Let $A$ be a point on a circle with center $O$ and $B$ be the midpoint of $[OA]$. Let $C$ and $D$ be points on the circle such that they lie on the same side of the line $OA$ and $\widehat{CBO} = \widehat{DBA}$. Show that the reflection of the midpoint of $[CD]$ over $B$ lies on the circle.
let $M'$ be the reflection of $M$ over $B$. it's enough to prove $OM'=OA=r$ but since $M'AMO$ is a parallelogram it's enough to prove $AM=r$ if $CD\cap AO=R$ and $RA,DB$ meet $(O)$ again at $K,L$ then since $\angle ABL=\angle DBO=\angle CBA$ points $L,C$ are symmetric wrt $OA$. so $AC=AL$ which means $AD$ bisects $\angle BDC$ so $\angle BDC=2*\angle ADC=\angle AOC=\angle BOC$ meaning that $BODC$ is cyclic. By power of $R$ to circles $(O)$ and $BODC$ $RA*RK=RC*RD=RB*RO$(*) but $RK=RA+2r$ $RB=RA+r/2$ and $RO=RA+r$ putting these into (*) we get $RA=r$ since $OC=OD$ and $M$ is the midpoint of $CD$ $\angle OMR=90$ so since $A$ is the midpoint of $OR$ we get $AM=AO$ Q.E.D
