You did not define $K$ well. [WakeUp: this has been edited]

Let $KL,KM$ meet $BC,BA$ at $L',M'$ respectively. Let us prove that $L'$ and $M'$ lie on the circle $(BLM)$. Now $BK=BM$ since $M$ is the reflection of $K$ through $BC$ but $L$ is the reflection of $K$ through $AB$ so clearly we have $BK=BL$ too. Thus $BK=BL=BM$ and so $B$ is the circumcentre of $\triangle KLM$. Then an angle chase suffices. If $K$ lies between the antipode of $A$ on $(ABC)$ and $C$ for example (as in the diagram), then: $B$ is the circumcentre of $\triangle MLK$ and so $\angle MBK=2\angle MLK$ but $\angle MBK=2\angle MBC$ and thus $\angle MLK=\angle MBC=\angle MBL'$. So $L'$ lies on $(BKL)$. Also, if $\angle M'ML=\alpha$ then $\angle LMK=180^{\circ}-\alpha$ so $\angle LBK=2\alpha$ since the chord $LK$ subtends a angle of $\alpha$ using directed angles mod 180. But then $2\alpha =\angle LBK=2\angle LBA$ and hence $\alpha =\angle LBA=\angle LBM'$ and so it follows $M'$ lies on $(BKL)$ too. Now, suppose $EM$ and $AH$ meet at $A'$. Note $AH$ and $MK$ are both perpendicular to $BC$, and so $MK||AH$. Therefore $\angle AA'M=\angle A'MK=\angle M'ME=\angle EBM'=\angle EBA$. Therefore $\angle AA'E=\angle ABE$, so $A'$ lies on the circumcircle of $\triangle ABC$. It is well known that if $AH$ meets $(ABC)$ at $P$ then $P$ is the reflection of $H$ through $BC$ (it's easy to prove, note that $\angle BHC=180^{\circ}-A$ by an easy angle chase, but also $\angle BPC=180^{\circ}-A$ so by symmetry it follows $BC$ is the perpendicular bisector of $HP$). Therefore, $BC$ is both the perpendicular bisector of $MK$ and $HA'$. So $HA'KM$ is an isosceles trapezoid. Therefore, its diagonals meet on the perpendicular bisector, i.e. $BC$. The angles here assume that $K$ lies on the arc $A_1C$ of $(ABC)$ not containing $A$, where $A_1$ is the antipode of $A$ on $(ABC)$. In this case, $HM$ is the smaller side of the isosceles trapezoid $HA'KM$ and so $HM,KA'$ and $BC$ concur at a point on $BC$ closer to $H$ than $M$. If $K=A_1$ then $HA'KM$ is a rectangle, and if $A_1$ is on the arc $(BA_1)$ not containing $A$ then $HA'KM$ is an isosceles trapezoid with $HA'$ the larger parallel side. Remark: You can also prove $L,M,H$ are collinear.

Let ${c_2}$ is the reflection of $\Gamma $ in $AB$, ${c_3}$ is the reflection of $\Gamma $ in $BC$, then the orthocentre $H$ is on the ${c_2}$ and ${c_3}$. Let $AH \cap \Gamma = P$, then $P$ is the reflection of $H$ in the line $BC$, thus $D = HK \cap PM$ is on the line $BC$, suppose $PM \cap \Gamma = E'$. By symmetry we have $\angle BMH = \angle PKB = \angle PAB = \angle BLH = 90^\circ - \angle ABC$. Because $\angle LBM = \angle LBA + \angle ABM = \angle KBA + \angle ABM = 2\angle ABC$, $\angle LBM + \angle BMH + \angle BLH = 180^\circ $, so $L,H,M$ are collinear . Then have $\angle ME'B = \angle PE'B = \angle PAB = \angle BLM$, so $EMBL$ is cyclic, hence $E = E'$, $HK \cap EM \cap BC = D$.

yunxiu wrote: 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.) Solution using complex numbers: Let $\Gamma$ be the unit circle, let $BC\cap HK=\{F\}$ and $BC\cap ME=\{D\}$. It is enough to show that $D\equiv F$. We can easily get that: \[m=\frac{k(b+c)-bc}{k};\bar{m}=\frac{b+c-k}{bc};l=\frac{k(a+b)-ab}{k};\bar{l}=\frac{a+b-k}{ab}(1)\] Because $\angle MEB=\angle MLB$ we have $\frac{e-m}{\bar{e}-\bar{m}}:(-be)=\frac{l-m}{\bar{l}-\bar{m}}:\frac{l-b}{\bar{l}-\bar{b}}$ (2) so together with (1) we get: \[e=\frac{k(ab+bc+ca)-abc}{k(a+b+c-k)}=\frac{abc(\bar{h}-\bar{k})}{h-k}\] In above equation I used $h=a+b+c$. From (2) we get $\frac{e-m}{\bar{e}-\bar{m}}=\frac{bce}{a}=\frac{b^2c^2(\bar{h}-\bar{k})}{h-k}$ (3). Because $D\in\widehat{BC}$ we have $\bar{d}=\frac{b+c-d}{bc}$ (4) and because $E,M$ and $D$ are collinear we have $\frac{e-m}{\bar{e}-\bar{m}}=\frac{m-d}{\bar{m}-\bar{d}}$ but together with (3),(4) we get $\frac{b^2c^2(\bar{h}-\bar{k})}{h-k}=\frac{m-d}{\bar{m}-\frac{b+c-d}{bc}}$ which is with (1) equivalent with: \[d=\frac{(h-k)(k(b+c)-bc)+kbc(k\bar{h}-1)}{bc(\bar{h}k-1)+(h-k)k}\] On the other side $K,F,H$ are collinear and $F\in\widehat{BC}$ so $\frac{f-k}{\bar{f}-\bar{k}}=\frac{h-k}{\bar{h}-\bar{k}}$ and $\bar{f}=\frac{b+c-f}{bc}$, therefore: \[f=\frac{(b+c)(h-k)k+bc(k^2\bar{h}-h)}{bc(\bar{h}k-1)+(h-k)k}\] We see that $d=f$ so $D\equiv F.\square$

