First, two lemmas. First lemma: Let $D$ be the midpoint of $BC$ in the triangle $ABC$ which has orthocentre $H$. $AH,BH,CH$ meet the opposite sides at $A_1,B_1,C_1$. Then the power of $D$ with respect to $(AB_1HC_1)$ is $DB^2$. Proof Let $M$ be the midpoint of $AH$ and therefore the centre of $(AB_1HC_1)$. Clearly $D$ is the centre of the circle $(BC_1B_1C)$ since $BC$ is the diameter. Now $\angle DB_1H=\angle DB_1B=\angle DBB_1=90^{\circ}-C$. But Clearly $\angle B_1AH=90^{\circ}-C$. Hence by the converse of the alternate segment theorem, $DB_1$ is tangent to $(AB_1HC_1)$ and hence the power of $D$ with respect to $(AB_1HC_1)$ is $DB_1^2=DB^2=DC^2$. Second lemma: If $HD$ meets $(ABC)$ at $H'$ then $HAH'C$ is a parallelogram. This is well known, http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=455274 is just one of many threads that prove it. Note that this means $DH=DH'$. Now back to the solution. Let $P$ be on $DH$ such that $\angle DPA=90^{\circ}$. Clearly $P$ lies on $(AB_1HC_1)$. So the power of $D$ with respect to this circle is $DH\cdot DP=DH'\cdot DP$ and hence $DB^2=DH'\cdot DP\implies DB\cdot DC=DH'\cdot DP$ hence $PBH'C$ is cyclic and therefore $P$ lies on $(ABC)$. Now $A_2$ becomes the radical centre of the three circles $(ABC),(AB_1HC_1)$ and $(BC_1B_1C)$. Therefore $A_2$ lies on $AP$ and hence the line perpendicular to $AA_2$ through $D$ passes through the orthocentre $H$ of triangle $ABC$. Hence so do the others and thus they are concurrent.

Now for the solution I feel the problem is begging for (for me, anyway..) Again, start with a lemma: After an inversion with centre $X$ and arbitrary radius, a circle $\Gamma$ with centre $Y$ that doesn't pass through $X$ is mapped to $\Gamma'$ with centre $Y'$. Then $X,Y$ and $Y'$ are collinear. I'll refrain from posting a proof from this, as I have proposed a problem to a particular Mathematics Journal where exactly this lemma was used to kickstart the solution. Nonetheless, I will edit/post later with a proof Anyway, thanks to the lemma at http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2180447&#p2180447, after inverting with respect to $H$ we can re-word the problem to the equivalent: Let $ABC$ be a triangle with incentre $I$ and excentres $X,Y,Z$. Let $(IYZ)$ and $(IBC)$ intersect at $D$. Let $IP$ be the diameter of $(IDP)$. Prove that the centre of the circle passing through $B,C,Y,Z$ lies on the line $IP$. We will in fact prove that $P$ is the centre of this circle. Note that $B,C,Y,Z$ lie on a circle because $I$ is the orthocentre of triangle $XYZ$ and so $\angle ZBY=\angle ZCY=90^{\circ}$. Now, $XC\cdot XY=XB\cdot XZ=XI\cdot XA=XD\cdot XP$ (because $BCYZ$ is cyclic, $AIBZ$ is cyclic, $AIDP$ is cyclic, respectively). Therefore $CDYP$ is cyclic. Therefore $\angle PCY=\angle PDY=\angle IDY- 90^{\circ}=(180^{\circ} -\angle IZP)-90^{\circ}=90^{\circ}-\frac{B}{2}$. But $AICY$ is cyclic so $\angle AYC=180^{\circ}-\angle AIC=90^{\circ}-\frac{B}{2}$. So we conclude $\angle PCY=\angle PYC$ hence $PY=PC$. Similarly $PB=PZ$. Thus $P$ lies on the perpendicular bisectors of two chords of the circle $(BCYZ)$ and so must be the centre. Hence proven.

$AA_2$ is the polar of $H$ wrt $\odot B_1C_1BC$ and similar for others. Hence those lines concur at the point $H$, orthocenter of $\triangle ABC$.

$AA_2H, BB_2H , CC_2H $ are polar triangles .So$ DH\perp AA_2 $.And so done.

