Triangle with two medians

Question

I need help with this proof:

AD and BE are medians in triangle ABC, meeting at point F. Point G is the middle of segment BF. The continuation of the segment AG intersects the side BC at point H. Prove that BC=5BH.

Here's what I could find:

• AD=DC
• BE=EC
• BG=GF=FE
• ED || AB

However I don't see how these facts help me prove the claim. Any help would be appreciated.

Answer 1

Hint: Line $AH$ is a transversal of $\triangle BCE.$ Now apply Menelaus' Theorem.

$$\frac{CA}{EA}\cdot \frac{EG}{GB}\cdot \frac{BH}{HC}=1\implies 2\cdot 2\cdot \frac{BH}{BC-BH}=1\implies 5\cdot BH=BC$$

Answer 2

Here is a solution using the fact that the cross-ratio of four aligned points is preserved under a perspective (one of the fundamental properties of "projective geometry"), here the perspective sending $B \to B, G \to H, F \to D, E \to C$ :

$$\underbrace{\frac{BE}{BG}\times\frac{FG}{FE}}_{= 3 \times 1}=\frac{BC}{BH}\times\frac{DH}{DC}\tag{1}$$

(The value of the LHS is a consequence of relationship $BG=GF=FE$ you have remarked).

Let us take coordinates on line $BC$ such that :

$$ B=0, \ \ H=h, \ \ D=1, \ \ C=2.\tag{2}$$

Therefore, (1) becomes :

$$3=\frac{2}{h}\frac{1 - h}{1}$$

a first degree equation giving $h=\frac25$

As a consequence, using (2) again :

$$\frac{BC}{BH}=\frac{2}{h}=5.$$

Answer 3

Notice: points are renamed.

Considering figure:

$FG=\frac12 AG$

and $HI||GO||FN\Rightarrow AI=IO=ON$

Also:

$EG=\frac12 GB$

and $EO||GN||LM$

where L is mid point of GB, so we have:

$ON=NM=MB$

Therfore:

$AI=IO=ON=NM=MB$

that is $AI=\frac15 AB$