$\color{blue} \boxed{\textbf{SOLUTION}}$ Let the Semi-circle be $\omega_1$ $\angle KDC=90°$, Draw a circle With Diameter $KC$, So it will pass through $D$, call is $\omega_2$, Denote $I$ be the center of $\omega_2$ Now, $KD$ is the radical axis of $\omega_1$ and $\omega_2$ So, $OI \perp KD \equiv J$ and as $OK=OD$, $J$ is the mid point of $KD$ $OJ \parallel CD$ $\angle BCD=\angle BOJ=\angle BOK +\angle KOJ=2\angle BEK +\frac{1}{2} \angle KOD=2\angle BEK+ \angle DEK...(1)$ $\color{red} \boxed {\textbf{Claim}}$ $\angle OEK =2\angle BEK$ $\color{black} \textbf{proof : }$ Let, $F$ on $AB$ such that, $\angle FEK=2\angle BEK,$ We want to show $F \equiv O$ $(1) \implies \angle FCD= \angle BCD=\angle FEK+ \angle DEK=\angle FED$ So, $F,C,E,D$ concyclic. $\angle FED=\angle FCD=\angle ACE=\angle FDE$ $FE=FD$ Let $E',D'$ be the reflection of $E,D$ over $AB$ So, $E',D'$ lies on $\omega_1$ and $FE=FE'=FD=FD'$ Draw a circle $\omega_3$ with center $F$ and radius $FE$ Here, $\omega_1 \cap \omega_3 \equiv E,E',D,D'$ where $E,E',D,D'$ are different. But we know a circle can intersect another circle in a maximum of two points, So, we get $\omega_3 \equiv \omega_1 \implies F \equiv O$ So, $\angle OEK=2\angle BEK$ $\implies \angle OEB=\angle BEK$ Now, $OE=OB \implies \angle OEB=\angle OBE$ $\implies \angle OBE= \angle BEK$ $\implies$ $KE \parallel AB$ $\blacksquare$

Let $\omega$ be the circle with diameter $\overline{AB}$ and $D' = \overline{DC} \cap \omega$. Then $\overline{KD'}$ is a diameter and passes through $O$, and $D'$ and $E$ are reflections across $\overline{AB}$. Thus, $AEBD'$ is harmonic, so $$-1 = (A, B; E, D') \overset{K}{=} (A, B; \overline{KE} \cap \overline{AB}, O).$$Since $O$ is the midpoint of $\overline{AB}$, this implies that $\overline{KE} \parallel \overline{AB}$.