[asy][asy] import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.250413030748777, xmax = 9.900436046904861, ymin = -2.944494629373918, ymax = 5.464993173930597; /* image dimensions */ pen zzttqq = rgb(0.6000000000000006,0.2000000000000002,0.000000000000000); /* draw figures */ draw((1.660000000000002,4.460000000000005)--(-0.9000000000000010,1.400000000000002), zzttqq); draw((-0.9000000000000010,1.400000000000002)--(3.660000000000004,-0.2000000000000002), zzttqq); draw((3.660000000000004,-0.2000000000000002)--(1.660000000000002,4.460000000000005), zzttqq); draw((-4.054173742651869,2.506727629000656)--(-0.9000000000000010,1.400000000000002)); draw((-4.054173742651869,2.506727629000656)--(2.370821446903827,2.803786028714091)); draw((1.660000000000002,4.460000000000005)--(-4.054173742651869,2.506727629000656)); draw((3.660000000000004,-0.2000000000000002)--(0.1900570452089564,2.702958811851334)); draw((2.370821446903827,2.803786028714091)--(-0.9000000000000010,1.400000000000002)); draw((0.2278782388448655,3.970458730526274)--(1.380000000000002,0.6000000000000006)); draw((1.660000000000002,4.460000000000005)--(0.4233872293779123,0.9356536037270493)); /* dots and labels */ dot((1.660000000000002,4.460000000000005),dotstyle); label("$A$", (1.733025663218286,4.559879707057765), NE * labelscalefactor); dot((-0.9000000000000010,1.400000000000002),dotstyle); label("$B$", (-0.8265288129037566,1.491695815808908), NE * labelscalefactor); dot((3.660000000000004,-0.2000000000000002),dotstyle); label("$C$", (3.718321122261664,-0.09982203152338612), NE * labelscalefactor); dot((2.370821446903827,2.803786028714091),dotstyle); label("$B_1$", (2.438543884200644,2.902732257773623), NE * labelscalefactor); dot((0.1900570452089564,2.702958811851334),dotstyle); label("$C_1$", (0.2563596193017228,2.804287854845852), NE * labelscalefactor); dot((-4.054173742651869,2.506727629000656),dotstyle); label("$A_2$", (-3.993157107080386,2.607399048990311), NE * labelscalefactor); dot((1.380000000000002,0.6000000000000006),dotstyle); label("$D$", (1.437692454434973,0.7041405923867421), NE * labelscalefactor); dot((0.2278782388448655,3.970458730526274),dotstyle); label("$A_3$", (0.2891744202776464,4.067657692418911), NE * labelscalefactor); dot((0.8498235971988313,2.150997252016667),dotstyle); label("$H$", (0.9126556388201952,2.246436238255151), NE * labelscalefactor); dot((0.4233872293779123,0.9356536037270493),dotstyle); label("$A_1$", (0.4860632261331882,1.032288602145978), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $H$ be the orthocenter of $\triangle ABC.$ I claim that $H$ is the desired point of concurrency. Let $A_3$ be the foot of perpendicular from $D$ on $AA_2.$ Since $AA_1 \perp BC$ and $DA_3 \perp AA_2,$ quadrilateral $A_3A_1DA$ is cyclic. By PoP, we have $A_2C_1 \times A_2B_1 = A_2A_3 \times A_2A.$ Again, by PoP wrt the nine point circle of $\triangle ABC,$ $A_2A_1 \times A_2D = A_2C_1 \times A_2B_1,$ so combining them, $A_2C_1 \times A_2B_1= A_2A_3 \times A_2A,$ implying quadrilateral $A_3C_1B_1A$ is cyclic. But $H$ lies on this circle too, since $HC_1 \perp AB$ and $HB_1 \perp AC.$ It follows that $\angle HA_3A= 180^{\circ} - \angle HB_1A = 90^{\circ},$ so $D,H,A_3$ are collinear. Defining $B_3, C_3$ analogously, similar arguments show that $E,H,B_3$ and $F,H,C_3$ are also collinear, so the lines in the problem are concurrent at $H.$ $\blacksquare$

Dear Mathlinkers, this is a consequence of http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=72094 incerely Jean-Louis

