General problem Euclidean geometry

Question

Let $\square ABCD$ be a square of side-length $a$, and consider also the circle with diameter $\overline{AC}$. Let $E$ be a point on $\stackrel{\frown}{BC}$ of the circle, and let $K$ a point on $\overrightarrow{EC}$ such that $|\overline{AE}|=|\overline{EK}|$.

$i)$ Why does $|\overline{DK}|=a$?
$ii)$ Let $J=\overleftrightarrow{AE}\cap\overleftrightarrow{DC} $. Prove that $\triangle ADJ$ and $\triangle CEJ$ are similar.
$iii)$ Let $ L= \overleftrightarrow{AK}\cap\overleftrightarrow{DE} $. Prove that $L$ is in an arc that you will determine.

I've tried doing it , and the only thing that I've got is that $\angle AEK =90 $, but I can't move foward. Edit:I figured it out $ii)$, but I´m still stucked in $i)$ and $iii)$

Answer 1

$\alpha = \alpha’$ (because AD = DC).

By SAS, $\triangle EAD \cong \triangle EKD$. Hence, $DK = DA = a$.

AS for (iii), we have sufficient info from the above to say that $\angle ELK = 90^0$. Hence, L is part of a circle (ELK, say).

Answer 2

Let $AE\cap BD=\{G\}$, $AE\cap BC=\{F\}$, $AK\cap BD=\{M\}$ and $AK\cap DC=\{N\}$.

Thus, since $$\measuredangle GAN=\measuredangle FBM=\measuredangle LDN=\measuredangle CKM=45^{\circ},$$ we see that $ABFM$, $AGND$ and $CMDK$ they are cyclic.

From the first and the second quadrilateral we obtain $$\measuredangle FMA=\measuredangle FGN=90^{\circ},$$ which says that $GFCNM$ is cyclic.

Now, let $\measuredangle ECB=\alpha$.

Thus, $$\measuredangle DCK=90^{\circ}-\alpha,$$ $$\measuredangle BAG=\measuredangle ECB=\alpha,$$ $$\measuredangle AGM=\measuredangle ABG+\measuredangle BAG=45^{\circ}+\alpha,$$ which says $$\measuredangle MGN=\measuredangle AGM-\measuredangle AGM=90^{\circ}-(45^{\circ}+\alpha)=45^{\circ}-\alpha.$$ Thus, $$\measuredangle MKD=\measuredangle MCD=\measuredangle MGN=45^{\circ}-\alpha,$$ which says $$\measuredangle CKD=\measuredangle CKM+\measuredangle DKM=45^{\circ}+45^{\circ}-\alpha=90^{\circ}-\alpha=\measuredangle DCK,$$ which gives $$DK=a.$$

ii) is true because $$\measuredangle CEJ=\measuredangle ADJ=90^{\circ}.$$

iii) Since $$DK=a=AD,$$ we obtain that $L$ is a midpoint of $AK$ and $ED\perp AK$.