In the triangle ABC, take A as the origin, and write b = overrightarrowAB and c = overrightarrowAC…

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Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

In the triangle $ABC$ , take $A$ as the origin, and write $\vec b = \overrightarrow{AB}$ and $\vec c = \overrightarrow{AC}$ . The point $D$ lies on $BC$ with $BD : DC = 2 : 3$ , and $E$ is the middle of $AD$ . The line $BE$ is extended until it meets $AC$ at the point $F$ . Find $\overrightarrow{AD}$ and $\overrightarrow{AE}$ in terms of $\vec b$ and $\vec c$ . Then find the ratio $AF : FC$ .

Need a reminder? The key ideaShow

If $AP : PB = m : n$ , then $\vec p = \dfrac{n\vec a + m\vec b}{m + n}$ . For an external division use $\dfrac{m\vec b - n\vec a}{m - n}$ . The centroid of a triangle is $\dfrac{\vec a + \vec b + \vec c}{3}$ , and three points are collinear when $\overrightarrow{AC}$ is a number times $\overrightarrow{AB}$ .

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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