My solution: $O$ is the midpoint of $AB$ and also the center of the circle $(ABCD)$ $EM\cap{AB} = N$ It’s easy to see that $\angle{MNB} = \angle{CBN}$ $\Rightarrow CMNB$ is an isoceles trapezium. Moreover: $BO.BN = BE^2 = BK^2$ $\Rightarrow \frac{BO}{BK} = \frac{BK}{BN}$ $\Rightarrow \triangle{BOK} \sim \triangle{BKN}$ $\Rightarrow KN = KB (\because OK = OB)$ $\Rightarrow KL$ is the perpendicular bisector of $BN$ $\Rightarrow M = R_{KL}(C)$ $\Rightarrow M$ is the orthocenter of $\triangle{DKL}$ and the conclusion follows. Q.E.D

Why do we have $KN=KB$? I don´t think that that´s obvious. Actually, $BO \cdot BN=BK^2$ holds because the triangles $\triangle BEO$ and $\triangle BEN$ are similar ($\angle BOE=\angle NEB=90^\circ$), and therefore $BK^2=BE^2=BO \cdot BN$. From this it follows that $KN=KB$.

Another more ugly solution: Again we show that $M$ is the orthocenter of $\triangle DKL$. Because of $KL \perp DM$, it suffices to prove that $DM=AB \cos \angle LDK$, because $AB$ is two times the circumradius of $\triangle DKL$. Because of $ME \parallel AD$ and $AB \parallel CD$, we have $DM = \frac{AE \cdot CD}{AC}=\frac{AB \cdot CD}{AB + CD}$, therefore the claim is equivalent to $\cos \angle LDK = \frac{CD}{AB+CD}$. We have $\angle LDK = 180^\circ - \angle KBL=180^\circ -2\angle KBO$. Because of $OK=OB$, we get $\cos \angle KBO=\frac{BK}{2BO}=\frac{BE}{AB}$, and we have $BE=\frac{BD \cdot AB}{AB+CD}$. This implies $$\cos \angle LDK = 1 - 2 \cos^2 \angle KBO=1-2\left( \frac{BE}{AB} \right) ^2 = \frac{(AB+CD)^2-2BD^2}{(AB+CD)^2}$$, and this should be equal to $\frac{CD}{AB+CD}$, which is equivalent to $AB^2+AB \cdot CD = 2BD^2$. But, by ptolemy, $AB \cdot CD = BD^2-AD^2$ and by pythagoras $AB^2=AD^2+BD^2$. This implies the desired statement.

My solution. Let $O$ be the center of the circle k. Then$EO$$\perp$$BA$ $\implies$ $BCEO$ cyclic. Observe that the centers of the circles $\odot$$BCEO$ and $\odot$$BCDA$ lies on the line$BE$ $\implies$ $ME$ is the radical axis of these two circles. Hence by applying radical axis theorem to circles $\odot$$BCEO$,$\odot$$BCDDA$ and $\odot$$KEL$ $\implies$ $BC,LK,EM$ are concurrent. Assume that they meets at $P$. Since $PE$ $\parallel$$DE$ $\implies$ $\angle$$PMC$$=$$\angle$$DAB$$=$$\angle$$ABC$$=$$\angle$$MCP$ by using this together with the fact $PL$$\parallel$$MC$ $\implies$ $PK$ is the perpendicular bisector of $CM$. Rest is easy angle chasing If $M'$$=$$KM$$\cup$$DL$ $\implies$ $\angle$$EMM'$$=$$\angle$$KMP$$=$$KCP$$=$$\angle$$BLK$$=$$\angle$$BKL$$=$$\angle$$BDL$ Hence $MEM'D$ are cyclic and we are done.

