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#### Pranav

##### Well-known member

- Nov 4, 2013

- 428

The median AD of the $\Delta$ ABC is bisected at E. BE meets AC in F. Find AF:AC.

Attempt:

Let point E divide BF in the ratio $\mu : 1$ and let F divide the line AC in the ratio $\lambda : 1$.

I take A as the origin. Then,

$$\vec{AD}=\frac{1}{2}(\vec{AB}+\vec{AC})$$

$$\vec{AF}=\frac{\lambda \vec{AC}}{\lambda+1}$$

$$\vec{AE}=\frac{\mu\vec{AF}+\vec{AB}}{\mu+1}$$

Also,

$$\vec{AE}=\frac{1}{2}\vec{AD}=\frac{1}{4}(\vec{AB}+\vec{AC})$$

I can substitute the above in the 3rd equation but I don't see what to do with $\vec{AB}$ as I have to find the ratio AF:AC.

Any help is appreciated. Thanks!