图中显示了一个三角形 ABC。E 是 AC 的中点。$\overline{AB}=3a$ 且 $\overline{AC}=3c$。K 在 EB 上，使得 $EB=3EK$。

1. Express $\overrightarrow{BC}$ using the path $B \to A \to C$.

$\overrightarrow{BC}=\overrightarrow{BA}+\overrightarrow{AC}$
Since $\overrightarrow{AB}=3a$, we have $\overrightarrow{BA}=-3a$.
Also, $\overrightarrow{AC}=3c$.

Hence,
$\overrightarrow{BC}=-3a+3c=3c-3a.$

1. Find the vector from $E$ to $A$ using the midpoint information.

Since $E$ is the midpoint of $AC$,
$\overrightarrow{AE}=\dfrac{1}{2}\overrightarrow{AC}=\dfrac{1}{2}(3c)=\dfrac{3}{2}c$.

Hence
$\overrightarrow{EA}=-\dfrac{3}{2}c$.

2. Express $\overrightarrow{EB}$ by adding vectors along $E \to A \to B$.

$\overrightarrow{EB}=\overrightarrow{EA}+\overrightarrow{AB}$

$=-\dfrac{3}{2}c+3a$

$=3a-\dfrac{3}{2}c$

1. Find $\overrightarrow{AE}$ and $\overrightarrow{EK}$ using the midpoint and ratio information.

Since $E$ is the midpoint of $AC$,
$\overrightarrow{AE}=\frac{1}{2}\overrightarrow{AC}=\frac{1}{2}(3c)=\frac{3}{2}c.$
From part (b), $\overrightarrow{EB}=3a-\frac{3}{2}c$.

Since $EB=3EK$,
$\overrightarrow{EK}=\frac{1}{3}\overrightarrow{EB}=\frac{1}{3}\left(3a-\frac{3}{2}c\right)=a-\frac{1}{2}c.$

2. Express $\overrightarrow{AK}$ in terms of $a$ and $c$.

Using the path $A \to E \to K$,
$\overrightarrow{AK}=\overrightarrow{AE}+\overrightarrow{EK}.$
Thus
$\overrightarrow{AK}=\frac{3}{2}c+\left(a-\frac{1}{2}c\right)=a+c.$

3. Compare $\overrightarrow{KP}$ with $\overrightarrow{AK}$ and conclude collinearity.

Given
$\overrightarrow{KP}=\frac{1}{2}a+\frac{1}{2}c=\frac{1}{2}(a+c).$
Since $\overrightarrow{AK}=a+c$,
$\overrightarrow{KP}=\frac{1}{2}\overrightarrow{AK}.$
Therefore $\overrightarrow{KP}$ is a scalar multiple of $\overrightarrow{AK}$, so the two vectors are parallel. Since both involve the point $K$, $A$, $K$ and $P$ lie on a straight line.

1. Relate the areas using the common base $AK$ and the given division of $EB$.

Since $EB=3EK$,
$KB=EB-EK=3EK-EK=2EK.$

Triangles $AEK$ and $AKB$ have the same base $AK$, and $E$, $K$, $B$ are collinear, so their areas are in the ratio of their corresponding heights:
$\frac{\text{area of }\triangle AEK}{\text{area of }\triangle AKB}=\frac{EK}{KB}=\frac{EK}{2EK}=\frac{1}{2}.$

1. Compare the area of $\triangle AEB$ with the area of $\triangle ABC$, then use the subdivision of $\triangle AEB$ by $AK$.

Since $E$ is the midpoint of $AC$, $AE=\dfrac{1}{2}AC$.

$\triangle AEB$ and $\triangle ABC$ have bases $AE$ and $AC$ on the same straight line, and the same height from $B$.

So
$\frac{\text{area of }\triangle AEB}{\text{area of }\triangle ABC}=\frac{AE}{AC}=\frac{1}{2}.$

From part (d),
$\frac{\text{area of }\triangle AEK}{\text{area of }\triangle AKB}=\frac{1}{2},$
so if $\text{area of }\triangle AEK=x$, then $\text{area of }\triangle AKB=2x$.

Thus
$\text{area of }\triangle AEB=x+2x=3x,$
so
$\frac{\text{area of }\triangle AEK}{\text{area of }\triangle AEB}=\frac{x}{3x}=\frac{1}{3}.$

Therefore