[asy][asy]size(8cm); pointpen=black; pathpen=black; pair A = Drawing("A", dir(50), dir(50)); pair B = Drawing("B", dir(209), dir(180)); pair C = Drawing("C", dir(331), dir(0)); Drawing(A--B--C--cycle); pair H = orthocenter(A,B,C); pair Ka = foot(A,B,C); pair Kc = foot(C,A,B); Drawing("H", H, dir(45)); pair Ha = Drawing("H_A", 2*Ka-H, dir(-70)); pair Hc = Drawing("H_C", 2*Kc-H, dir(135)); pair K = Drawing("K", dir(-64), dir(-90)); pair M = Drawing("M", reflect(B,C) * K, dir(180)); pair L = Drawing("L", reflect(B,A) * K, dir(90)); pair E = Drawing("E", 2*foot(origin,M,Ha)-Ha, dir(70)); Drawing(A--Ha--B--Hc--C); Drawing(L--Hc); Drawing(Hc--E, dotted); draw(unitcircle); draw(Ha--E); draw(B--H--K); draw(L--K--M,dashed); [/asy][/asy] In what follows, all angles are directed. Let $H_A$ and $H_C$ be the reflections of $H$ across $\overline{BC}$ and $\overline{BA}$, which lie on $\Gamma$. Let $E'$ be the second intersection of line $H_AM$ with $\Gamma$. By construction, lines $E'M$ and $HK$ concur on $\overline{BC}$. First, we claim that $L$, $H_C$, and $E'$ are collinear. Notice that \[ \measuredangle LH_CB = -\measuredangle KHB = \measuredangle MH_AB \]by reflections, and that \[ \measuredangle MH_AB = \measuredangle E'H_AB = \measuredangle E'H_CB \]as desired. Now, \[ \measuredangle LE'M = \measuredangle H_CE'H_A = \measuredangle H_CBH_A = 2\measuredangle ABC \]and \[ \measuredangle LBM = \measuredangle LBK + \measuredangle KBM = 2\measuredangle ABK + 2\measuredangle KBC = 2\measuredangle ABC \]so $B$, $L$, $E'$, $M$ are concyclic. Hence $E=E'$ and we are done.

My solution is essentially the same as yunxiu's, but it is less diagram dependent, and perhaps has technical advantages through the use of directed angles.

Let $H'$ be the reflection of $H$ in $BC$, which lies on $\Gamma$. Let $\Gamma_A$ be the reflection of $\Gamma$ in $BC$; analogously define $\Gamma_C$. Define $X \equiv HK \cap H'M$; which lies on $BC$, the axis of reflection. Finally, let $E' \equiv H'M \cap \Gamma$. Using the reflections, we have $\measuredangle HMB = \measuredangle BKH' = \measuredangle BAH' = 90^{\circ} - \measuredangle CBA$; similarly, $\measuredangle BLH = 90^{\circ} - \measuredangle CBA$. Now \begin{align*}\measuredangle MBL = \measuredangle ABL + \measuredangle MBA &= \measuredangle KBA + \measuredangle MBA \\ &= (\measuredangle MBA + \measuredangle CBM + \measuredangle KBC) + \measuredangle MBA \\ &= 2\measuredangle CBA. \end{align*} $\therefore \measuredangle MBL + \measuredangle HMB + \measuredangle BLH = 180^{\circ}$, so $L, M, H$ are collinear. $\therefore \measuredangle BE'M = \measuredangle BE'H' = \measuredangle BAH' = \measuredangle BLM$, so $E'MBL$ is a cyclic quadrilateral, and $E'$ lies on the circumcircle of $\triangle BLM$ and also on $\Gamma$; it follows that $E' \equiv E$. $\therefore KH \cap EM \cap BC \equiv X$, so these lines are concurrent, as required.

Dear Mathlinkers, 1. according to Carnot, the symmetric U od H wrt BC is on gamma 2. HUKM is a isoceles trapeze with BC as axis of symmetry And we are done… Sincerely Jean-Louis

@jayme, aren't you suppose to prove E,M,U are collinear? Sorry if I'm missing something..

My proof is the same as that of v_Enhance so i wouldnot post it. But one observation: $ L,M,H$ are collinear.

It's pretty evident that L,M,H are collinear by homothety about K, but is there a way to use that to prove the problem statement? I thought this fact seemed interesting and nontrivial, which suggeted that it might be useful for solve the problem. However, I couldn't actually put it into use.

