BMO Short List

$$\angle EPD=90^{\circ }-\angle EDC=90^{\circ }-\angle ACB=\angle EAC$$ So the points $E$, $A$, $C$, $P$ are concyclic. It follows that $\angle CPA=90^{\circ }$, therefore the triangle $APZ$ is right-angled. Since also $B$ is the midpoint of $AZ$, then $PB=AB=BZ$.

We have $$\angle BPE=\angle ABC-\angle BPZ=\angle ABC-\angle PZB$$ and $$\angle PAE=\angle PCE=90^{\circ }-\angle ACB-\angle CZA=\angle ABC-\angle CZA=\angle ABC-\angle PZB$$ Therefore $\angle BPE=\angle PAE$.

Since also $\angle EPD=\angle EAC=\angle EBA$, then $PEBZ$ is a cyclic quadrilateral and we get $\angle BPE=\angle EZB$. Therefore $\angle PAE=\angle EZB$, i.e. $PA$ is the tangent of $(c)$ at $A$.



Since $AZ$ is parallel to $MN$ then $TB\perp AZ$ and $TO\perp MN$.

The quadrilateral $KBOM$ is a rectangle. Consider the circumcircle of the rectangle and a tangent of this at $K$. Let $X$ be a point on this tangent. So $XK\perp KO$. Since the triangle $OBA$ and $OTA$ are similar then $O{A}^2=OT\cdot OB$. Since $OA=OM$ and $KM=OB$ we get $O{M}^2=OT\cdot KM$ so $OT/OM=OM/KM$. Since also $\angle KMO=\angle TOM=90^{\circ }$, the triangles $TOM$ and $OMK$ are similar. Therefore $$\angle MTO=\angle KOM=\angle XKM$$ Since $KM$ is parallel to $TO$ we have $\angle MTO=\angle KMT$. Therefore $\angle XKM=\angle KMT$. I.e. the tangent at point $K$ is parallel to $MT$ and since $XK\perp KO$ we get $OK\perp TM$.