Let $ABC$ be an acute-angled triangle with circumcircle $\Gamma$ and orthocentre $H$. Let $K$ be a point of $\Gamma$ on the other side of $BC$ from $A$. Let $L$ be the reflection of $K$ in the line $AB$, and let $M$ be the reflection of $K$ in the line $BC$. Let $E$ be the second point of intersection of $\Gamma $ with the circumcircle of triangle $BLM$. Show that the lines $KH$, $EM$ and $BC$ are concurrent. (The orthocentre of a triangle is the point on all three of its altitudes.)
Surely a very hard problem ! It took me 6 hrs to solve without any hints ! But this question is very diagram dependent, so if possible can someone verify my proof ? Thanks in advance.
Plus you can send your solution too, It helps me a lot.
My Proof: Now, let $X$ be the reflection of $H$ over side $BC$. It is well known that $ABCX$ is cyclic.
Claim: $BMHC$ and $LBHA$ are cyclic quads.
Proof: Note that $ABCK$ is cyclic ( It is given) . So $\angle BMC=\angle BKC= \angle BXC= \angle BHC$ and hence $BMHC$ is cyclic. Similarly, we can prove it for $LBHA$.
Let $MX\cap HK=Y$ . Note that by angle chase, we have $Y \in BC$.
So it is enough to show that $E,M,X$ are collinear .
Now, since $BC$ is the perpendicular bisector of $MK$ and $AB$ is the perpendicular bisector of $LK$, note that $B$ is the circumcentre of $\Delta KLM$ .
Define $I=MK\cap BC$ and $G= LK\cap AB$. Note that $BIGK$ is cyclic.
Now, we move to our next claim. ( Note: the Proof might look simple but it took me 4 hrs)
Claim: $L,M,H$ are collinear
Proof: Since $BIGK$ is cyclic, we get $\angle ABC=\angle GKI=\frac {1}{2} \angle LBM \implies \angle BML=90-\angle ABC$ .
So it is enough to show that $\angle HMB= 90+\angle ABC $ or $\angle HCB=90-\angle ABC$ (which is true by angle chase , $HC \perp AB$)
Now,the main proof .
Claim: $E,M,X$ are collinear
Proof: Note that by using the cyclic quads $(BMHC)$, $(LBHA)$, $(ABCEX)$ and $(LBME)$, we note that
$\angle BEM=\angle MLB=\angle BLH=\angle BAH=\angle BAX= \angle BEH =\angle BEX$ .
Hence $\angle BEM=\angle BEX$. Hence $EMX$ are collinear.
And we are done!
This proof relies on a following lemma (which is easy to prove):
Lemma: Reflection of $H$ across side of a triangle lies on circumcircle of that triangle.
Let $H'$ and $H''$ be respectively reflections of $H$ across $BC$ and $AB$. If we prove $E,M,H'$ are collinear we are done since $H'M$ and $HK$ meets in side $BC$.
Let $H'M$ and $H''L$ meet at point $F$. If we prove that $F$ lies on both circles we are done.
Let $\angle H'HC = x$, $\angle H'HK = y$ and $\angle MKB =z$.
Circle $ABC$:
Clearly $\angle HCB = 90-x$ and so $\angle BCH' = 90-x$. Also $\angle HH'F = y$ and $\angle H''HK = 180-x-y$ and thus $\angle FH''H = 180-x-y$. Since the sum of all angles in quadrilateral is $360$ we have (look at $H''HKF$) $\angle H''FK = 2x$ and thus $F$ is on circle $ABC$ (since $\angle H''CH' +\angle H''FH' =180$).
Circle $MBL$:
Since reflection preserves angles we have $\angle H'MB = y+z$ and $\angle BLF = y+z$ and we are done.