Let $A'$ be reflection of $A$ wrt $BC$ and $T$ be foot of altitude from $A$ to $BC$.Then \[AM\cdot AA'=\tfrac{AH}{2}\cdot 2\cdot AT=AH\cdot AT=AD\cdot AC\].Hence $\odot{MDA'C}$ is cylic.Also clearly $\odot{ADEA'}$ is a cylic trapezium.Thus \[\angle{FCD}=\angle{MCD}=\angle{MA'D}=\angle{AED}=\angle{FED}\].Hence $\odot{FDEC}$ is cylic.Hence $F\in \odot{BCDE}$.Thus $\angle{BFC}=\angle{BDC}=90^{\circ}$ as desired.$\blacksquare$ [asy][asy] import olympiad; import geometry; size(10cm); pair O=origin; pair A,B,C,D,E,T,AT,H,M,K,F; A=dir(110); B=dir(210); C=dir(330); D=foot(B,A,C); T=foot(A,B,C); AT=2*T-A; H=orthocenter(A,B,C); M=(A+H)/2; K=foot(D,B,C); E=2*K-D; dot(A^^B^^C^^D^^E^^H^^AT^^T^^M); draw("$A$",A,N); draw("$B$",B,W); draw("$C$",C,NE); draw("$D$",D,NE); draw("$E$",E,SW); draw("$T$",T,NE); draw("$H$",H,NE); draw("$M$",M,NE); draw("$A'$",AT,SW); draw(segment(A,AT)); draw(circumcircle(A,B,C)); draw(circumcircle(B,D,C)); draw(circumcircle(M,D,C),dashed); draw(circumcircle(A,D,E),dashed); draw(A--B--C--A); [/asy][/asy]

Nice problem

Some other nice properties in the configuration. Property 1: wrote: $BP,CI,AO$ are concurrent. Proof:(by amar_04) $M$ is the centre of $\odot (ADHI)$. Brokard on $ADHI$ gives the desired result.$\square$ Property 2: wrote: $\overline{I-O-E}$ collinear. Proof: Just the fact that $DC=CE$ along with some trivial angle chase.$\square$.

Let AE intersect the circumcircle of BDC at X. Notice that BX is perpendicular to CX. Now if we prove CX intersect AH at M. Then we are done. Let say the feet of C-altitude on AB be K. Claim-(K,D;X,C) is harmonic Project (K,D;X,C) through E on AH. Let say EC intersect AH at T. K maps to L, where L is the feet of A-altitude on BC, X maps to A, D maps to point at infinite, and C maps to T, and clearly AL=LT. So (K,D;X,C)=-1 Now project (K,D;X,C) Through C on AH. D maps to A, K maps to H, C maps to point at infinity, and X maps to Y, the because of harmonic AY=HY, so Y coincide with M. So done

Is this correct? Let $(ADM)$ intersect $CM$ at $F'$. Note that $F'DCB$ is cyclic.$MD$ is tangent to $(DBC)$. Hence, $\angle{DCM}=\angle{F'DM}=\angle{F'AM}$. Let $Y$ be the orthogonal projection of $D$ onto $BC$. Extend $AF'$ and $DE$ till they meet each other at $E'$. It is $\angle{DE'F'}=\angle{F'AM}=\angle{DCM}$. Hence, $E'$ $\in (DF'BC)$. Note though that $E \in (DF'BC)$ too and they both belong on $DY$. Hence, $E'=E$ $\Rightarrow$ $F'=F$. Done.

neyft wrote: Is this correct? Let $(ADM)$ intersect $CM$ at $F'$. Note that $F'DCB$ is cyclic.$MD$ is tangent to $(DBC)$. Hence, $\angle{DCM}=\angle{F'DM}=\angle{F'AM}$. Let $Y$ be the orthogonal projection of $D$ onto $BC$. Extend $AF'$ and $DE$ till they meet each other at $E'$. It is $\angle{DE'F'}=\angle{F'AM}=\angle{DCM}$. Hence, $E'$ $\in (DF'BC)$. Note though that $E \in (DF'BC)$ too and they both belong on $DY$. Hence, $E'=E$ $\Rightarrow$ $F'=F$. Done. It's correct. But I think you might do justifications about your arguments...

The point naming is weird. Let me change it. Quote: Let $H$ be the orthocenter of $\triangle ABC$ and $M$ be the midpoint of $AH$. Let $E=BH\cap AC$ and $E'$ be the reflection of $E$ across $BC$. If $P = CM\cap AE'$, prove that $BP\perp CM$. Redefine $P$ as the intersection of $AE'$ and $\odot(BC)$. Let $F$ be the foot of altitude from $C$ to $AB$. Then we have $$-1 = (BC;EE') \stackrel{A}{=} (EF;CP) \stackrel{C}{=} (AH;\infty M')$$where $M'=CP\cap AH$. Thus $M'=M$ so $C,P,M$ are colinear as desired.