yunxiu wrote: 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.) Luxembourg (Pierre Haas) Here is my solution. First, we present two simple lemmas, Lemma $(1).$ Let $A,B,C,D$ be four distinct points such that, the quadrilateral $ABCD$ is convex. If there exists a line $\ell$ such that: the reflection of the points $A,D$ in $\ell$ are the points $B,C,$ respectively. Then, the intersection of $AC$ and $BD$ lies on $\ell.$ Proof. Let $X$ be the intersection of $AC$ and $\ell,$ and let $D_1$ be the intersection of $BX$ and $CD.$ We deduce from thales' theorem that $XC=XD_1,$ but $XC=XD$ and the points $C,D,D_1$ are collinear, hence $D\equiv D_1$ and this complete the proof. Lemma $(2).$ the reflections of the orthocenter of any triangle $XYZ$ about its sides lie on its circumcircle. Proof. Just angle chasing. Let $H_1$ be the reflection of $H$ in the line $AB,$ from $(1)$ we deduce that the points of intersection of the lines $HL$ and $KH_1$ lies on $AB,$ and from $(2)$ we deduce that $H_1$ lies on the circle $(\Gamma),$ hence, we get $\angle KLH=\angle KH_1H=\angle KH_1C.$ Because $M,L$ are the reflection of $K$ in the lines $BC,AB$ respectively, we have $BM=BK=BL$ and $B$ is the circumcircle of the triangle $LMK,$ hence, we have \[\angle KLM=\frac{1}{2}\angle KBM=\angle KBC=\angle KH_1C.\]We deduce that the points $L,H,M$ are collinear. Now, because \begin{align*}\angle BMC=\angle CKB=180^\circ-\angle BAC=180^\circ-\left[ (90^\circ-\angle ACB)+(90^\circ-\angle CBA)\right]=\angle BHC,\end{align*}we deduce that the quadrilateral $BHMC$ is cyclic. Therefore, \begin{align*}\angle BEM=\angle BLM=\angle LMB=\angle HMB=\angle HCB=\angle BAH. \end{align*}Beacuse the triangle $ABC$ is acute-angled, then $H$ lies inside it, and since $\angle BAH=\angle BEM$ we deduce that the intersection of $AH$ and $EM$ is a point $H_2$ lies on the circle $(\Gamma).$ From $(2)$ we deduce that $H_2$ is the reflection of $H$ in the line $AB$ and since, $M$ is the reflection of $K$ in this line we deduce from $(1)$ that the point of intersection of $MH_2$ and $HK$ lies on the line $AB.$ We conclude that the lines $KH$, $EM$ and $BC$ are concurrent.

Solution: Claim: $L $, $H $ and $M $ are collinear. Proof of the claim: Let $LM\cap AH = I $ and let $H'$ be the reflection of $H $ in $BC $ (which lies on $\Gamma $). Observe that $K$ is the center of the spiral similarity that takes $LM $ to $AC $. So, $\measuredangle MKL = \measuredangle CKA $. Since, $B$ is the circumcenter of $\Delta MKL $, so, $\measuredangle BLM = 90 - \measuredangle MKL = 90 - \measuredangle CKA = \measuredangle BAH $. Thus, $B$, $L$, $A$ and $I $ are concyclic. Also, $\measuredangle AHB = \measuredangle BCA= \measuredangle ALB $. So, $B$, $L$, $A$ and $H $ are also concyclic. Hence, $H = I $. Back to the problem: Since, $HMKH'$ is an isosceles trapezoid, so, $BC $, $HK $ and $H'M $ are concurrent. Let $H'M\cap \Gamma = N $. It would be enough to prove that $L$, $B$, $M$ and $N $ are concyclic, which follows directly, as $\measuredangle BNH' = \measuredangle BAH' = \measuredangle BLH $. P.S.- All angles above are directed modulo $180$°.

v_Enhance wrote: [asy][asy]size(8cm); pointpen=black; pathpen=black; pair A = Drawing("A", dir(50), dir(50)); pair B = Drawing("B", dir(209), dir(180)); pair C = Drawing("C", dir(331), dir(0)); Drawing(A--B--C--cycle); pair H = orthocenter(A,B,C); pair Ka = foot(A,B,C); pair Kc = foot(C,A,B); Drawing("H", H, dir(45)); pair Ha = Drawing("H_A", 2*Ka-H, dir(-70)); pair Hc = Drawing("H_C", 2*Kc-H, dir(135)); pair K = Drawing("K", dir(-64), dir(-90)); pair M = Drawing("M", reflect(B,C) * K, dir(180)); pair L = Drawing("L", reflect(B,A) * K, dir(90)); pair E = Drawing("E", 2*foot(origin,M,Ha)-Ha, dir(70)); Drawing(A--Ha--B--Hc--C); Drawing(L--Hc); Drawing(Hc--E, dotted); draw(unitcircle); draw(Ha--E); draw(B--H--K); draw(L--K--M,dashed); [/asy][/asy] In what follows, all angles are directed. Let $H_A$ and $H_C$ be the reflections of $H$ across $\ol{BC}$ and $\ol{BA}$, which lie on $\Gamma$. Let $E'$ be the second intersection of line $H_AM$ with $\Gamma$. By construction, lines $E'M$ and $HK$ concur on $\ol{BC}$. First, we claim that $L$, $H_C$, and $E'$ are collinear. Notice that \[ \measuredangle LH_CB = -\measuredangle KHB = \measuredangle MH_AB \]by reflections, and that \[ \measuredangle MH_AB = \measuredangle E'H_AB = \measuredangle E'H_CB \]as desired. Now, \[ \measuredangle LE'M = \measuredangle H_CE'H_A = \measuredangle H_CBH_A = 2\measuredangle ABC \]and \[ \measuredangle LBM = \measuredangle LBK + \measuredangle KBM = 2\measuredangle ABK + 2\measuredangle KBC = 2\measuredangle ABC \]so $B$, $L$, $E'$, $M$ are concyclic. Hence $E=E'$ and we are done. why does showing those 4 four points being concyclic imply that $E=E'$, and that the other three points are collinear?

bump? /8char

By definition $E'$ lies on $(ABC)$ and once we show those points are concyclic then we know $E'$ lies on $(BLM)$, thus $E=E'$, as $E$ is the points on both $(BLM)$ and $(ABC)$.

claserken wrote: By definition $E'$ lies on $(ABC)$ and once we show those points are concyclic then we know $E'$ lies on $(BLM)$, thus $E=E'$, as $E$ is the points on both $(BLM)$ and $(ABC)$. ahh okay thanks, that makes sense. having trouble wrapping my head around on how to be rigorous for some reason lol

