Find the locus of point $E'$.

Question

$E$ is a point of side $BC$ of $\square ABCD$. $DE \cap AB = \{D'\}$, $AE \cap CD= \{A'\}$ and $A'B \cap C'D = \{E'\}$. Find the locus of point $E'$.

This question was in a competition in 1999. Things have been more difficult since.

Answer 1

Using the intercept theorem with $\triangle ABE$ and $\triangle CDE$, we have $\dfrac{AB}{A'C} = \dfrac{BE}{CE}$ and $\dfrac{CD}{BD'} = \dfrac{CE}{BE}$.

$$\implies \dfrac{AB}{A'C} \cdot \dfrac{CD}{B'D} = \dfrac{BE}{CE} \cdot \dfrac{CE}{BE} = 1 \implies AB \cdot CD = A'C \cdot B'D \implies BC^2 = A'C \cdot B'D$$

$$\implies \triangle A'CB \sim \triangle CDB' \implies \widehat{CBA'} + \widehat{BCD} = 90^\circ \implies \widehat{CE'B} = 90^\circ$$

That means $E'$ lies on the semicircle diameter $BC$.