Notice EM//BC,AB//MC so KL bisect CM=F so FM.FD=FK.FL but KL perp CD so done

If $KM\perp DL$ then why not $LM\perp DK$ since assumptions are the same for both points? That's why we will prove both which means $M$ is the orthocenter of triangle $DLK$. Complex coordinates: $$a=-1,b=1,|c|=1,d=-\frac1c$$Then $$e=\frac{ab(c+d)-cd(a+b)}{ab-cd}=\frac{c-1}{c+1}$$$C,M,D$ are collinear: $$\frac{c-m}{c-d}=\overline{\left(\frac{c-m}{c-d}\right)}$$$$\overline{m}=m+\frac{1}{c}-c$$$EM\perp BE$ yields $$\frac{b-e}{e-m}=-\overline{\left(\frac{b-e}{e-m}\right)}$$as we know relationship between $m$ and $\overline m$ we deduce $$m=\frac{2c^3-c^2-1}{c(c+1)^2}$$$k,l$ are all solutions $x$ of two equations: $$|x|=1\wedge |x-1|=\left| 1-\frac{c-1}{c+1}\right|$$Squaring both sides of the second gives $$0=x^2-x\cdot\frac{2(c+1)^2-4c}{(c+1)^2}+1$$By Viete formulas $$k+l+d=k+l-\frac{1}{c}=\frac{2(c+1)^2-4c}{(c+1)^2}-\frac{1}{c}=\frac{2c^3-c^2-1}{c(c+1)^2}=m$$Because $|k|=|l|=|d|=1$ (they all lie on circle $k$) we have that $M$ is orthocenter of triangle $KLD$.

Balkan MO 2014 G5 wrote: Let $ABCD$ be a trapezium inscribed in a circle $k$ with diameter $AB$. A circle with center $B$ and radius $BE$,where $E$ is the intersection point of the diagonals $AC$ and $BD$ meets $k$ at points $K$ and $L$. If the line ,perpendicular to $BD$ at $E$,intersects $CD$ at $M$. Prove that $KM\perp DL$. Solution: Note that Proving the following Lemma directly implies the question: Equivalent Lemma wrote: Let $\Delta ABC$ be a right triangle at $B$. Let $M,M'$ be midpoints of arc $BC$ and $BAC$. Let $AM$ meet $BM'$ at $E$. Let $\omega$ be circle $(M,ME)$. Let $\omega$ meet $\odot (ABC)$ at $L$ and $N$. Let tangent to $\omega$ at $E$ meet $AB$ at $K$. Prove, $B$ is reflection of $K$ over $LN$ Solution #1: (AlastorMoody, MathPassionForever & Naruto.D.Luffy) $E$ is the incenter WRT $\Delta ALN$ and also $EO||LN||BC$. Then, using the following lemma solves the problem: Lemma#: In $\Delta ABC$, If $OI || BC$ and $H$ is orthocenter WRT $\Delta ABC$. Then, $AI \perp IH$ Proof: Let $D$ be $A-$intouch point and $D'$ be reflection of $D$ over $I$. Let $M$ be midpoint of $BC$. Let $T_A$ be $A-$extouch point and $H'$ be reflection of $H$ over $BC$. $\angle D'OI$ $=$ $\angle ODC$ $=$ $\angle OT_AD$ $=$ $180^{\circ}-\angle IOT_A$ $\implies$ $A-D'-O-T_A$. Now, $\angle MOA'$ $=$ $\angle MOD$ $=$ $\angle HAA'$ $=$ $\angle HOM$ $\implies$ $H-D-O$ $\implies$ $HD||AO$. Now, $HD=H'D=AD'=D'I=DI$ $\implies$ $D$ is center of $\odot (IHH')$ $$\angle AHI=180^{\circ}-\angle H'HD-\angle DHI=90^{\circ}-\angle H'AA'+\angle AH'I=90^{\circ}-\angle HAI \implies AI \perp HI \qquad \blacksquare$$ Solution #2: (e_plus_pi) Let $F$ = $MB \cap EE$. Also let $O$ be the circumcenter of $\odot(ABC)$. Consider the following claim: Claim: $MOEB$ is cyclic. Proof: Firstly, $\angle MOE = 90^{\circ}$. Next, note that $ABMM^{\prime}$ is a isosceles trapezoid which means that $EM = EM^{\prime} \implies EO \perp MO$. Hence the claim. So by radical axis theorem $F \in LN$. Consider the inversion about $\omega$. Clearly, $\odot(ABC) \mapsto LN$ and so $F \mapsto B$. Also, $\overline{EE} \mapsto \odot (MOEB)$. Thus we have, $$\angle BFK = \angle MFE = \angle MEB = \angle MOB = \angle A$$And, $$\angle FBK = 90^{\circ} - \angle KBE = \angle ABM^{\prime} = \frac{1}{2} \cdot \angle (90^{\circ} + \angle C)$$These two imply the desired result $\qquad \blacksquare$ Solution #3: (Naruto.D.Luffy) Let $KE \cap LN=I$. Applying Radical Axes Theorem on $\odot (ABC)$, $\odot (MOEB)$ and $\odot (LEN) $ $\implies$ $B-M-I$ Now, $EA=EB$ and $\Delta MEI$ $\sim$ $\Delta EBI$. Hence, $$\angle BIK=\angle MIE=\angle BEM=2\angle BAE=\angle BAC$$And, $$\angle BKI=\angle EKA =90^{\circ}-\frac{A}{2} \qquad \blacksquare $$ Solution #4: (Pluto1708) Redefine $K$ as reflection of $B$ over $LN$. Let $U=AM\cap BC$.Note that it suffices to show $\odot{KEBU}$ concyclic.Equivalently since $EA=EB=EU=\dfrac{c}{2\cos \dfrac{A}{2}} \Rightarrow AU.AE=2AE^2=\dfrac{c^2}{2\cos^2 \dfrac{A}{2}}=\dfrac{c^2}{1-\cos A}=\dfrac{bc^2}{b-c}$.Because of POP it suffices to show $AK.AB=\dfrac{bc^2}{b-c} \Rightarrow AK=\dfrac{bc}{b-c} \implies BK=\dfrac{c^2}{b-c} \implies BT=\dfrac{c^2}{2(b-c)}$.By Los $$\dfrac{BN}{\sin{\dfrac{A}{2}+\angle NAC}}=2R=\dfrac{BM}{\sin{\dfrac{A}{2}}}=\dfrac{c}{\sin C}=\dfrac{MN}{\sin{\dfrac{A}{2}+\angle NAC}}=\dfrac{ME}{\sin{\dfrac{A}{2}+\angle NAC}}$$. Recall that $ME=\dfrac{R}{\cos \dfrac{A}{2}}$.Hence $BN=\dfrac{R\sin{A+\angle NAC}}{\cos{\dfrac{A}{2}}(\sin{\dfrac{A}{2}}+\angle NAC})$.Now $BT=BN\sin NAC=\dfrac{R\sin{A+\angle NAC}\sin NAC}{\cos{\dfrac{A}{2}}\sin{\dfrac{A}{2}}+\angle NAC}$.Now note that $2R=\dfrac{ME}{\sin{\dfrac{A}{2}+\angle NAC}} \implies \sin{\dfrac{A}{2}+\angle NAC}=\dfrac{1}{2\cos \dfrac{A}{2}}$.Thus $$BT=\dfrac{R\sin{((\dfrac{A}{2}+\angle NAC)+\dfrac{A}{2})}\sin{((\dfrac{A}{2}+\angle NAC)-\dfrac{A}{2}})}{\cos \dfrac{A}{2} \sin{(\dfrac{A}{2}+\angle NAC})}=2R(\dfrac{1}{4\cos^2\dfrac{A}{2}}-\sin^2\dfrac{A}{2})$$.This after some easy simplifications gives $\dfrac{c^2}{b-c}$ as desired. $\qquad \blacksquare$

