Triangles and Vectors.

Question

In triangle $ABC$, $D$ is the point on $BC$ such that $AD$ bisects $∡CAB$ , and $M$ is the midpoint of $BC$. $E$ is the point on the extension of $BA$ such that $ME$ is parallel to $AD$ and intersects $AC$ at $F$.

Prove that $$|BE|=|CF|=\frac{|AB|+|AC|}{2}$$

I got the following equations : $v(BA)+v(AC)=v(BC)$ ---> equation 1

$v(BM)=v(BE)+v(EM)$ ---> equation 2

$v(CF)=v(CM)+v(MF)$ ---> equation 3

$v() =>$ notation for vector for e.g. $v(AB) => \vec{AB}$

I am not able to proceed further for the proof, please advice.

Answer 1

I'll help you prove @String's claim:

Firstly, notice that by angle chasing we have $|AE|=|AF|$ - use the fact that we have angle bisector and parallel lines.

Then use Menelaus' Theorem for $\triangle ABC$ and the line $EFM$:

$$\frac{|CF|}{|AF|}\frac{|AE|}{|EB|}\frac{|BM|}{|MC|}=1$$

Use that $M$ is the midpoint and the above to get that $|CF|=|BE|$

Now you have $|CF|=|BE|=|BA|+|AE|=|BA|+|AF|=|BA|+|AC|-|CF|$ which implies

$$|CF|=\frac{|BA|+|AC|}{2}$$

Note that for the last part we used the positioning of the points which requires a bit more reasoning since there is more than one possibility

Answer 2

In the standard notation by Thales we obtain: $$\frac{BE}{c}=\frac{\frac{a}{2}}{BD},$$ which gives $$BE=\frac{ac}{2BD}.$$ Also, by Thales again $$\frac{CF}{\frac{a}{2}}=\frac{b}{CD},$$ which gives $$CF=\frac{ab}{2CD}.$$ Thus, $BE=CF$ it's $$\frac{ac}{BD}=\frac{ab}{CD}$$ or $$\frac{c}{b}=\frac{BD}{CD},$$ which is true because $AD$ is a bisector of $\Delta ABC.$

Also, we need to prove that $$BE=\frac{b+c}{2}$$ or $$\frac{ac}{2BD}=\frac{b+c}{2}$$ or $$BD=\frac{ac}{b+c},$$ which is true because $AD$ is a bisector of $\Delta ABC.$