Let $H$ be the orthocenter of $ABC$.Note that the radical axis of $\odot{AA_1D}$ and the nine point circle is $A_1D$ and the radical axis of the nine pont circle and $\odot{BC_1B_1C}$ is $B_1C_1$.So $A_2$ lies on the radical axis of $\odot{AA_1D}$ and $\odot{BC_1B_1C}$.We also have $AH \cdot AA_1=BH \cdot BB_1$ so the power of $H$ w.r.t both the circles are same.Thus $A_2H$ is the radical axis of the two circles.Also note that the center of $\odot{BC_1B_1C}$ is $D$ and that of $\odot{AA_1D}$ lies on $AD$.So we have $A_2H \perp AD$.This means that $H$ is also the orthocenter of $DAA_2$ so the perpendicular from $D$ to $AA_2$ passes through $H$.Analogously the perpendiculars from $E$ to $BB_2$ and $F$ to $CC_2$ also pass through $H$ and this finishes the problem.

Hello, see here Problem 16 is essentially the same.

Brocard Theorem in the circle of diameter $BC$, $\Rightarrow D$ is orthocenter of $\Delta AHA_2$, $\Rightarrow DH \bot AA_2$

Notice that $(A_2C, A_2H, A_2B_1, A_2A) = -1$. This means that $A_2H$ is the polar of $A$ with respect to the circle passing through $B_1, C_1, B, C$. Thus, $AD \perp A_2H$ so $H$ is the orthocenter of $ADA_2$. Therefore, $DH \perp AA_2$. Similarly $EH \perp BB_2$ and $FH \perp CC_2$ so all three lines concur at $H$ as desired.

Invert about $H$ (the orthocenter) with radius $\sqrt{HA\cdot HA_1}$, where the lengths are directed, so that $A$ swaps with $A_1$. Then $A_2$ maps to $A_2H \cap AD$, so that $\angle HA_2'A=90^{\circ}$. Now let $DH$ hit $AA_2$ at $E'$, and we want to prove that $E'$ lies on the circumcircle, since $HD$ passes through the antipode of $A$, so that $AED=90^{\circ}$. $E' \equiv (A_2'A_1H) \cap EH$, which is clearly $D$, as $\angle HA_2'D=HA_1D=90^{\circ}$. Yet $D$ lies on the nine point circle, so inverting back gives that $E$ lies on the circumcircle, hence $H$ is the desired concurrency point.

What a cute problem. Let $H$ be the orthocenter of $\triangle ABC$. By Brokard's theorem, $F$ is the orthocenter of $CC_2H$, so the perpendicular from $F$ to $CC_2$ passes through $H$. Analogously, so do the the perpendicular's from $E$ to $BB_2$ and $D$ to $AA_2$. $\Box$.

My solusions

Let $H$ be the orthocenter of $\triangle ABC$. Consider the circumcircle of the quadrilateral $BC_1B_1C$, obviously the point $D$ is its circumcenter. Note that the line $AA_2$ is the polar of the point $H$ wrt this circle, we conclude that the perpendicular line from $D$ to $AA_2$ passes through $H$. Analogously, so do the the perpendiculars from $E$ to $BB_2$ and $F$ to $CC_2$ so we're done. $\Box$

