<s>[INST] Find the area of ​region triangular BPM For reference: In triangle $ABC$, the external angle bisector $BE$ and the median $AM$ intersect at $P$. Determine the area of ​​triangular region $BPM$; if $AB =3BC$ and $S_{ABC}=40\ \mathrm{m^2}$.

My progress:
$S_{ABM}=S_{AMC}=\frac{40}{2} = 20\\
\frac{S_{ABP}}{S_{APE}}=\frac{BP}{PE}\\
\frac{S_{BPM}}{S_{BCE}}=\frac{BM.BP}{BC.BE}=\frac{BP}{2BE}\\
\frac{S_{ABM}}{S_{ABP}}=\frac{AB.AM}{AB.AP}\implies \frac{20}{S_{ABP}} = \frac{AM}{AP}\\
\frac{S_{ABC}}{S_{BCE}}=\frac{AC}{CE}$
...?
 [/INST] $S_{\triangle MBP} = \frac 12 \cdot \frac{BC}{2} \cdot BP \sin{(90^\circ - \frac{\angle B}{2})} = \frac{BC \cdot BP}{4} \cos (\frac{\angle B}{2})$
$S_{\triangle ABP} = \frac1 2 \cdot 3 BC \cdot BP \sin{(90^\circ + \frac{\angle B}{2})} = 6 \cdot  S_{\triangle MBP}$
$S_{\triangle ABM} = S_{\triangle ABP} - S_{\triangle MBP} = 5 \cdot S_{\triangle MBP} = 20$
$S_{\triangle MBP} = 4$

Or applying external angle bisector theorem in $\triangle ABM$,
$\frac{AB}{BM} = \frac{AP}{MP} \implies AP = 6 \cdot MP$ or $AM = 5 \cdot MP$
$S_{\triangle ABM} = 5 S_{\triangle MBP} = 20$
$ \therefore S_{\triangle MBP} = 4$
</s>