MarkBcc168 wrote: The point naming is weird. Let me change it. Quote: Let $H$ be the orthocenter of $\triangle ABC$ and $M$ be the midpoint of $AH$. Let $E=BH\cap AC$ and $E'$ be the reflection of $E$ across $BC$. If $P = CM\cap AE'$, prove that $BP\perp CM$. Redefine $P$ as the intersection of $AE'$ and $\odot(BC)$. Let $F$ be the foot of altitude from $C$ to $AB$. Then we have $$-1 = (BC;EE') \stackrel{A}{=} (EF;CP) \stackrel{C}{=} (AH;\infty M')$$where $M'=CP\cap AH$. Thus $M'=M$ so $C,P,M$ are colinear as desired. $FCEE'$ is the harmonic. $$-1 = (FC;EE') \stackrel{A}{=} (FE;CP) \stackrel{C}{=} (HA;\infty M')$$

$AH$ and $DE$ are both perpendicular to $BC$, which implies $\angle HAC = \angle EDC$. Also $CD=CE$ and $MA=MD$ which leads to: \[DAM \sim DEC \implies DAE \sim DMC \implies \angle AED = \angle MCD \implies FDCE \text{ is cyclic} \implies BF \perp CM. \]

Solution. Let $\varphi$ the inversion centered at $A$ with radius $\sqrt{AD\cdot AC}$. Because $BDCE$ is cyclic with diameter $\overline{BC}$, we prove that $\varphi(F)=E$, which clearly suffices. Let $M'$ be the reflection of $M$ across $BC$. Since $$\angle MDA=\angle MAD=\angle M'AC=\angle CM'A$$we have $\bigtriangleup AMD\sim\bigtriangleup ACM'$, thus $\varphi(CM)= (ADM')$. By symmetry, $E$ lies on $(ADM')$. Moreover, $AF$ coincides with its image, so $E=(ADM')\cap \overline{AF}=\varphi(F)$, as required.

Well, he asks to prove that the green line makes an angle of 90 ° c the red line. first extends CM until it touches AB at point A '. as H is the orthocenter, CH when touching the AB side forms a 90 ° angle with it. thus we draw A'K ', being perpendicular to BC. So, to prove that m <BFC is 90 °, just use the tactics for orthocenter demonstration. thus, we will start using quadrilateros inscritiveis (I denote "#" to represent quadrilaterals): we see that # JK'CA 'is inscribable, because m <CK'A' = m <CJA ', because they are looking at the same arc of circumference. if we denote the smallest angle measurement in A 'of alpha, we fear that the angle measurement JCA' is 90-alpha, because m <CJA '= 90 °. thus, since # JK'CA 'is inscribable, m <JCA' = m <JK'A '= 90-alpha. denoting the meeting of A'K 'and CJ of H', we now look at # JBK'H ', which is also inscribable, as the opposite angles are supplementary (in this case 90 and 90). therefore, m <JK'H '= m <JBH' = 90-alpha. Now we look at the triangle A'BF, if the angle at A 'is alpha and the angle at B is 90-alpha, we have left that the angle at F must be 90 °, cqd

Pluto1708 wrote: Let $A'$ be reflection of $A$ wrt $BC$ and $T$ be foot of altitude from $A$ to $BC$.Then \[AM\cdot AA'=\tfrac{AH}{2}\cdot 2\cdot AT=AH\cdot AT=AD\cdot AC\].Hence $\odot{MDA'C}$ is cylic.Also clearly $\odot{ADEA'}$ is a cylic trapezium.Thus \[\angle{FCD}=\angle{MCD}=\angle{MA'D}=\angle{AED}=\angle{FED}\].Hence $\odot{FDEC}$ is cylic.Hence $F\in \odot{BCDE}$.Thus $\angle{BFC}=\angle{BDC}=90^{\circ}$ as desired.$\blacksquare$ [asy][asy] import olympiad; import geometry; size(10cm); pair O=origin; pair A,B,C,D,E,T,AT,H,M,K,F; A=dir(110); B=dir(210); C=dir(330); D=foot(B,A,C); T=foot(A,B,C); AT=2*T-A; H=orthocenter(A,B,C); M=(A+H)/2; K=foot(D,B,C); E=2*K-D; dot(A^^B^^C^^D^^E^^H^^AT^^T^^M); draw("$A$",A,N); draw("$B$",B,W); draw("$C$",C,NE); draw("$D$",D,NE); draw("$E$",E,SW); draw("$T$",T,NE); draw("$H$",H,NE); draw("$M$",M,NE); draw("$A'$",AT,SW); draw(segment(A,AT)); draw(circumcircle(A,B,C)); draw(circumcircle(B,D,C)); draw(circumcircle(M,D,C),dashed); draw(circumcircle(A,D,E),dashed); draw(A--B--C--A); [/asy][/asy] Excellent construction I did a complex bash to this problem

