Vector ratios on parallelogram

Question

There is a parallelogram ABCD.

On length CD there is point E with statement |CE|=2|ED|

On length AD there is point F with statement |AF|=1/4|AD|

Point S is intersection of length BE and CF.

In which ratio does point S divide length BE and FC?

The answer to this problem is:

|FS| : |SC| = 5 : 4

|ES| : |SB| = 1 : 2

I have been stuck on this problem for some time now and any guidance would be helpful.

Answer 1

Consider the line $l$ which is parallel to $AB$ and passes through $F$. Let $G$ denote the intersection of $l$ and line $EB$, and $H$ be the intersection with side $BC$. This introduces some useful similar triangles in the construction. Notice that:

• Triangle $BGH$ is similar to $BEC$.
• Triangle $SFG$ is similar to $SCE$.

Notice that the first ratio is straightforward since $|AF| = 1/4 |AD|$. For the ratio on the second similarity, just notice that $|DC| = 3/2 |EC|$ implies that $$|FG| = |FH| - |GH| = 3/2 |EC| - 1/4 |EC| = 5/4 |EC|$$ and thus the ratio for the second similarity is $5/4$. From here, you can deduce the ratio $|SE|/|SB|$. Denoting $|SE|$ by $x$, we can write $$ |SG| = 5/4 x \quad \text{and} \quad |GE| = x + 5/4 x = 9/4 x.$$ To obtain $|GB|$ in terms of $x$ we apply Tales theorem to get $$ \frac{|GB|}{|GE|} = \frac{1/4}{3/4} = \frac{1}{3} \implies |GB| = 3/4 x.$$ Finally, we have $$ \frac{|SE|}{|SB|} = \frac{|SE|}{|SG| + |GB|} = \frac{x}{5/4x + 3/4x} = \frac{1}{2}.$$ A similar reasoning can be aplied for the other ratio.

Answer 2

Here's a solution using vectors . . .

Place the origin at $A$.

Letting $\mathbf{u}=\vec{D}$, and $\mathbf{v}=\vec{B}$, we get \begin{align*} \vec{C}&=\vec{D}+\vec{B}=\mathbf{u}+\mathbf{v}\\[4pt] \vec{F}&={\small{\frac{1}{4}}}\vec{D}={\small{\frac{1}{4}}}\mathbf{u}\\[4pt] \vec{E}&=\vec{D}+{\small{\frac{1}{3}}}(\vec{C}-\vec{D})=\mathbf{u}+{\small{\frac{1}{3}}}\mathbf{v}\\[4pt] \end{align*} Also, since $S$ is on the line segments $FC$ and $EB$, we get \begin{align*} \vec{S}&=\vec{F}+x(\vec{C}-\vec{F})=\Bigl(\frac{1+3x}{4}\Bigr)\mathbf{u}+x\mathbf{v}\\[4pt] \vec{S}&=\vec{E}+y(\vec{B}-\vec{E})=(1-y)\mathbf{u}+\Bigl(\frac{1+2y}{3}\Bigr)\mathbf{v}\\[4pt] \end{align*} for some $x,y\in\mathbb{R}$, with $0 < x < 1$ and $0 < y < 1$.

Since the vectors $\mathbf{u},\mathbf{v}$ are linearly independent, we must have \begin{align*} \frac{1+3x}{4}&=1-y\\[4pt] x&=\frac{1+2y}{3}\\[4pt] \end{align*} which yields $x={\large{\frac{5}{9}}}$ and $y={\large{\frac{1}{3}}}$.

It follows that

• $S$ is ${\large{\frac{5}{9}}}$ of the way from $F$ to $C$, so $|FS|{\;:\,}|SC|=5{\,:\,}4$.$\\[10pt]$
• $S$ is ${\large{\frac{1}{3}}}$ of the way from $E$ to $B$, so $|ES|{\;:\,}|SB|=1{\,:\,}2$.