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

$\color{blue}{\text{It is not necessary to use vectors, indeed it is possible}\\\text{to get a proof by applying Thales’ Theorem and}\\\text{Menelaus’s Theorem.}}$

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:

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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

$\cfrac{AB}{AE}+\cfrac{AD}{AG}=2+2\left(\cfrac{AO}{AF}-1\right)=\cfrac{AC}{AF}\;.$

Tags:

Geometry

Euclidean Geometry

Quadrilateral

Ratio

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