Let $O$ be the circumcenter of triangle $ABC$. Line $CO$ intersects the altitude from $A$ at point $K$. Let $P,M$ be the midpoints of $AK$, $AC$ respectively. If $PO$ intersects $BC$ at $Y$, and the circumcircle of triangle $BCM$ meets $AB$ at $X$, prove that $BXOY$ is cyclic.
My Progress: Since this was from a spiral similarity handout, I am using spiral similarity.
Note that $XM$ is anti parallel to $BC$. Now since $AK$ and $AO$ are isogonals and $AK\perp BO \implies AO \perp XM$.
Now define $AO\cap XM=U$ , $M'$=reflection of $O$ wrt $M$ , $A'$= antipode of $A$ wrt $(ABC)$ , $W=AC\cap A'K$ , $Z=A'M' \cap (ABC) .$
I observed that there is a spiral similarity (say $\gamma $ ) centred at $X$ taking $M'M$ to $AK$ , and hence $O$ to $P$ . So I am trying to prove that observation. Here is what I got.
- $ACA'Z$ is a rectangle : as $A-O-A'$ and $Z-O-C$ are collinear
- $MM'A'C$ is a rectangle : as $M$ and $M'$ are midpoint of $AC$ and $ZA'$
- $MUM'A'C$ is cyclic : Note that $\angle MUA'=\angle MM'A'=90$
So by the above claims , we get $\angle XAK =\angle MAO=\angle OA'M'= \angle UA'M'= \angle UMO=\angle XMM'$.
Now after this I want to prove
$\angle MXM'= \angle AXK $
After this we will be done, because then we will have $\gamma : M'M \rightarrow KA \implies \gamma : O \rightarrow P \implies \Delta POX \sim \Delta AMX \implies \angle POX = \angle AMX=\angle ABC $
Thanks in advance!
Nice observations and almost there!
So, as you have already shown that $\angle XAK = \angle XMM'$, its enough to show that $\angle AXP =\angle MXO$ (Notice that instead of $\gamma: MM'\mapsto AK$, I'm trying to show $\gamma: MO\mapsto AP$ and the reason for this is because as you observed $XM\perp AO$, showing angles $AXP$ and $MXO$ same would be equivalent of showing center of $\odot(AXO)$ lies on $XP$ and thus, this is a more natural way of approaching the problem.)
Anyways, lets move on...
Note that as $\angle AUM = \angle AMO =90,$ we get, $\angle UMO =\angle OAM=\angle OCA\overset{Thales'}{=}\angle AMP$. Further note that $\{AP,AO\}$ are isogonal lines with respect to $\triangle XAM$ and by the fact that $\angle UMO =\angle AMP$ we conclude that points $\{P,O\}$ are isogonal with respect to $\triangle XAM$ and thus, $\angle PXA =\angle OXM$ and we are done!$$\tag*{$\blacksquare$ ::&}$$