What's the distance from the midpoint of $AE$ to $BD$ in the pentagon below?

Question

For reference: In a convex pentagon $ABCDE$: $AB=BC$ and $CD=DE$ ($CD > BC$); If: $BD=K$ and angle $B$ is equal to the measure of angle $D$ which measures $90°$. Calculate the distance from the midpoint of $AE$ to $BD$. (Answer:$\frac{k}{2}$)

My progress..

Here is my figure and my considerations:

Fill in the angles implies $\angle BAC = \angle BCA = \angle DEC=\angle DCE=45^o$

ABDE is a isosceles trapezoid

$\angle ACE =45^o$ but I couldn't demonstrate

therefore $\angle CAE = \angle AEC = 67,5^o$

Law of Cosines$\triangle BCD: k^2=2l^2-2l^2.cos135\implies k^2 = l^2(2+\sqrt2)\therefore k = l\sqrt{2+\sqrt2}\\ AC = l\sqrt2$

(by property) $HJ^2 = k.AE\implies HJ=\sqrt{k.AE}$

...???

CORRECTED DRAWING

Van Aubel's theorem $\angle DJB = 90^0, DJ=JB\\ \triangle DJB: JL^2 = DH.BH = \frac{k}{2}.\frac{k}{2} \therefore JL = \sqrt\frac{k^2}{4}=\frac{k}{2}$

Answer 1

Draw $CG \perp AC$ and $CF \perp CE$. By midpoint theorem,
$JB \parallel EG$ and $JB = \frac 12 EG$
$JD \parallel AF$ and $JD = \frac 12 AF$

We also see that $\triangle GCE~$ is $~\triangle ACF$ rotated $90^\circ$ clockwise about point $C$. So, $GE \perp AF$ and $GE = AF$.

That shows $JD \perp JB$ and $JD = JB$

So, $JL = BL = \dfrac k2$