Showing that a segment is the geometric mean of two segments in a parallelogram by using Intercept theorem

Question

We have a parallelogram $ABCD$ and a line $a$ such that $A\in a$ and $a$ intersects the lines $BD,BC$ and $DC$ in $E,F$ and $K$, respectively. I should show that $AE$ is the geometric mean of $EF$ and $EK$.

I should show $\dfrac{EF}{AE}=\dfrac{AE}{EK}$. We can look at $\angle AED$ and $AD\parallel BC$. Intercept theorem tells us that $\dfrac{EF}{EA}=\dfrac{EB}{ED}=\dfrac{BF}{AD}=\dfrac{BF}{BC}$. How can I continue from here?

Answer 1

To conclude your path:

1. Intercept Theorem on $AD\parallel BC$ with intercept lines $AK$ and $DB$, gives you, as you found out,$$\frac{\overline{EF}}{\overline{AE}}=\frac{\overline{EB}}{\overline{ED}}.\tag{1}\label{1}$$
2. Intercept Theorem on $AB\parallel DK$ with intercept lines, again, $AK$ and $DB$, gives you $$\frac{\overline{EB}}{\overline{ED}}=\frac{\overline{AE}}{\overline{EK}}.\tag{2}\label{2}$$

The thesis comes from equating \eqref{1} and \eqref{2}.