Calculate the value of $\frac{AN}{AD}$ and $\frac{ME}{MP}$. 
$M, N$ 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.
 A: 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 


*

*to $BC$ through $A$, 

*to $AB$ through $C$, 


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,
$\displaystyle
\frac{DN}{DM}\cdot
\frac{EM}{EZ}\cdot
\frac{BZ}{BN}
=
\frac31\cdot
\frac19\cdot
\frac31
=1$ .