Tumon2001 wrote: Solution: Claim: $L $, $H $ and $M $ are collinear. Proof of the claim: Let $LM\cap AH = I $ and let $H'$ be the reflection of $H $ in $BC $ (which lies on $\Gamma $). Observe that $K$ is the center of the spiral similarity that takes $LM $ to $AC $. So, $\measuredangle MKL = \measuredangle CKA $. Since, $B$ is the circumcenter of $\Delta MKL $, so, $\measuredangle BLM = 90 - \measuredangle MKL = 90 - \measuredangle CKA = \measuredangle BAH $. Thus, $B$, $L$, $A$ and $I $ are concyclic. Also, $\measuredangle AHB = \measuredangle BCA= \measuredangle ALB $. So, $B$, $L$, $A$ and $H $ are also concyclic. Hence, $H = I $. Back to the problem: Since, $HMKH'$ is an isosceles trapezoid, so, $BC $, $HK $ and $H'M $ are concurrent. Let $H'M\cap \Gamma = N $. It would be enough to prove that $L$, $B$, $M$ and $N $ are concyclic, which follows directly, as $\measuredangle BNH' = \measuredangle BAH' = \measuredangle BLH $. P.S.- All angles above are directed modulo $180$°. How to prove $K$ is the center of a spiral similarity that takes $LM $ to $AC $?

Let $H'$ be the reflection of $H$ wrt $\overline{BC}$, $P = \overline{KL} \cap \overline{AB}$, and $Q = \overline{MK} \cap \overline{BC}$. Since $\angle BPK = \angle BQK = 90^\circ$ (this follows from reflections), $BPQK$ is cyclic. Then, \begin{eqnarray*} \angle LKM = \angle PKQ = \angle PBQ = B \end{eqnarray*}(where $B = \angle ABC$). It follows from reflections that $BK = BM = BL$. Therefore, $B$ is the circumcenter of $\triangle LMK$. Hence, \begin{eqnarray*} \angle LBM = 2\angle LKM = 2B. \end{eqnarray*}Since $LBM$ is isosceles, \begin{eqnarray*} \angle BLM & = & \frac{180^\circ - \angle LBM}{2} \\ & = & \frac{180^\circ - 2B}{2} \\ & = & 90^\circ - B. \end{eqnarray*}Therefore, \begin{eqnarray*} \angle BEM = \angle BLM = 90^\circ - B. \end{eqnarray*} It is well known that $H'$ lies on $(ABC)$. Noting that $\overline{AH'} \perp \overline{BC}$, we have \begin{eqnarray*} \angle BEH' = \angle BAH' = 90^\circ - B. \end{eqnarray*} Therefore, \begin{eqnarray*} \angle BEM = \angle BEH', \end{eqnarray*}which is enough to imply that $E, M, H'$ are collinear. Hence, $\overline{EM} = \overline{MH'}$ is the reflection of $\overline{HK}$ wrt $\overline{BC}$. This implies that $\overline{HK} \cap \overline{BC} = \overline{EM} \cap \overline{BC}$. Thus, $\overline{HK}, \overline{EM},$ and $\overline{BC}$ are concurrent, as desired. $\Box$

Let $F, N$ be the reflections of $H$ in $BC, AB$ respectively, it's well known that both $F,N$ lie on $\Gamma$. By their definition we have: $FM,BC,HK$ are concurrent in a point $X$ and $LN,AB,HK$ are also concurrent in a point $Y$. Let $E'\equiv (FM)\cap (LN)$; we claim that $E=E'$ which is equivalent to $E\in (MX)$ which is what we want to prove. $\textbf{Claim\ 1.}\ E'\in \Gamma.$ $\textbf{Proof.}\ \angle NE'F=\angle YE'X=180-\angle E'YX-\angle E'XY=180-(180-2\angle XYB)-(180-2\angle YXB)$ $=180-2\angle YBX=180-2\angle YBH -2\angle HBX=180-\angle NBF$. $\textbf{Claim\ 2.}\ E'\in (LBM).$ $\textbf{Proof.}\ \angle LBM=\angle LBA+ \angle ABM=\angle KBA+ \angle ABM$ $=\angle KBC+\angle CBA +\angle CBA -\angle CBM=2\angle CBA=\angle NBF=180-\angle NE'F.$ By Claim 1 and 2 we deduce that $E'=E$ as wanted.

