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 $AE:AF = 2:1$

Question

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 :-

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.

The figure dosen't matter by the way, I know $AB$ and $AC$ seems almost equal although they should be in $4:3$ ratio. — Anonymous, Oct 31 '20 at 11:29
@Anonymous - a small hint : $\Delta AEG$ and $\Delta AGF$ will have the same height, and so will $\Delta AMC$ and $\Delta AMB$. — Spectre, Oct 31 '20 at 11:37
@Spectre, I don't think what you are saying is right. The heights may not be same at all. — Anonymous, Oct 31 '20 at 11:44
@Anonymous, the heights are going to be the same, since triangles that lie between the same parallel lines, in spite of having different base lengths, will have the same height. — Spectre, Oct 31 '20 at 13:37
@Spectre I get it, you said that with respect to $EF$ and $BC$ as bases, I thought about $AG$ and $AM$ as bases. — Anonymous, Oct 31 '20 at 13:38

nonuser · Accepted Answer · 2020-10-31T13:46:59.497

4

Rename the points like on a picture: $AE:AD =2:1$

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 $x=54$.

edited Oct 31 '20 at 13:46

answered Oct 31 '20 at 11:57

nonuser

91,557

Right, I understood, E is arbitrary in this solution. – cosmo5 Oct 31 '20 at 12:56
I suggest you not to accept the answer yet. Perhaps you can get better or shorther or... solution. – nonuser Oct 31 '20 at 13:46
This solution actually looks the easiest to me. – Anonymous Oct 31 '20 at 15:08

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 $AE:AF = 2:1$

1 Answers1