$M, N$ and $P$ are respectively the midpoint of $BC$, the centroid of $\triangle ABC$ and a point on $CA$ such that $NP \parallel BC$. A line passing through point $B$ intersects $AM$ and $PM$ respectively at $D$ and $E$ such that $BD = 5DE$. Calculate the value of $\dfrac{AN}{AD}$ and $\dfrac{ME}{MP}$.
We have that $BD = 5DE \implies 5\overrightarrow{BE} = 6\overrightarrow{BD} \implies 5 \cdot \left(\overrightarrow{BM} - \overrightarrow{EM}\right) = 6 \cdot \left(\overrightarrow{BM} - \overrightarrow{DM}\right)$
$\implies \overrightarrow{BM} = 6\overrightarrow{DM} - 5\overrightarrow{EM} \implies - \dfrac{1}{2}\overrightarrow{AB} + \dfrac{1}{2}\overrightarrow{AC} = 6\overrightarrow{DM} - 5\overrightarrow{EM}$
Let $\dfrac{\overrightarrow{DM}}{\overrightarrow{AM}} = x$ and $\dfrac{\overrightarrow{EM}}{\overrightarrow{PM}} = y \implies - \dfrac{1}{2}\overrightarrow{AB} + \dfrac{1}{2}\overrightarrow{AC} = 6x\overrightarrow{AM} - 5y\overrightarrow{PM}$
Furthermore, we have that $\overrightarrow{AM} = \dfrac{1}{2}\overrightarrow{AB} + \dfrac{1}{2}\overrightarrow{AC}$
and $\overrightarrow{PM} = \dfrac{1}{3}\cdot \left(\overrightarrow{PA} + \overrightarrow{AM}\right) + \dfrac{2}{3} \cdot \left(\overrightarrow{PC} + \overrightarrow{CM}\right)$
$ = - \dfrac{2}{9}\overrightarrow{AC} + \left(\dfrac{1}{6}\overrightarrow{AB} + \dfrac{1}{6}\overrightarrow{AC}\right) + \dfrac{2}{9}\overrightarrow{AC} + \left(\dfrac{2}{3}\overrightarrow{AB} - \dfrac{2}{3}\overrightarrow{AC}\right) = \dfrac{5}{6}\overrightarrow{AB} - \dfrac{1}{2}\overrightarrow{AC}$
So we have the solve the following equation $$- \dfrac{1}{2}\overrightarrow{AB} + \dfrac{1}{2}\overrightarrow{AC} = 6x \cdot \left(\dfrac{1}{2}\overrightarrow{AB} + \dfrac{1}{2}\overrightarrow{AC}\right) - 5y \cdot \left(\dfrac{5}{6}\overrightarrow{AB} - \dfrac{1}{2}\overrightarrow{AC}\right)$$
$\left\{ \begin{align} -\dfrac{1}{2} = 3x - \dfrac{25}{6}y\\ \dfrac{1}{2} = 3x + \dfrac{5}{2}y \end{align} \right. \implies (x, y) = \left(\dfrac{1}{24}, \dfrac{3}{20}\right) \implies \dfrac{AN}{AD} = \dfrac{16}{23}$, $\dfrac{ME}{MP} = \dfrac{3}{20}$
I wrote this solution almost at midnight and did all the calculations by hand so there might have been multiple mistakes. Please help me find them.
I will go an other way, it is the simplest way to check. We try to apply the theorems of Ceva and/or Menelaus. The situation is as follows.
Let $Q$ be the mid point of $AC$. Let $Z$ be the intersection of the parallels
so that $ABCZ$ is a parallelogram. $Q$ is the mid point of the one diagonal $AC$, then also the mid point of the other diagonal, and the centroid $N$ is on the median $BQ$. So $B,N,Q,Z$ colinear.
The point $Z$ is also on $MP$, because of the match of the proportion $$ \frac{NP}{AZ}= \frac{NP}{BC}= \frac{QN}{QB}= \frac13= \frac{MN}{MA}\ . $$
$BDE$ with $AC$. Then using Menelaus for $\Delta EBZ$, intersected by the line $NDM$, we get: $$ 1 = \frac{DB}{DE}\cdot \frac{ME}{MZ}\cdot \frac{NZ}{NB} = \frac{5}{1}\cdot \frac{ME}{MZ}\cdot \frac{2}{1} \ . $$ This leads to $$ \frac{ME}{MP}= \frac{ME}{MZ}\cdot \frac{MZ}{MP} =\frac 1{2\cdot 5}\cdot\frac31 =\color{blue}{\frac 3{10}}\ . $$ We can use this or independently "do the same", and apply Menelaus for $\Delta DBN$, intersected by the line $MEZ$, when we get: $$ 1 = \frac{ED}{EB}\cdot \frac{ZB}{ZN}\cdot \frac{MN}{MD} = \frac{1}{6}\cdot \frac{3}{2}\cdot \frac{MN}{MD} = \frac{1}{4}\cdot \frac{MN}{MD} \ . $$ So $\displaystyle MD=\frac 14 MN=\frac 14 \cdot \frac 13 AD=\frac 1{12}AD$. This leads to $$ \frac{AN}{AD}= \frac{2/3}{11/12}= \color{blue}{\frac{24}{33}} \ . $$
$\square$
I wrote this solution hours after mid night, it is 02:15 in Germany, hope things are all right, and easy to check in case errors came in. At any rate, $MD:MN=1:4$, and $ME:MZ=1:10$. One last check is done by using the theorem of Menelaus in $\Delta MNZ$ w.r.t. the line $BDE$, and indeed, $\displaystyle \frac{DN}{DM}\cdot \frac{EM}{EZ}\cdot \frac{BZ}{BN} = \frac31\cdot \frac19\cdot \frac31 =1$ .