Ratios of lengths in a triangle

Question

Problem I'm trying to solve:

My Attempt:

To be honest, I don't have any clue on how to solve it and the method I'm about to give is severely overcomplicated. We first isolate triangle $\triangle BEG$ to get a triangle where $BF:EF=3:2$ and $BD:DG=2:3$ and project this triangle onto a cartesian plane with $EG$ as the base. Then let each of the points be given by the coordinates $E=(0,0)$, $G=(c,0)$, $B=(a,b)$. Then it follows that $D=(a+\frac{2}{5}(c-a), \frac{3}{5}b)$ and $F=(\frac{2}{5}a,\frac{2}{5}b)$. The line passing through the points $E,D$ is then given by $$y=\frac{\frac{3}{5}b}{a+\frac{2}{5}(c-a)}x=\frac{3b}{3a+2c}x$$ and the line going through the points $F,G$ is $$y=\frac{\frac{-2}{5}b}{c-\frac{2}{5}a}(x-c)=\frac{-2b}{5c-2a}(x-c)$$

From there it's just a matter of finding their points of intersection and determine each length and their ratio however this is very tedious and long.

I'v also tried other methods for example constucting lines parallel to the altitude of the base $EG$ crossing points $D$ and $F$, however I wasn't sure where to continue from there. I've also tried looking for similar triangles but that hasn't led me anywhere either.

Any suggestions?

P.S. The correct answer is apparently $9:10$.

Answer 1

Hint: Apply Menelaus's Theorem in $\triangle BDE$, where side $BD$ is extended to $G$ and $F,H,G$ are collinear.

Answer 2

Rough sketch. Let $\vec a = \vec{BA}$ and $\vec b = \vec{BC}$. We have $$\vec 0 = \vec{EH}+\vec{HF}+\vec{FE}.$$ Now $$\begin{align} \vec{EH}&=x\vec{ED}=x\left(-\frac56\vec a+\frac29\vec b \right),\\ \vec{HF}&=y\vec{GF}=y\left(-\frac59\vec b+\frac12\vec a \right),\\ \vec{FE}&=\frac13\vec a.\\ \end{align} $$ Substitute these in $\vec 0 = \vec{EH}+\vec{HF}+\vec{FE}$, sort for $\vec a$ and $\vec b$ and use that $\vec a$ and $\vec b$ are linearly independent to get the linear system $$\begin{align}-\frac56x+\frac12y&=-\frac13\\ \frac29x-\frac59y&=0 \end{align}$$ from which $x=10/19$ follows.

Answer 3

If you don't know Menelaus' Theorem, then notice that the problem statement places no constraints on the triangle ABC, and the choices imply that there aren't any.

Make $B$ a right angle, set $BF, FE = 3, 2$ and $BD, DG = 2, 3$, and then finding the intersection point H is easy.

Find the projection $x$ Along $EB$: $$\frac{2x}{5} = \frac{5(x-2)}{3}$$ $$x = \frac{50}{19}$$ $$\frac{DH}{HE} = \frac{5-x}{x} = \frac{9}{10}$$

If you want to derive the freedom to specify these values without using the problem statement, then you just have to know that all affine transformations preserve the ratio of parallel line segments.