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

Question

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?

Answer 1

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

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

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

Answer 2

It is evident from the diagram that $\vec{AB}=AB\frac{\vec{AE}}{AE}$. $\vec{AD}$ and $\vec{AC}$ can be found similarly. Let the vector along the line cutting the three sides be $\vec n$.

$$\begin{align}\vec{AB}+\vec{AD}&=\vec{AC}\\ AB\frac{\vec{AE}}{AE}+AD\frac{\vec{AG}}{AG}&=AC\frac{\vec{AF}}{AF}\\\\ \text{Now, cross multiply by $\vec n$. Finally,we get}\\\\ \frac{AB}{AE}(AE\sin\alpha)+\frac{AD}{AG}(AG\sin\beta)&=\frac{AC}{AF}(AF\sin\gamma)\\ \frac{AB}{AE}(AH)+\frac{AD}{AG}(AH)&=\frac{AC}{AF}(AH)\\ \frac{AB}{AE}+\frac{AD}{AG}&=\frac{AC}{AF}\end{align}$$

Answer 3

Assuming $A$ as the origin of coordinates, I will denote the points $B$ and $D$ by $\vec b$ and $\vec d$. Other points will be denoted by their small lettered vectors similarly.

$$\vec b+\vec d=\vec c$$

$$|\vec b|\frac{\vec e}{\left|\vec e\right|}+|\vec d|\frac{\vec g}{\left|\vec g\right|}=|\vec c|\frac{\vec f}{\left|\vec f\right|}$$

As $\vec e, \vec f, \vec g$ are collinear, then, by the necessary and sufficient condition for collinearity of three vectors (the link has a discussion on the equation)

$$\frac{\left|\vec b\right|}{\left|\vec e\right|}+\frac{\left|\vec d\right|}{\left|\vec g\right|}=\frac{\left|\vec c\right|}{\left|\vec f\right|}$$

Your proof ends here...

Answer 4

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!

Answer 5

This exercise refers purely to matters of incidence and does not at all involve the euclidian structure (i.e. the inner product and related notions of orthogonality, perpendiculars, angles etc) of the usual geometric plane, so in order to do it full justice may I present a solution in the most general frame in which it can be given.

Consider an arbitrary field $\mathbf{K}$, a left $\mathbf{K}$-vector space $\mathbf{V}$ of dimension $2$ and an affine space $\mathscr{P}$ whose space of translations (or director space) is $\mathrm{Dir}\mathscr{P}=\mathbf{V}$. Since $\mathscr{P}$ is of dimension $2$ we are entitled to refer to it is an affine plane. First, given arbitrary vectors $y \in \mathbf{V}^{\times}=\mathbf{V} \setminus \{0_{\mathbf{V}}\}$ and $x \in \mathbf{K}y$ we denote the unique scalar $\lambda \in \mathbf{K}$ such that $x=\lambda y$ by $\frac{x}{y}\colon=\lambda$ (the existence of such a scalar follows from the condition $x \in \mathbf{K}y$ whereas the uniqueness from the fact that $y \neq 0_{\mathbf{V}}$). Let us note that if we also have $x \neq 0_{\mathbf{V}}$ the relations $x \in \mathbf{K}y$ and $y \in \mathbf{K}x$ are equivalent and we have $\frac{y}{x}=\left(\frac{x}{y}\right)^{-1}$.

For arbitrary points $X, Y \in \mathscr{P}$ such that $X \neq Y$ the affine line generated by the subset $\{X, Y\}$ will be denoted by $XY$.

Next, consider a parallelogram $(A, B, C, D) \in \mathscr{P}^4$, which specifically means that:

• $\{A, C\} \cap \{B, D\}=\varnothing$
• since the pairs $(A, B)$, $(B, C)$, $(C, D)$, $(D, A)$ consist of distinct points they each generate an affine line and we have $AB \parallel CD$ and $BC \parallel AD$.
• $AB \neq AD$ (the two lines are distinct).

With this general axiomatic definition of a parallelogram, let us make the following elementary remarks:

