In $\Delta ABC$, $AB:AC = 4:3$ and $M$ is the midpoint of $BC$ . $E$ is a point on $AB$ and $F$ is a point on $AC$ such that $AE:AF = 2:1$

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In $\Delta ABC$, $AB:AC = 4:3$ and $M$ is the midpoint of $BC$ . $E$ is a point on $AB$ and $F$ is a point on $AC$ such that $AE:AF = 2:1$. Also $EF$ and $AM$ intersect at $G$ with $GF = 36$ cm, $GE = x$ cm. Find $x$ .

What I Tried: Here is a picture :-
enter image description here

No idea for this. I am not getting any idea on how to use these facts together :-
$CM = MB$ and $AB:AC = 4:3$ and $AE:AF=2:1$ .

I don't see any similar triangles unfortunately. That's why I cannot proceed with this, angle-chasing is anyways not going to help. As some ideas I think I have to make an extra construction, but I don't know where. I also think areas can help, but did not find a way of it.

Can anyone help me? Thank You.

Best Answer

Rename the points like on a picture: $AE:AD =2:1$ enter image description here

Draw a parallel through $B$ to $DE$ then if $AB = 4y$ then $AF = 2y$ and $AC = 3y$ so $AF:CF = 2:1$.

Draw a parallel through $F$ to $AM$ then $CI:IM = 1:2$, so if $MB = 3z$ then $MI = 2z$.

So $IM:MB = 2:3$ and thus $FH:HB = 2:3$. But $DG:GE = FH:HB$ so $x=54$.

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