# Thread: [SOLVED] Triangle &amp; vectors

1. ## [SOLVED] Triangle &amp; vectors

In triangle ABC there is point D on line segment AB, so that $|AD|:|DB|=2:1$, and there is point E on BC, so that $|BE|:|EC|=1:4$. Point F is the point where line segments AE and DC intersect. Express the relationship (ratio) between $|AF|:|FE|$ ( $\vec{AB}=\vec{a}$, $\vec{BC}=\vec{b}$).

Exoneration: I'm reposting my thread, as it seems less precarious than bumping the old one. I also can't come up with any new idea.

2. Originally Posted by courteous
In triangle ABC there is point D on line segment AB, so that $|AD|:|DB|=2:1$, and there is point E on BC, so that $|BE|:|EC|=1:4$. Point F is the point where line segments AE and DC intersect. Express the relationship (ratio) between $|AF|:|FE|$ ( $\vec{AB}=\vec{a}$, $\vec{BC}=\vec{b}$).

Exoneration: I'm reposting my thread, as it seems less precarious than bumping the old one. I also can't come up with any new idea.
According to your sketch you get:

$\overrightarrow{AE}=\vec a + \frac15 \cdot \vec b$ and

$\overrightarrow{CD}=-\vec b - \frac13 \cdot \vec a$

Now let $\overrightarrow{AF}= k\cdot \overrightarrow{AE} = k \cdot \left( \vec a + \frac15 \cdot \vec b \right)$ and

$\overrightarrow{CF}= r\cdot \overrightarrow{CD} = r \cdot \left( -\vec b - \frac13 \cdot \vec a \right)$ ............. then the vector $\overrightarrow{AF}$ could be expressed by:

$\overrightarrow{AF}=\vec a + \vec b + r\cdot \overrightarrow{CD} =\vec a + \vec b + r \cdot \left( -\vec b - \frac13 \cdot \vec a \right)$

You'll get the equation:

$k \cdot \left( \vec a + \frac15 \cdot \vec b \right) = \vec a + \vec b + r \cdot \left( -\vec b - \frac13 \cdot \vec a \right)$ ............. Expand the brackets and collect like terms:

$k\cdot \vec a + \frac15 k \vec b = \left(1-\frac13r\right)\vec a + (1-r) \vec b$ ............. From this equation you can extract a system of simultaneous equations:
$\left| \begin{array}{lcr}k&=&1-\dfrac13 r \\\dfrac15 k&=&1-r\end{array}\right.$ ............. I've got $k=\frac57~\wedge~r=\frac67$

Therefore

$\overrightarrow{AF} = \frac57 \overrightarrow{AE}$ and

$\overrightarrow{FE} = \frac27 \overrightarrow{AE}$

The ratio $\frac{\overrightarrow{AF}}{\overrightarrow{FE}}= \frac{\frac57 \overrightarrow{AE}}{\frac27 \overrightarrow{AE}} = \boxed{\bold{\frac52}}$