• it follows at once that $A \neq C$ and $B \neq D$, in other words the four points $A, B, C, D \in \mathscr{P}$ are pairwise distinct. Indeed, assuming for instance that $A=C$ would mean that the two parallel lines $AB$ and $CD$ intersect (in $A=C$) and therefore they are equal (in any affine space, two affine subspaces which have the same director subspace and intersect are in fact equal; the general notion of parallelism for affine subspaces applied in particular to affine lines means precisely that their director vector lines are equal). We could therefore infer that $AB=CD=AD$ (since $A, D \in AB=CD$ are distinct points on the same line they generate it), which contradicts one of the conditions of our definition.
• The vectors $a\colon=\overrightarrow{AB}$ and $b\colon=\overrightarrow{AD}$ form a basis of the director vector plane $\mathbf{V}$. Indeed, since $A \notin \{B, D\}$ we clearly have $\overrightarrow{AB}, \overrightarrow{AD} \neq 0_{\mathbf{V}}$ and furthermore the two vectors must be linearly independent, since otherwise a relation of the type $b=\alpha a$ for a certain nonzero scalar $\alpha \in \mathbf{K}^{\times}$ would mean that $D \in AB$ and subsequently that $AD=AB$, contradicting the definition. Thus, $\{a, b\}$ is a linearly independent set of cardinality $2$ within a vector space of dimension $2$, meaning that it must be a basis (since any linearly independent subset can be extended to a basis).
• it holds that $\overrightarrow{AB}=\overrightarrow{DC}$ and $\overrightarrow{BC}=\overrightarrow{AD}$, which easily entail the additional relations $\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{AD}$ respectively $\overrightarrow{BA}+\overrightarrow{BC}=\overrightarrow{BD}$ (Stevin's parallelogram law). Indeed, in the notations introduced in the previous paragraph, since $CD \parallel AB$ and $AD \parallel BC$ we have equivalently $\mathbf{K}\left(\overrightarrow{AB}\right)=\mathbf{K}\left(\overrightarrow{CD}\right)$ and $\mathbf{K}\left(\overrightarrow{AD}\right)=\mathbf{K}\left(\overrightarrow{BC}\right)$, which means that $\overrightarrow{CD}=\alpha a$ and $\overrightarrow{BC}=\beta b$ for certain scalars $\alpha, \beta \in \mathbf{K}$. We then have the relation: $$\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DA}=\left(\alpha+1_{\mathbf{K}}\right)a+\left(\beta-1_{\mathbf{K}}\right)b=0_{\mathbf{V}}.$$ By virtue of the linear independence of $\{a, b\}$ argued above, we infer that $\alpha=-1_{\mathbf{K}}$ and $\beta=1_{\mathbf{K}}$ which entail right away the stated relations.

We are now ready to state the general version of the problem.

Let $\mathscr{D} \subset \mathscr{P}$ be an affine line such that $A \notin \mathscr{D}$ and such that $\mathscr{D} \cap AB=\{M\}$, $\mathscr{D} \cap AC=\{N\}$ and $\mathscr{D} \cap AD=\{P\}$. Then $A \neq M, N, P$ and the following scalar relation holds: $$\frac{\overrightarrow{AB}}{\overrightarrow{AM}}+\frac{\overrightarrow{AD}}{\overrightarrow{AP}}=\frac{\overrightarrow{AC}}{\overrightarrow{AN}}.$$

Proof. Since by hypothesis $A$ is not on the line $\mathscr{D}$ which passes through all the points $M$, $N$, $P$, it is clear that $A \notin \{M, N, P\}$ (in a more formal perspective, $\{M, N, P\} \subseteq \mathscr{D}$ so $A \in \mathscr{P} \setminus \mathscr{D} \subseteq \mathscr{P} \setminus \{M, N, P\}$). We also have $\overrightarrow{AM} \in \mathbf{K}\left(\overrightarrow{AB}\right)$ and the analogues for $N$ and $P$. For simplicity we introduce: $$\begin{align*} \lambda\colon&=\frac{\overrightarrow{AB}}{\overrightarrow{AM}}\\ \mu\colon&=\frac{\overrightarrow{AD}}{\overrightarrow{AP}}\\ \nu\colon&=\frac{\overrightarrow{AC}}{\overrightarrow{AN}}. \end{align*}$$ Keeping the notation introduced in previous paragraphs, we obtain the relations: $$\begin{align*} \overrightarrow{AM}&=\lambda^{-1}a\\ \overrightarrow{AP}&=\mu^{-1}b\\ \overrightarrow{AN}&=\nu^{-1}(a+b). \end{align*}$$ Let us note that $M \neq P$, for otherwise we would have $M=P \in AB \cap AD=\{A\}$ (in any affine space, two distinct concurrent lines intersect in just one point), a contradiction. It follows that $N \in MP=\mathscr{D}$ and therefore there exists $\theta \in \mathbf{K}$ such that $N=\theta M+\left(1_{\mathbf{K}}-\theta\right)P$.

Considering vectors with origin at $A$, the above relation signifies that $\overrightarrow{AN}=\theta \overrightarrow{AM}+\left(1_{\mathbf{K}}- \theta\right)\overrightarrow{AP}$, which on the grounds of the above relations leads to: $$\nu^{-1}(a+b)=\theta\lambda^{-1}a+\left(1_{\mathbf{K}}- \theta\right)\mu^{-1}b.$$ By once again appealing to the linear independence of $\{a, b\}$, we gather that $\nu^{-1}=\theta\lambda^{-1}=\left(1_{\mathbf{K}}- \theta\right)\mu^{-1}$, which entails $\theta=\nu^{-1}\lambda$ respectively $1_{\mathbf{K}}- \theta=\nu^{-1}\mu$. Adding these latter relations leads to the desired conclusion that $\lambda+\mu=\nu$. $\Box$