Triangle Formed By Intercepts

Our aim is to find $\dfrac{NM}{BP}$ . Once we do that, we are done.How it will be clear later.

Apply Menelaus Theorem in $\triangle BPC$ with $AME$ as transversal. We have $\dfrac{CA}{AP} \times \dfrac{PM}{MB} \times \dfrac{BE}{EC} = 1$ .

Putting $BE = EM$ and $\dfrac{CA}{AP} = \dfrac{15}{6} = \dfrac{5}{2}$ , we have $\dfrac{BM}{MP} = \dfrac{5}{2}$ ..... (i) As $AD$ is the angle bisector of $\angle A$ , we have $\dfrac{BN}{NP} = \dfrac{10}{6}$ .......(ii)

Now assume $BN=x$ , $NM=y$ and $MP=z$ .

From (i) we have $\dfrac{x+y}{z} = \dfrac{5}{2} \implies 2x=5z-2y...(v) \implies 6x=15z-6y....(iii)$

From (ii) we have $\dfrac{x}{y+z} = \dfrac{10}{6} \implies 6x=10Y + 10z.....(iv)$

Equating (iii) and (iv) we have $10z+10y = 15z-6y \implies 5z=16y=k$ . hence $z=\dfrac{k}{5},$ $y=\dfrac{k}{16}$ .

Putting these values of $y$ and $z$ in (v) we have $x=\dfrac{7k}{16}.$ Hence $\large{x:y:z=35:5:16}$

Now $\dfrac{[ANM]}{[ABP]} = \dfrac{NM}{BP} = \dfrac{y}{x+y+z} = \dfrac{5}{35+5+16} = \dfrac{5}{56}$ . [ when two triangles have a common vertex, ratio of their areas is equal to ratio of their bases]

Hence $[ANM] = \dfrac{5}{56} \times [ABP]$ .

Again, $\dfrac{[ABP]}{[ABC]} = \dfrac{AP}{AC} = \dfrac{2}{5} \implies [ABP] = \dfrac{2}{5} \times [ABC]$

therefore, $[ANM] = \dfrac{5}{56} \times \dfrac{2}{5} \times [ABC] = \dfrac{1}{28} \times [ABC]$

Hence, $\dfrac{[ABC]}{[ANM]} = \boxed{28}$

Applying Menelaus theorem in triangle AEC, with transversal PMB, you'll get $\frac{AM}{ME}=\frac{4}{3}$ Further, applying Menaluas in ADE with MNB transversal, we get $\frac{AN}{ND}=\frac{5}{3}$ The internal bisector theorem says that $\frac{AB}{AC}=\frac{BD}{DC}=\frac{2}{3}$ , thus $BD:DE:EC::4:1:5$ (since E is the midpoint of BC) Thus $\frac{[ANB]}{[NBD]}=\frac{5}{3}$ Let it be 5x and 3x. Thus [ADB]=8x. Thus [ADE]= 2x (since, BD:DE=4:1) now $\frac{[ABM]}{[ABE]}=\frac{MA}{AE}=\frac{4}{7}=\frac{(5x + [ANM])}{(8x+2x)}$ . Thus $[ANM]= \frac{5x}{7}$ and $[ABC] = 2\times[ABD]=2\times10x = 20x$ Thus $\frac{[ABC]}{[ANM]}=\frac{20x}{\frac{5x}{7}}=28$

APB = $\frac { 2 }{ 5 }$ ABC

After much Calculation I Got AMB = $\frac { 2 }{ 7 }$ ABC

Also ANP = $\frac { 3 }{ 20 }$ ABC

So ANP + AMB - APB =( $\frac { 2 }{ 7 }$ + $\frac { 3 }{ 20 }$ ) ABC - $\frac { 2 }{ 5 }$ ABC

= ANM = $\frac { 5 }{ 140 }$ ABC

= $\frac { ABC }{ ANM }$ = $\frac { 140 }{ 5 }$

                                            = 28

apply Menalaus theorem 3 times to 3 triangles AEC with transversal MBP, ADC with transversal NBP and lastly, ANM with transversal BDE we can also find MN/BD. Note that suppose K is between A &B then ìf AK/KB=m then Ak/KB=m/m+1 and KB/AB=1/m+1 for faster calculation and to be beautiful geomatrics :D