xeroxia wrote: Let $O$ be the center and $[AB]$ be the diameter of a semicircle. $E$ is a point between $O$ and $B$. The perpendicular to $[AB]$ at $E$ meets the semicircle at $D$. A circle which is internally tangent to the arc $\overarc{BD}$ is also tangent to $[DE]$ and $[EB]$ at $K$ and $C$, respectively. Prove that $\widehat{EDC}=\widehat{BDC}$. Let $(I)$ be the circle which is internally tangent to the arc $\overarc{BD}$ is also tangent to $[DE]$ and $[EB]$ at $K$ and $C$, respectively. And $M$ is intersecttion of $AK$ and $(O)$. We have $IK || OA$ ( both of them is perpendicular to $DE$ ), so $\widehat{MKI} = \widehat{MAO} \rightarrow \widehat{MIK} = \widehat{MOA}$. Thus M, I, O are collinear. So $ADC$ is isosceles triangle in $A$. So $\widehat{ACD}=\widehat{ADC} \rightarrow 90 - \widehat{ACD}= 90 - \widehat{ADC} \rightarrow \widehat{BDC}=\widehat{ICD} = \widehat{EDC}$ . QED

My solution: Let $F$ the reflection of $D$ in $E$ $\Longrightarrow$ $ABDF$ is cyclic $\Longrightarrow$ by Sawayama's lemma we get $KC$ through incenter of $\triangle BDF$ but $BC$ is bisectot of $\angle DBF$ $\Longrightarrow$ $C$ is the incenter of $\triangle BDF$ $\Longrightarrow$ $\angle CDE=\angle CDB$...