menelaus' theorem 2 In triangle ABC, AD is the angle bisector of $\angle A$ with E the midpoint of BC. Through E, draw a parallel to AD which intersects AC in G and the extension of AB at F. Show that BF = CG. 
So far from the angle bisector theorem, I have that $\frac{BD}{DC} = \frac{AB}{AC}$. And Since $AD\parallel FE$, I have that $\triangle AFG$ similar to $\triangle EGC$ so $\frac{AF}{EC}=\frac{FG}{GE}=\frac{AG}{GC}$. I did menelaus's theorem in $\triangle BCA$ with transversal E-F-G to obtain that $\frac{BE}{CE}=\frac{CG}{AG}=\frac{AF}{BF} = 1$. Since BE=CE the first fraction cancels to 1. 
I am stuck from this point. I would appreciate any assistance to finish this problem. 
 A: We have $\measuredangle AFC=\measuredangle BAD=\measuredangle DAC=\measuredangle EGC=\measuredangle AGF$, therefore the triangle AFG is isosceles so $AF=AG$. Moreover by Menelaus' theorem in ABC
$$\frac{AF}{FB}\frac{EB}{EC}\frac{GC}{GA}=\frac{AF}{GA}\frac{GC}{FB}=\frac{GC}{FB}=1$$
From which the statement follows.
A: Can we use trig?
Law of sin says:
$\frac {CG}{\sin CEG} = \frac {CE}{\sin CGE}$
and
$\frac {BE}{\sin BFE} = \frac {BF}{\sin BEF}$
$BE = CE$ (because E is the midpoint)
$\angle BAD = \angle BFE = \angle CAD = \angle CGE$ (parallel lines and angle bisectors)
$\frac {CE}{\sin CGE} = \frac {BE}{\sin BFE}$
hence:
$\frac {CG}{\sin CEG} = \frac {BF}{\sin BEF}$
Since $\angle CEG$ and $\angle BEF$ are supplementary the sines of their respective angles are equal.
$CG = BF$
No trig but more constructions...
Construct triangle $GEQ$ such that it is congruent to $GEC$ and reflected across $GE$
$FGQB$ is a parallelogram.
A: You can avoid Menelaus all together and use properties of parallel lines and midsegments only. 

Let point $B^*$ be the intersection point of $CA$ and the line parallel to $AD$ through vertex $B$. Then by $AD \, || \, BB^*$ $$\angle \, AB^*B = \angle \, CAD = \angle \, BAD = \angle \, ABB^*$$ which means that triangle $ABB^*$ is isosceles and $AB = AB^*$. Analogously, because $AD \, || \, EG$ one can prove that triangle $AFG$ is isosceles and $GA = FA$. Hence
$$GB^* = GA + AB^* = FA + AB = FB = BF$$
Finally, $E$ is the midpoint of $BC$ and $EG \, || \, BB^*$ so point $G$ is the midpoint of $CB^*$. Hence
$$CG = GB^* = BF$$ 
