# Finding the ratio of a side of $\triangle ABC$ and its segment where one cevian line from the opposite vertex intersect the side in any point

In $$\triangle ABC$$, $$L$$ and $$M$$ are two points on $$AB$$ and $$AC$$ such that $$AL = \frac{2AB}{5}$$ and $$AM = \frac{3AC}{4}$$. $$BM$$ and $$CL$$ intersect at the point $$P$$ and the extension line of $$AP$$ and the side $$BC$$ intersect $$BC$$ at the point $$N$$. What is the value of $$\frac{BN}{BC}$$?

My Attempt:

Let denote the area of $$\triangle ALP$$, $$[ALP] = x$$ and $$[APM] = y$$.

So, from $$\triangle APB$$, showing the relation of the base of both the triangle $$\triangle APL$$ and $$\triangle LPB$$, I got $$[LPB] = \frac{3x}{2}$$.

Similarly, from $$\triangle APC$$, doing the above likewise approach, I got $$[MPC] = \frac{y}{3}$$. After that, getting two triangle $$\triangle ACL$$ and $$\triangle BCL$$ and showing their relation of area with their particular base, I got

$$[BPC] = \frac{5x}{6}$$....(given that $$AL:AB =2:5$$)

Again from $$\triangle ABM$$ and $$\triangle CBM$$,

$$[BPC] = 2y$$.....($$AM:AC = 3:4$$)

So, $$2y = \frac{5x}{6}$$ $$\implies$$ $$x =\frac{12y}{5}$$

And then expressing the area of all the triangle by $$y$$, I got $$[ABC] = \frac{28y}{3}$$

But, I can't anyhow relate the area of $$\triangle BPN$$ with $$y$$. Here, I got stuck. What should I do to find out the area of $$\triangle BPN$$. I need really some help Thank you.

• I think you can directly get your result using Ceva's Theorem. Have you tried? – Matteo Feb 20 at 18:20
• @Matteo Nope. I haven't tried. How can I do that? – Anirban Niloy Feb 20 at 18:29
• have a look here: en.wikipedia.org/wiki/Ceva%27s_theorem – Matteo Feb 20 at 18:31
• Good! By the way, if I remember correctly, the simplest demonstration of the theorem proceeds, as you were doing, equating areas. – Matteo Feb 20 at 18:47
• Yes, it could be intersting to get to the result independently of Ceva's Theorem, just for didactic purposes... – Matteo Feb 20 at 19:22

$$\frac{BN}{NC}\cdot\frac{CM}{MA}\cdot\frac{AL}{LB}=\frac{S_{\Delta PBN}}{S_{\Delta PCN}}\cdot\frac{S_{\Delta PCM}}{S_{\Delta PAM}}\cdot\frac{S_{\Delta PAL}}{S_{\Delta PBL}}=$$ $$=\frac{\frac{1}{2}PB\cdot PN\sin\measuredangle BPN}{\frac{1}{2}PC\cdot PN\sin\measuredangle CPN}\cdot\frac{\frac{1}{2}PC\cdot PM\sin\measuredangle CPM}{\frac{1}{2}PA\cdot PM\sin\measuredangle APM}\cdot\frac{\frac{1}{2}PA\cdot PL\sin\measuredangle APL}{\frac{1}{2}PB\cdot PL\sin\measuredangle BPL}=1.$$ Thus, $$\frac{BN}{NC}\cdot\frac{1}{3}\cdot\frac{2}{3}=1.$$ Can you end it now?
I got $$\frac{BN}{BC}=\frac{9}{11}.$$
Hint: Consider using Ceva's Theorem. Since the lines $$AN, CL$$ and $$BM$$ concur at $$P$$ $$\frac{CM}{MA}·\frac{AL}{LB}·\frac{BN}{NC}=\frac{1}{3}·\frac{2}{3}·\frac{BN}{NC}=1$$ Thus
$$\frac{BN}{NC}=4,5\iff \frac{BN}{NC}=\frac{4,5}{4,5+1}=\frac{9}{11}$$