$AA' , BB'$ are two perpendicular diameters of circle C(O).Consider arbitrary point $M$ on C between $A'$ and $B'$.Chord $BM$ intersects diameter $AA'$ at $N$.Draw a perpendicular line $d$ to $AA'$ through $N$.If tangent to the circle at $M$ intersects line $d$ at $K$ prove that: $KO||BM$(preferably DO NOT use point $B'$).
It's seen that $K,M,N,O$ are concyclic. I drew radius $OM$ and now I must show that $\angle OBM=\angle KOM$, but don't know how?
Proof involving $B'$.
Let $c_1$ be the circumcircle of triangle $ONM$.
$ONMB'$ is inscribed in $c_1$ because $ \angle \, B'MB = \angle \, B'MN = 90^{\circ} = \angle \, B'ON$
$ONMK$ is inscribed in $c_1$ and $OK$ is a diameter of $c_1$ because $\angle \, ONK = 90^{\circ} = \angle OMK$ ($MK$ tangent to $C(O)$).
$OK$ diameter of $c_1$ implies $OB'K = 90^{\circ}$. Hence $ONKB'$ is a rectangle.
Consequently, $\angle \, KON = \angle \, B'NO = \angle \, BNO$ which measn that $KO$ is parallel to $BM$.
Proof not involving $B'$.
$\angle \, OBN = \angle \, BMO = \angle \, NMO$ because triangle $BOM$ isosceles ($BO=MO$);
$ONMK$ is cyclic because $\angle \, ONK = 90^{\circ} = \angle \, OMK$ because $KM$ is tangent to circle $C(O)$ at point $M$;
$\angle \, NKO = \angle \, NMO$ because $ONMK$ is cyclic;
Triangles $ONK$ and $NOB$ are congruent because $\angle \, ONK = \angle \, BON = 90^{\circ}$ and $\angle \, NKO = \angle \, OBN$ and $ON$ common side;
Hence $BOKN$ is a parallelogram (for instance because $NK = OB$ and $OK = BN$) and therefore $KO$ is parallel to $BN$ and thus to $BM$.
