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?
Best Answer
$\color{blue}{\text{It is not necessary to use vectors, indeed it is possible}\\\text{to get a proof by applying Thales’ Theorem and}\\\text{Menelaus’s Theorem.}}$
Theorem:
If $\;ABCD\;$ is a parallelogram and a straight line $\;r\;$ meets the segments $\;AB$, $\;AC$, $\;AD\;$ respectively at the points $\;E$, $\;F$, $G\;,\;$ then
$\cfrac{AB}{AE}+\cfrac{AD}{AG}=\cfrac{AC}{AF}$.
Proof:
Let $\;O\;$ the intersection point of the diagonals of the parallelogram.
There are two possibilities:
$1)\quad r\parallel BD$
In this case, by applying Thales’ Theorem to the parallel lines $\;r\;$ and $\;BD\;$ cutting $\;AB\;$ and $\;AC\;,\;$ we get that
$\cfrac{AB}{AE}=\cfrac{AO}{AF}\;.\quad\color{blue}{(*)}$
Analogously, by applying Thales’ Theorem to the parallel lines $\;r\;$ and $\;BD\;$ cutting $\;AD\;$ and $\;AC\;,\;$ we get that
$\cfrac{AD}{AG}=\cfrac{AO}{AF}\;.\quad\color{blue}{(**)}$
And, from $\;(*)\;$ and $\;(**)\;$, it follows that
$\cfrac{AB}{AE}+\cfrac{AD}{AG}=2\cfrac{AO}{AF}=\cfrac{AC}{AF}\;.$
$2)\quad r \nparallel BD$
In this case, by extending the diagonal $\;BD\;$ to meet the line $\;r\;$, we get the intersection point $\;X\;.$
Moreover, by applying Menelaus’s Theorem on the triangle $\;AOB\;$ with the line $\;r\;,\;$ it follows that
$\cfrac{EB}{AE}=\cfrac{FO}{AF}\cdot\cfrac{XB}{XO}\;,$
$\cfrac{AB}{AE}-1=\left(\cfrac{AO}{AF}-1\right)\cdot\left(1-\cfrac{BO}{XO}\right)\;,$
$\cfrac{AB}{AE}=1+\left(\cfrac{AO}{AF}-1\right)\cdot\left(1-\cfrac{BO}{XO}\right)\;.\quad\color{blue}{(***)}$
Analogously, by applying Menelaus’s Theorem on the triangle $\;AOD\;$ with the line $\;r\;,\;$ it follows that
$\cfrac{GD}{AG}=\cfrac{FO}{AF}\cdot\cfrac{XD}{XO}\;,$
$\cfrac{AD}{AG}-1=\left(\cfrac{AO}{AF}-1\right)\cdot\left(1+\cfrac{OD}{XO}\right)\;,$
$\cfrac{AD}{AG}=1+\left(\cfrac{AO}{AF}-1\right)\cdot\left(1+\cfrac{OD}{XO}\right)\;.\quad\color{blue}{(****)}$
Since $\;BO\cong OD\;,\;$ from $\;(***)\;$ and $\;(****)\;,\;$ it follows that
$\cfrac{AB}{AE}+\cfrac{AD}{AG}=2+2\left(\cfrac{AO}{AF}-1\right)=\cfrac{AC}{AF}\;.$