anantmudgal09 wrote: My proof is the same as that of v_Enhance so i wouldnot post it. But one observation: $ L,M,H$ are collinear. vsathiam wrote: It's pretty evident that L,M,H are collinear by homothety about K, but is there a way to use that to prove the problem statement? I thought this fact seemed interesting and nontrivial, which suggeted that it might be useful for solve the problem. However, I couldn't actually put it into use. Here is an outline for a solution that uses this fact. Let $N$ be the reflection of $K$ over $AC$, so $L,H,M,N$ are collinear. Let $X$, $Y$, and $Z$ be the feet of the perpendiculars from $K$ to $AB$, $AC$, and $BC$, respectively. Let $KL$ and $KN$ meet $BC$ at $R$ and $S$, respectively. Define $P=HK \cap BC$. We wish to show that $E, M, P$ are collinear. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.5) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -15.221172766627305, xmax = 9.021251475796944, ymin = -5.927701692247144, ymax = 7.209370326643048; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw(circle((-3.62,0.33), 4.8845880071916), linewidth(1) + wrwrwr); draw(circle((-10.834010068032772,5.993741903880838), 8.682595696196357), linewidth(1) + wrwrwr); draw(circle((-0.750226398326001,1.446911282614519), 3.8459692000793124), linewidth(1) + wrwrwr); draw((-6.185603077820917,-0.4826952164090483)--(-4.38781227002869,-4.493864044310681), linewidth(1) + wrwrwr); draw((-4.38781227002869,-4.493864044310681)--(2.832183361700406,2.846132279893439), linewidth(1) + wrwrwr); draw((-11.96255112469488,-2.6151989779602896)--(2.832183361700406,2.846132279893439), linewidth(1) + linetype("4 4") + wrwrwr); draw((-8.175181697361785,-3.554531511135485)--(-0.7778144541641416,-0.8238658822086206), linewidth(1) + linetype("4 4") + wrwrwr); draw((-6.18,4.49)--(-8.175181697361785,-3.554531511135485), linewidth(1) + wrwrwr); draw((-6.18,4.49)--(0.5816300849103664,-2.161085030579269), linewidth(1) + wrwrwr); draw((-13.859135172380286,-2.144813626841671)--(0.5816300849103664,-2.161085030579269), linewidth(1) + wrwrwr); draw((-2.20778230183693,5.005985583061074)--(-5.4363942488225545,-2.154304108523395), linewidth(1) + linetype("4 4") + wrwrwr); draw((-13.859135172380286,-2.144813626841671)--(-4.38781227002869,-4.493864044310681), linewidth(1) + wrwrwr); /* dots and labels */ dot((-6.18,4.49),dotstyle); label("$A$", (-6.515938606847688,4.650810704447064), NE * labelscalefactor); dot((-7.827233162731285,-2.151610185829781),dotstyle); label("$B$", (-8.39983116883116,-2.070952380952381), NE * labelscalefactor); dot((0.5816300849103664,-2.161085030579269),dotstyle); label("$C$", (0.7253049980322824,-2.464498229043683), NE * labelscalefactor); dot((-6.185603077820917,-0.4826952164090483),linewidth(3pt) + dotstyle); label("$H$", (-6.515938606847688,-0.2606414797323897), NE * labelscalefactor); dot((-4.38781227002869,-4.493864044310681),dotstyle); label("$K$", (-4.390791027154653,-4.942998819362454), NE * labelscalefactor); dot((-8.175181697361785,-3.554531511135485),linewidth(3pt) + dotstyle); label("$X$", (-8.404958677685942,-4.040377804014154), NE * labelscalefactor); dot((-13.859135172380286,-2.144813626841671),linewidth(3pt) + dotstyle); label("$R$", (-14.150728059818961,-2.653014561196379), NE * labelscalefactor); dot((-4.385177461520879,-2.1554885868011855),linewidth(3pt) + dotstyle); label("$Z$", (-4.427823691460045,-2.0237268791814245), NE * labelscalefactor); dot((-0.7778144541641416,-0.8238658822086206),linewidth(3pt) + dotstyle); label("$Y$", (-0.8803620621802328,-0.6699291617473442), NE * labelscalefactor); dot((-2.0902102447239272,-2.1580744842577064),linewidth(3pt) + dotstyle); label("$S$", (-2.1239669421487495,-2.608949232585596), NE * labelscalefactor); dot((-11.96255112469488,-2.6151989779602896),linewidth(3pt) + dotstyle); label("$L$", (-12.072805981896883,-3.1312042502951586), NE * labelscalefactor); dot((2.832183361700406,2.846132279893439),linewidth(3pt) + dotstyle); label("$N$", (2.8976780794962735,2.96643447461629), NE * labelscalefactor); dot((-4.382542653013067,0.18288687070831036),linewidth(3pt) + dotstyle); label("$M$", (-4.1389216843762195,-0.07173947264856458), NE * labelscalefactor); dot((-2.20778230183693,5.005985583061074),linewidth(3pt) + dotstyle); label("$E$", (-2.1397087760724016,5.1388075560802795), NE * labelscalefactor); dot((-5.4363942488225545,-2.154304108523395),linewidth(3pt) + dotstyle); label("$P$", (-5.913105076741431,-2.5857890594254227), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] 1. Show that $R \in (BLM)$ and $S \in (CNM)$. This is angle chasing using the fact that $XY \parallel LN$ and cyclic quads $KZYC$ and $KZBX$. 2. Let $E'$ be the second point of intersection of $(BLRM)$ and $(CNSM)$. Show that $E'=E$, or equivalently that $E' \in \Gamma$. This is also angle chasing using cyclic quads and the fact that $MSKR$ is a kite. 3. Note that since $BH \parallel KS$ and $CH \parallel KR$, we have similar figures $BHCP \sim SKRP$. From this it follows that $$\frac{PB}{PC}=\frac{PS}{PR} \implies PB \cdot PR=PC \cdot PS$$Thus $P$ lies on the radical axis of $(BME)$ and $(CME)$, so $E,M,P$ are collinear as desired. $\blacksquare$

Let $H'$ be the reflection of $H$ over $BC$. It is enough to show that $E, M, H'$ are collinear, since $HK$ and $MH'$ intersect on $BC$. Since \begin{align*} \angle LBM &= \angle LBA + \angle ABM\\ &= \angle ABK + \angle B - \angle CBK\\ &= 2\angle B \end{align*}and $BL = BK = BM$, $EB$ bisects $\angle LEM$, and in particular, $\angle BEM = 90 - \angle B$. Also, $\angle BEH' = \angle BAH' = 90 - \angle B$, so $E, M, H'$ are collinear, as desired.

yunxiu wrote: 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.) Luxembourg (Pierre Haas) A projective solution: $1.$ Let the intersection of HK with BC be S.Now fix $A,B,C$ and move $K$ along $\Gamma$ $2.$ look at this projective map : $ S \mapsto K \mapsto L \mapsto M \mapsto E$ $3.$ now we need to determine $3$ special cases $(i)$ $K$ on $B$ $(ii)$ $K$ on $C$ $(iii)$ $K$ on $AH \cap \Gamma$ finished!

I think it's different Solution. Claim 1. $L,M,H$ is collinear. Proof. Take a homothety at $K$ with factor of $\frac 1 2$, then by Simson Line Bisection, the three points are collinear. Let $H'$ be the reflection of $H$ over $BC$, it's well-known that $H' \in (ABC)$. Let $KH \cap BC = D$. Let $H'D \cap (ABC)=E$. It's enough to show that $LBME$ is cyclic or $\angle BLM = \angle BLH = \angle BEH' = \angle BAH$ which is equivalent to proving $ALBH$ cyclic. This is true since $\angle ALB = \angle AKB = \angle C$ and by easy angle chasing $\angle AHB = \pi-\angle C$.

