# Alternative method of finding a ratio in a parallelogram via composition of two homotheties

Let $$ABCD$$ be a parallelogram and let $$E\in\overline{AD},\ F\in\overline{CD}$$ such that: $$\frac{|AE|}{|ED|}=\frac{|DF|}{|FC|}=\frac12.$$ Find the ratio in which the line segment $$\overline{EF}$$ divides the diagonal $$\overline{BD}$$.

One approach I'm familiar with:

$$\overrightarrow{AB}=\overrightarrow{DC}\ \&\ \overrightarrow{AD}=\overrightarrow{BC}$$

Let $$S$$ be the intersection point of $$\overline{EF}$$ and diagonal $$\overline{BD}$$, then:

$$\overrightarrow{ES}=\lambda\overrightarrow{EF}\ \&\ \overrightarrow{DS}=\mu\overrightarrow{DB}$$

$$\overrightarrow{ES}=\lambda\overrightarrow{EF}=\lambda\left(\overrightarrow{ED}+\overrightarrow{DF}\right)=\lambda\left(\frac23\overrightarrow{AD}+\frac13\overrightarrow{DC}\right)=\frac{2\lambda}3\overrightarrow{AD}+\frac{\lambda}3\overrightarrow{AB}$$

On the other hand,

$$\overrightarrow{ES}=\overrightarrow{ED}+\overrightarrow{DS}=\frac23\overrightarrow{AD}+\mu\overrightarrow{DB}=\frac23\overrightarrow{AD}+\mu\left(\overrightarrow{DA}+\overrightarrow{AB}\right)=\left(\frac23-\mu\right)\overrightarrow{AD}+\mu\overrightarrow{AB}$$

$$\overrightarrow{AD}$$ and $$\overrightarrow{AB}$$ can form a basis, so we have obtain the following system:

$$\begin{cases}\frac{2\lambda}3&=\frac23-\mu\\\frac{\lambda}3&=\mu\end{cases}\implies \lambda=\frac23\implies\mu=\frac29$$

So we get that $$\overline{EF}$$ divides the diagonal $$\overline{BD}$$ in the ratio $$2:7$$

My question:

How can we solve this problem using the following theorem about the composition of two homotheties ( found here, in the answer by Aqua):

If $$\mathcal{H}_{M,k_1}$$ and $$\mathcal{H}_{N,k_2}$$ are homotheties then their compostion $$\mathcal{H}_{M,k_1}\circ \mathcal{H}_{N,k_2}$$ is again some homothety $$\mathcal{H}_{S,k}$$ with $$k=k_1k_2$$ (if $$k\ne 1$$) and it center $$S$$ lies on a line $$MN$$.

I thought I could do the following:

\begin{aligned}\mathcal H_{E,-2}&:A\mapsto D\\\mathcal H_{F,-2}&:D\mapsto C\end{aligned} so that the center of the homothety $$\mathcal H_{E,-2}\circ\mathcal H_{F,-2}$$ lies on the line $$EF$$, but this doesn't lead me to the right result.

Picture:

Thank you very much!

Let $$G\in FC$$ such that $$AG||EF$$, $$I\in AB$$ such that $$CI||AG,$$
$$AG\cap BD=\{Q\}$$, $$CI\cap BD=\{R\}$$ and $$FG=y$$.
Thus, $$DF=2y,$$ $$FC=4y,$$ $$GC=3y,$$ $$AI=3y$$ and since $$AB=2y+y+3y=6y,$$ we obtain $$IB=3y,$$ which gives $$QR=RB=3SQ$$ and $$DS=2SQ.$$ Thus, $$\frac{DS}{SB}=\frac{2SQ}{SQ+3SQ+3SQ}=\frac{2}{7}.$$
Let $$AB = 6a$$. Use similarity (or homothety if you wish): $${GA\over DF} = {EA\over ED} ={1\over 2} \implies GA =a$$ so $${DS\over SB} = {DF\over GB} = {2a\over 6a+a} = {2\over 7}$$