<s>[INST] Prove that $\frac{AB}{AE} + \frac{AD}{AG} = \frac{AC}{AF}$ in parallelogram $ABCD$, where $E$, $F$, $G$ are points on a line intersecting the sides 
Let $ABCD$ be a parallelogram. A line meets segments $AB$, $AC$, $AD$ at points $E$, $F$, $G$, respectively. Prove that $\frac{AB}{AE} + \frac{AD}{AG} = \frac{AC}{AF}$.

So recently I've been assigned a few problems, and this is one of them. So far, I've thought of extending line $EG$ and diagonal $BD$ to meet at a point, which we can call point $X$ and making point $O$ as the intersection of the diagonals in the paralellogram. And from here, I thought that maybe using Menelaus's theorem on triangle $BEX$ with line $AC$, which gets us $\frac{BA}{AE} \cdot \frac{EF}{FX} \cdot \frac{XO}{OB} = 1$. And similarly on triangle $DGX$ with line $AC$, we would get $\frac{DA}{AG} \cdot \frac{GF}{FX} \cdot \frac{XD}{OD} = 1$. But I'm not sure how to proceed from here and to relate these back to the original problems. Does anyone have any ideas on how I could do so?
 [/INST] Let $K\in DO$ and $M\in BO$ such that $GK||AO$ and $EM||AO.$
Also, let $AO=a$, $GK=x$, $FO=y$ and $EM=z$.
Let $x>z$ and $L\in GK$ such that $LKME$ be parallelogram, $LE\cap FO=\{N\}.$
Thus, $$\frac{x-y}{y-z}=\frac{x-z}{y-z}-1=\frac{EL}{EN}-1=\frac{KM}{OM}-1=\frac{KO+OM}{OM}-1=\frac{KO}{MO}=$$
$$=\frac{\frac{KO}{DO}}{\frac{MO}{DO}}=\frac{\frac{DO-DK}{DO}}{\frac{BO-BM}{BO}}=\frac{1-\frac{x}{a}}{1-\frac{z}{a}}=\frac{a-x}{a-z},$$
which gives $$(x+z-2y)a+xy+yz-2xz=0.$$
Now, we need to prove that: $$\frac{AB}{AE}+\frac{AD}{AG}=\frac{2AO}{AF}$$ or
$$\frac{AB}{AB-BE}+\frac{AD}{AD-DG}=\frac{2a}{a-y}$$ or
$$\frac{1}{1-\frac{BE}{AB}}+\frac{1}{1-\frac{DG}{AD}}=\frac{2a}{a-y}$$ or
$$\frac{1}{1-\frac{z}{a}}+\frac{1}{1-\frac{x}{a}}=\frac{2a}{a-y}$$ or
$$\frac{1}{a-z}+\frac{1}{a-x}=\frac{2}{a-y}$$ or
$$(x+z-2y)a+xy+yz-2xz=0$$ and we are done!
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