EGMO 2012 P7 wrote: 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.) Luxembourg (Pierre Haas) Collinearity of $L,H,M$ directly follows from the fact that $L$ and $M$ lie on the Steiner line and the Steiner line passes through $H$. Reflect orthocenter $H$ about $BC$ to $H_{A}$ which lies on $(ABC)$. It is clear that $HMKH_{A}$ is an isoceles trapezoid since $BC$ is a common perpendicular bisector. Then extend $MH_{A}$ to meet $(ABC)$ at $E_1$. By simple angle chasing it can be proven that $E_1$ lies on $(BLM)$ and we conclude $E_1= E$.

Clearly not a unique solution, but maybe a different approach: Claim. $L,M,H$ are collinear. We have $h=a+b+c$, $m=b+c-\frac{bc}{k}$ and $l=a+b-\frac{ab}{k}$. We have $$\frac{h-l}{\overline{h}-\overline{l}}=\frac{a+b+c-a-b+\frac{ab}{k}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a}-\frac{1}{b}+\frac{k}{ab}}=\frac{abc}{k}$$and $$\frac{h-m}{\overline{h}-\overline{m}}=\frac{a+b+c-b-c+\frac{bc}{k}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{b}-\frac{1}{c}+\frac{k}{bc}}=\frac{abc}{k}.$$Claim follows. Let $H'$ be the reflection of $H$ over $BC$. Claim. $E, M, H'$ are collinear. $$\measuredangle BEM=\measuredangle BLM=\measuredangle LMB=\measuredangle HMB=\measuredangle BKH'=\measuredangle BAH'=\measuredangle BEH'$$Claim follows. Now notice that since $MHH'K$ is an isosceles trapezoid, the perpendicular bisector of $MK$, $BC$ passes through the intersection of $KH$ and $MH'\equiv EM$.

Let $H_a$ be the reflection of $H$ across $BC$, and let $H_c$ be the reflection of $H$ across $AB$. Note that $MH_a,KH,BC$ concur. Ignore the defintion of point $E$ given and let $E=H_aM\cap\Gamma$. It suffices to show that $LBME$ is cyclic. Claim 1: (well-known) $H_a$ lies on $\Gamma$. Note that $\angle H_aBC=\angle CBH=90-\angle ACB$ while $\angle H_aCB=\angle BCH=90-\angle ABC$. Summing these, we have $180-\angle BH_aC=\angle BAC$, so $ABH_aC$ is cyclic. Claim 2: $L,E,H_c$ are collinear. This is true since $$\measuredangle LH_cB=-\measuredangle KHB=\measuredangle MH_aB=\measuredangle EH_aB=\measuredangle EH_cB.$$ Claim 3: $L,B,M,E$ are concyclic We have $$\measuredangle LEM=\measuredangle LEH_a=\measuredangle H_cEH_a=\measuredangle H_cBH_a=2\measuredangle ABC$$and $$\measuredangle LBM=\measuredangle LBK+\measuredangle KBM=2\measuredangle ABK+2\measuredangle KBC=2\measuredangle ABC,$$so the result holds.

ok I think it's right this time Let $H'$ denote the reflection of $H$ over $\overline{BC}$, $K'$ the midpoint of $\overline{MK}$ (which lies on $\overline{BC}$), and $L'$ the midpoint of $\overline{LK}$ (which lies on $\overline{AB}$). It suffices to show that $H', M, E$ are collinear, since $\overline{KH}$ and $\overline{MH'}$ intersect on $\overline{BC}$ by symmetry. Note that by reflections we have $BL=BM=BK$, so $B$ is the center of $(LMK)$. A homothety of scale factor $\tfrac{1}{2}$ at $K$ thus shows that $BL'M'K$ is cyclic. Further, from $BL=BM$ it follows that $\overline{EB}$ bisects $\angle LEM$. Thus we have \begin{align*} \measuredangle BEM&=\measuredangle LEB\\ &=\measuredangle LMB\\ &=90^\circ-\frac{\measuredangle LBM}{2}\\ &=90^\circ-\measuredangle LKM\\ &=90^\circ-\measuredangle L'KM'\\ &=90^\circ-\measuredangle L'BM'\\ &=90^\circ-\measuredangle ABC\\ &=\measuredangle BAH'\\ &=\measuredangle BEH', \end{align*}where we are allowed to write $\tfrac{\measuredangle LBM}{2}$ because we treat it as an arc measure (modulo 360 degrees) rather than a conventional directed angle modulo 180 degrees. This implies that $E,M,H'$ are collinear, so we're done. $\blacksquare$

Let $AH$ intersect $\Gamma$ again at $A'$, and let $KL$ intersect $BC$ at $P$. Moreover, let $KM$ and $KL$ intersect $BC$ and $BA$ at $X$ and $Y$, respectively, so that $BKXY$ is cyclic. Claim: $P$ lies on $(BLM)$. Proof: We have $$\measuredangle MBL = \measuredangle KBL - \measuredangle KBM = 2\measuredangle KBA - 2\measuredangle KBC = 2\measuredangle CBA = 2\measuredangle XBY = 2\measuredangle XKY = 2\measuredangle MKP = \measuredangle MPK = \measuredangle MPL$$ as desired. Claim: $A'$, $M$, and $E$ are collinear. Proof: We have $$\measuredangle A'EP = \measuredangle A'EB - \measuredangle PEB = \measuredangle A'AB - \measuredangle PLB = (90^\circ - \measuredangle ABC) - (90^\circ - \measuredangle ABK) = \measuredangle CBK = \measuredangle MBP = \measuredangle MEP$$ as desired. Since $HA'KM$ is an isosceles trapezoid symmetric about $BC$, $MA'$ and $HK$ concur on $BC$, as desired.