$\color{green} \textbf{Construction :}$ Let $$EM \cap AB \equiv Z, ZM \cap BC \equiv Y, KL \cap AB \equiv P, KL \cap CD \equiv Q$$$$AC \cap \omega_2 \equiv R, KM \cap DL \equiv S, LD \cap AB \equiv T$$ $\color{blue} \textbf{Claim 1 :}$ $BCMZ$ is a trapezoid $\textbf{Proof :}$ $KL$ is the radical axis of the circles, So $KL \perp AB \equiv CD$ We have $\angle ECB=\angle ACB=90°$ as $AB$ is a diameter, $ER$ is a chord of $\omega_2$ and $B$ is the center $$\implies \angle EBC=\frac{1}{2} \angle EBR=\angle MEC$$Again $$\angle EBC= \angle DBC=\angle DAC$$So, $$\angle DAC =\angle MCC \implies AD \parallel ME\equiv MZ \implies AD=MZ=BC \square$$ $\color{blue} \textbf{Claim 2 :}$ $KL$ passes through $Y$ $\textbf{Proof :}$ $$\triangle EYC \sim \triangle EBY \implies YE^{2}=YC.YB \implies Pow_{\omega_2} Y =Pow_{\omega_1} Y$$Hence the radical axis $KL$ passes through $Y\square$ As $BCMZ$ is a trapezoid $YBZ$ is isosceles and $YD$ is the perpendicular bisector of $BZ$ Hence, $$\angle SKD=\angle MKQ=\angle QKC=\angle LKC=\angle LDC=\angle LTD$$So, $KSTD$ cyclic $\implies \angle KST=180-\angle KDT=90 \blacksquare$ $\color{black} \textbf{Remark :}$ Really great problem!