proof: you immediately see that: $$\frac{\triangle ABP}{\triangle ABC}+ \frac{\triangle ACP}{\triangle ABC}+\frac{\triangle CBP}{\triangle ABC}=\frac{\triangle ABC}{\triangle ABC}=1.$$ Now we just have to prove that $\frac{\triangle ABP}{\triangle ABC}=\frac{PF}{CF},\; \frac{\triangle ACP}{\triangle ABC}=\frac{PE}{BE},\;\frac{\triangle CBP}{\triangle ABC}=\frac{PD}{AD}$ to get the thesis, i'll do only the first one (the other are analogous).

Let's trace heights of $ABP$,$ABC$ that passes through $P$, respectively $MP,\, MN$. Now because these triangles have the same base $AB$: $$\frac{\triangle ABP}{\triangle ABC}=\frac{MP}{MN}, $$ now by the Intercept theorem $$\frac{MP}{MN}=\frac{PF}{CF}, $$ and we are done (or you can also see the last equality by proving that triangles $MPF,\, NPC$ are similar by the $aa$ criterior).

second part: its obvious that: $$\frac{AD}{AD}+\frac{BE}{BE}+\frac{CF}{CF}=3, $$ and by subtracting the last equation $$\frac{PD}{AD}+\frac{PE}{BE}+\frac{PF}{CF}=1, $$ we get: $$2=\frac{AD-PD}{AD}+\frac{BE-PE}{BE}+\frac{CF-PF}{CF}=$$ $$=\frac{AP}{AD}+\frac{BP}{BE}+\frac{CP}{CF}$$

Answer 2

HINT.-(A way without areas) Stewart's theorem applied to triangles $\triangle{ABD},\space\triangle{ECB},\space\triangle{FBC}$ gives the three equations in six unknowns $X_1=\frac{PD}{AD},\space X_2= \frac{PE}{BE},\space X_3= \frac{PF}{CF},\space Y_1=\frac{AP}{AD} ,Y_2=\space \frac{BP}{BE},\space Y_3= \frac{CP}{CF}$. $$BD^2X_1+BA^2Y_1=BP^2+AP\cdot PD\\BC^2X_2+EC^2Y_2=CP^2+BP\cdot PE\\BC^2X_3+BF^2Y_3=BP^2+CP\cdot PF$$ It remains the data of three adicional equations in order to have a system with only one solution which are the obvious $$X_1+Y_1=1\\X_2+Y_2=1\\X_3+Y_3=1$$ In fact we have three, easier systems of two equation of two unknowns, one of them being, for example $$\begin{cases}BD^2X_1+BA^2Y_1=BP^2+AP\cdot PD\\X_1+Y_1=1\end{cases}$$ Solving, one have just to verified the proposed equalities for triangle $\triangle{ABC}$.