EGMO predicted BLM?! Let $H_A$ denote the reflection of $H$ over $BC$. It is sufficient to show that $EM$ passes through $H_A$. Observe that $AHBL$ and $BHMC$ are cyclic as they are reflections of $(ABC)$ over $AB,BC$, respectively. Thus, \[ \measuredangle LHB=\measuredangle LAB=\measuredangle BAK=\measuredangle BCK=\measuredangle MCB=\measuredangle MHB, \]which proves Steiner's theorem, which states that $L,H,M$ are collinear. Finally, \[ \measuredangle BEM=\measuredangle BLM=\measuredangle BLH=\measuredangle BAH=\measuredangle BAH_A=\measuredangle BEH_A, \]which proves the desired. $\blacksquare$ [asy][asy]import olympiad;import geometry; size(10cm);defaultpen(fontsize(10pt)); pair A,B,C,H,K,L,M,Ha,E; A=dir(120);B=dir(205);C=dir(335);H=orthocenter(A,B,C);K=dir(265);L=2*foot(K,A,B)-K;M=2*foot(K,B,C)-K;Ha=2*foot(A,B,C)-H; E=2*foot((0,0),M,Ha)-Ha; draw(A--B--C--cycle);draw(circumcircle(A,B,C));draw(L--M);draw(K--H^^Ha--E,dotted); draw(circumcircle(A,B,H)^^circumcircle(H,B,C)^^circumcircle(B,L,M),dashed);draw(A--Ha); clip((1.1,-1.1)--(-2,-1.1)--(-2,1.1)--(1.1,1.1)--cycle); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$H$",H,dir(H)); dot("$K$",K,dir(K)); dot("$H_A$",Ha,dir(Ha)); dot("$L$",L,dir(L)); dot("$M$",M,dir(M)); dot("$E$",E,dir(E)); [/asy][/asy]

Well $H$ has no usage in our problem so lets make it useful. Let $S,P,Q$ be reflections of $H$ across $BC,AB,AC$. Note that $HSKM$ is isosceles trapezoid so $S,KH,BE$ are concurrent so we need to prove $E,M,S$ are collinear. Let $SM$ meet $\Gamma$ at $E'$ we will prove $LE'MB$ is cyclic. Claim : $L,P,E'$ are collinear. Proof : Note that $LPHK$ is isosceles trapezoid so $\angle LPB = \angle BHK = \angle BHS + \angle SHK = \angle C + \angle E'SA = \angle E'CB \implies L,E',P$ are collinear. Note that $B$ lies on perpendicular bisector of $HS,KM$ so $\angle BMS = \angle BKH = \angle BLP = \angle BLE'$ so $LE'MB$ is cyclic as wanted. we're Done.

Let $X$ be the reflection of $H$ over $BC$ Note that the problem basically asks us to show that $X-M-E$ and also observe that $B$ is the centre of $\odot(KLM)$ Also notice that $L-H-M$ because $LM$ is the steiner line of $K$ Now $$\angle MEB = \angle MLB = 90^{\circ} - \angle LKM = 90^{\circ}-\angle ABC = \angle XAH = \angle XEB \implies KH \ \cap \ ME \in BC \ \ \blacksquare $$

Let $H_A$ be the reflection of $H$ in $\overline{BC}.$ Notice $H,M,$ and $K$ are collinear by the Simson Line Bisection Lemma. Also, $AHBL$ is cyclic as $$\measuredangle AHB=\measuredangle BCA=\measuredangle BKA=\measuredangle ALB.$$Thus, $$\measuredangle BEM=\measuredangle BLM=\measuredangle BAH=\measuredangle BCH_A=\measuredangle BEH_A$$and $H_A,M,$ and $E$ are collinear. Since $\overline{KH_A}$ is the reflection of $\overline{MH}$ in $\overline{BC},$ we are done. $\square$

Solved with L567, mxlcv and jelena_ivanchic! Add $H_A,H_C$, the reflection of $H$ in $BC,BA$ respectively. Due to reflections we have $KH, MH_A, BC$ concurrent. Hence we want to prove $H_A-M-E$. We have $L-H-M$ by Steiner Line. As $ABH_CK$ cyclic, reflection in $AB$ implies $ABHL$ cyclic. We have been given that $BMEL$ cyclic. Hence, $\angle MEB=\angle MLB=\angle HLB=\angle HAB=\angle H_AAB=\angle H_AEB$. This implies $H_A-M-E$, and we are done!

Let $H’$ be the reflection of $H$ across $BC$. $I=EM\cap HK$ We prove $B,I$ and $C$ are collinear or by letting $E’=MH’\cap \Gamma$ it suffices to prove $LE’MB$ cyclic (because $MH’$, $HK$ and $BC$ are obviously concurrent). Angle chasing the diagram we get $LAHB$ and $BHMC$ are both cyclic. Which implies $M,H$ and $L$ are indeed collinear. Now we’re ready to prove the main cyclicity , We have $\angle LME’=180^{\circ}-\angle HMH’$ and $\angle LBE’= \angle LBA+ \angle ABE= \angle ABK+ \angle HH’M$ They’re equal iff $$\angle HMH + \angle ABK+ \angle HH’M =180 ^{\circ} $$but $$\angle HMH+ \angle HH’M=180 ^{\circ}-\angle H’HM$$$$=180 ^{\circ}-\angle H’HC-\angle CHM=180 ^{\circ}-\angle FHA-\angle KBC$$$$90 ^{\circ} + \angle BAD-\angle KBC=180 ^{\circ}-\angle ABK.$$Where $F$ and $D$ are the foot of the altitudes.