I guess this solution is not posted yet... Let $X$ be a point on $CM$ such that $BMDC$ is cyclic,we intend to show that $X=F$ We prove that power of $M$ w.r.t.$(BXDC)$ is $MA^2$ Since.Let $N$ be the midpoint of $BC$.We have $MN=2R_{N_9}=R$ therefore $NM^2-NC^2=R^2\times (1-cos^2A)=R^2sin^2A= AM^2$ Thus $MA^2=MX\times MC\Rightarrow \angle MAX=\angle MCA=\angle MCD=\angle XCD=\angle XED$ Since $AM\parallel DE$ we definitely have $A,X,E$ collinear thus $X=F$

plagueis wrote: Let $H$ be the orthocenter of acute-angled triangle $ABC$ and $M$ be the midpoint of $AH$. Line $BH$ cuts $AC$ at $D$. Consider point $E$ such that $BC$ is the perpendicular bisector of $DE$. Segments $CM$ and $AE$ intersect at $F$. Show that $BF$ is perpendicular to $CM$. Let $AF \cap AC = F’$. Note that $\measuredangle{DFA}$ is bisected by $FM$. So it is enough to prove that $BF$ bisects $\measuredangle{DFA}$ which is equivalent to proving that $(A,F’;D,C) = -1$. This follows easily from $-1 = (E,B;D,C) \stackrel{F}{=} (A,F’;D,C)$ $\blacksquare$.

There exist a solution involving HM-point. P.S I am to lazy to post it

Please remove this post.

A different solution. Let $F'$ be a point in $CM$ such that $BF' \perp CM$. We need to show that $A, F'$ and $E$ are collinear (implying that $F'=F$). Let $\ell$ be the perpendicular line to $AB$ through $A$. Let $X, Y$ be the intersections of $\ell$ with $EC$ and $CM$, respectively. Since both $XA$ and $CH$ are orthogonal to $AB$, then $XY||CH$. Moreover, since $M$ is midpoint of $AH$, then $AYHC$ is a parallelogram. Thus, $CA||HY$. Now, by simple angle chasing $\angle YHD = \angle YHA + \angle AHD = (90 - \angle C) + \angle C = 90$. Therefore, $AHBY$ is cyclic and $\angle XYB = \angle C$. Since $\angle BCX = 180 - \angle C$, then $BCXY$ is cyclic. Finally, note that $A, F', E$ belong to a Simson line (perpendicular feet from $B$ to sides of triangle $CXY$), which concludes the proof.

Note that MA = MD and CD = CE and ∠DMA = 180 - 2∠MAD = 180 - 2∠CDE = ∠DCE so AMD and DCE are similar so DAE and DCM are similar. ∠DCM = ∠DCF = ∠DEA = ∠DEF so DCEF is cyclic, we also know DCEB is cyclic so DCBF is cyclic and ∠BFC = ∠90.

Let $J$ be the center of $(BCD)$. Notice that $E$ lies on $(BCD)$ because $BC$ is the perpendicular bisector or $DE$. Let $F'=AE\cap (BCD)$. It is well known that $JD$ is tangent to $(ADH)$, so $MD$ is also tangent to $(BCD)$. Then \[ \angle MAF=\angle MAE=\angle AED=\angle F'ED=\angle F'CD=\angle F'DM \]which means $AMF'D$ is cyclic. So $\angle CF'D=90^\circ-\angle C=\angle MAD$. That menas $C,F',M$ are collinear. So $F'\equiv F$.

$\measuredangle MAD=\measuredangle (BC,AD)+90^{\circ}=\measuredangle (ED,AD)=\measuredangle EDC$, so $\triangle MAD\sim\triangle CDE$, so $\triangle DAE\sim\triangle DMC$. Now we have $\measuredangle DCF=\measuredangle DCM=\measuredangle DEA=\measuredangle DEF$ so $DECF$ is cyclic and $B$ lies on $\odot(CDE)$, so $DBECF$ is cyclic and $\measuredangle BFC=\measuredangle BDC=90^{\circ}$.

Let $F'=\overline{CM} \cap (BCDE)$. Since $\measuredangle MF'D=\measuredangle CF'D=\measuredangle CBD=\measuredangle MAD$, $AMDF'$ is cyclic. Then, $$\measuredangle DFM=\measuredangle DFE \iff \measuredangle DMA=\measuredangle DCE \impliedby \measuredangle MAD=\measuredangle EDC$$since $MA=MD$ and $CD=CE$, so $F'$ lies on $\overline{AE}$ as well, i.e. $F=F'$. Then $\angle BFC=\angle BDC=90^\circ$, so we're done. $\blacksquare$