# Prove that $BXOY$ is cyclic with spiral similarity over a midpoint

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!

## 1 Answer

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!$$\tag*{\blacksquare ::&}$$

• Thank you so much :) actually point M' was looking good , that's why I chose this way .. Thanks again ::@ – Sunaina Pati Oct 4 '20 at 13:14
• @SunainaPati You're welcome :) and about the point M' its a really nice choice to introduce it :) ::& – Anand Oct 4 '20 at 18:05