First, note that $H,L,M$ are collinear $(\ast)$, because \[\angle BHL+\angle BHM\stackrel{(\ast)}{=}\angle BH_BK+\angle BH_AK\stackrel{(ABC)}{=}180^\circ.\]Let $H_A,H_C$ be the reflections of $H$ across $\overline{BC},\overline{AB}$ respectively, so that clearly $HH_AKM$ and $HH_BLK$ are isosceles trapezoids $(\dagger)$. Since $BC,HK,H_AM$ concur, it remains to prove that $H_A,M,E$ are collinear. Indeed, we claim that the intersection $E'$ of $H_AM$ and $(BLM)$ lies on $(ABC)$, which suffices since such a point is unique. Write \[\angle BE'H_A=\angle BE'M\stackrel{(BE'LM)}{=}\angle BLM\stackrel{(\ast)}{=}\angle BLH\stackrel{(\dagger)}{=}\angle BKH_B=90^\circ-B=\angle BAH_A,\]and we are done. $\square$

First, extend $\overline{EM}$, and have $(ABC) \cap \overline{EM}$ be $X$. Note that $\overline{BL}=\overline{BK}=\overline{BM}$, and that $BMHC$ is cyclic (by a reflection over $\overline{BC}$. Claim: $KH$, $EM$ and $BC$ are concurrent iff $X$ is the reflection of $H$ over $\overline{BC}$. Proof: We note that $MHH'K$ is an isosceles trapezoid with axis of symmetry $BC$, implying the claim. Claim: $\triangle{BHC}$ is congruent to $\triangle{BXC}$. Proof: We note that $\angle{BCX}=\angle{BEX}=\angle{BLM}=\angle{BML}=\angle{BCH}$. We also note that $\angle{BHC}=\pi-(\frac{\pi}{2}-\angle{C}+\frac{\pi}{2}-\angle{B})=\pi-\angle{A}=\angle{BXC}$. Therefore, the 2 triangles are congruent, implying the conclusion.

Let me provide a solution using Stiener line Let $H_a$ be the reflection of $H$ in the line $BC$ (It’s known that $H_a \in \Gamma$) Since $H$, $H_a$ and $M$, $K$ are symmetric to $BC$ So $HK \cap MH_a \in BC$ Therefore, we only have to prove $E$, $M$, $H_a$ collinear Let $N$ be the reflection of $K$ in the line $AC$ Then $LMH$ is the Stiener line of $K$ with respect to $\triangle ABC$ So $L$, $H$, $M$, $N$ are collinear Let $H_aM \cap \Gamma =E’$, through easy angle chasing $$\measuredangle BE’M=\measuredangle BE’H_a =\measuredangle BKH_a=\measuredangle BMH=\measuredangle BML=\measuredangle BLM$$$\therefore \odot(BLME’) \implies E=E’ \implies H_a$, $M$, $E$ collinear $\implies$ $KH$, $EM$ and $BC$ are concurrent

Note that $E'M$ intersects at $H_A$. By construction $E'M$ intersects $HK$ at $BC$. Now we will prove $LH_CE$ is collinear. It can be easily seen that $\triangle H_CBL \sim \triangle HBK \sim \triangle H_ABM$. We can prove $\measuredangle LH_CB=\measuredangle BHK=\measuredangle MH_AB$ and $\measuredangle EH_CB=\measuredangle E'AB =\measuredangle E'CB= \measuredangle E'KB=\measuredangle E'CB=\measuredangle MH_AB$ $\Rightarrow \measuredangle LH_CB=\measuredangle EH_CB$, so $LH_CE$ is collinear. Now we prove $LE'MB$ is concyclic. $\measuredangle LE'M=\measuredangle H_CBH_A= \measuredangle ABC$. $\measuredangle LBM = \measuredangle LBK =\measuredangle KBC=\measuredangle MBC$. $\measuredangle CBD =\measuredangle ABC$ this gives use $\measuredangle H_ABD=\angle ABM$ $\Rightarrow LE'MB$ is concyclic. $\therefore$ $E \equiv E'$ and we are done.

Redefine $E$ is Anti - Steiner point of $HK$ WRT $\triangle ABC$. Then $EL, EM$ must pass through the intersections of $AH, CH$ with $(ABC),$ let them be $D, F$. We have $\angle{LEM} = \angle{DEF} = \angle{DEB} + \angle{BEF} = \angle{DAB} + \angle{BCF} = 2\angle{OAC} = 180^{\circ} - \angle{AOC} = 180^{\circ} - 2\angle{ABC} = 180^{\circ} - 2\angle{LKM} = 180^{\circ} - \angle{LBM}$. So $E \in (LBM)$

Unlike many of the solutions my proof uses the fact that $L$ , $M$ and $H$ are collinear, instead of obtaining it as a corollary. Let $H'$ be the reflection of $H$ across side $BC$ which is well known to lie on $(ABC)$. First of all note that $\overline{ML}$ is the Steiner Line of $K$ with respect to $\triangle ABC$ which is well known to pass through its orthocenter $H$. Thus, points $L$ , $M$ and $H$ are collinear. With this in hand we prove the following key claim. Claim : Points $E$ , $M$ and $H'$ are collinear. Proof : First of all, since $H'$ is the reflection of $H$ across $BC$ and $M$ is the reflection of $K$ across $BC$ , the quadrilateral $KH'HM$ is in fact an isosceles trapezoid, and hence cyclic. Simply note that, \[\measuredangle H'EB = \measuredangle H'KB = \measuredangle BMH = \measuredangle BML = \measuredangle MLB = \measuredangle MEB \]which indeed implies that points $E$ , $M$ and $H'$ are collinear. Now we are done since $\overline{EM}$ and $\overline{KH}$ are the diagonals of isosceles trapezoid $KMHH'$ which intersect on the common perpendicular bisector of the parallel sides, $\overline{BC}$.