Oops solved this many months ago but didn't post. Nice problem! [asy][asy] import olympiad; unitsize(12); pair A, B, C, A1, B1, C1, A2, B2, C2, D, E, F, R, S, T; /* Points */ A = (1,9); B = (0,0); C = (18,0); A1 = foot(A,B,C); B1 = foot(B,A,C); C1 = foot(C,A,B); A2 = extension(B1,C1,B,C); B2 = extension(A1,C1,A,C); C2 = extension(A1,B1,A,B); D = midpoint(B--C); E = midpoint(C--A); F = midpoint(A--B); R = foot(D,A,A2); S = foot(E,B,B2); T = foot(F,C,C2); /* Draw */ draw(A--B--C--cycle); draw(A--A1); draw(B--B1); draw(C--C1); draw(B1--A2); draw(A1--B2); draw(A--B2); draw(B--A2); draw(B--C2); draw(B1--C2); draw(A--A2); draw(C--C2); draw(B--B2); draw(D--R, red); draw(E--S, red); draw(F--T, red); draw(circumcircle(A,A1,D), orange); draw(A--D); draw(A--R); draw(R--A1); draw(A1--D); draw(circumcircle(A1,B1,C1), orange); draw(circumcircle(A,B1,C1), orange); draw(B1--D); draw(C1--B1); /* Label */ dot("$A$", A, N); dot("$B$", B, SW); dot("$C$", C, SE); dot("$A_2$", A2, SW); dot("$B_2$", B2, N); dot("$C_2$", C2, SW); dot("$D$", D, S); dot("$E$", E, NE); dot("$F$", F, NW); dot(R, blue); dot(S, blue); dot(T, blue); [/asy][/asy] Define the projection of $D$ onto $AA_2$ as $R.$ Clearly from given conditions $\angle ARD = \angle AA_1D = 90^\circ.$ Quadrilateral $ARA_1D$ is thus cyclic. By Power of a Point on $(ARA_1D),$ we obtain\[A_2R \cdot A_2A = A_2A_1 \cdot A_2D.\]Now we show quadrilateral $B_1C_1A_1D$ is cyclic. Recall the nine point circle of triangle $ABC$ goes through the midpoints of the sides and the feet of the altitudes dropped from the vertices. Indeed, we see $(B_1C_1A_1D)$ is the nine point circle, and quadrilateral $B_1C_1A_1D$ is cyclic. By Power of a Point on $(B_1C_1A_1D),$ we obtain\[A_2A_1 \cdot A_2D = A_2C_1 \cdot A_2B_1.\]Therefore $A_2R \cdot A_2A = A_2C_1 \cdot A_2B_1.$ By the converse of Power of a Point, quadrilateral $RC_1B_1A$ is cyclic. Note by Lemma 1.14 of EGMO that $H$ also lies on $(RC_1B_1A).$ However, recall $\angle ARD = 90^\circ,$ and now we have $\angle ARH = 90^\circ,$ (note this is because $\angle HB_1A = 90^\circ,$ and opposite angles are supplementary). Therefore, $H$ lies on $DR.$ Define $S$ as the projection of $E$ onto $BB_2$ and $T$ as the projection of $F$ onto $CC_2.$ Constructing the desired circles symmetrically for these cases, we find $H \in ES, FT.$ Therefore, $DR \cap ES \cap FT = H,$ as required.

Now I return to this problem with projective geometry. Notice $D$ is the circumcenter of cyclic quadrilateral $BC_1B_1C,$ so by Brokard's Theorem, $D$ is the orthocenter of $A_2AH.$ Similarly, $E$ is the orthocenter of $B_2BH$ and $F$ is the orthocenter of $C_2CH.$ The result follows, as the lines concur at $H.$

Oh you could actually Brokard this Claim: $DH$ is the perpendicular from $D$ onto $AA_2$ Proof: Note that $DH$ intersects $(ABC)$ on the minor arc not containing $A$ at the antipode of $A$ Let $DH$ intersect $(ABC)$ at $X$. Note $A,B_1,C_1,X$ cyclic since $$\angle{AB_1H}=\angle{AXH}=90$$ Note that the radical center of $(AB_1C_1X),(C_1B_1CB),(ABC)$ must be at $A_2$, which implies that $A,C_1,A_2$ collinear Hence, using our claim, $H$ must pass through all the three perpendiculars

We claim that the lines concur at the orthocenter. By Brocard, the orthocenter of $\triangle AA_2H$ is the circumcenter of $CBC_1B_1$. However, this is just the midpoint of $BC$, which is $D$, since $\angle BC_1C=\angle BB_1C=90.$ Thus, $D$ is the orthocenter of $\triangle AA_2H$, so $DH\perp AA_2$, hence the perpendicular from $D$ to $AA_2$ passes through $H$. Similarly for the other lines, so we are done.

The lines concur at the orthocenter H. Let A_3 be the foot of perpendicular from D to AA_2, etc. I’ll show one part and the others analogously follow. Since AA_3A_1D is cyclic and so is C_1B_1A_1D (nine point circle), by pop we get A_2B_1*A_2C_1=A_2A_1*A_2D=A_2A_3*A_2A, so B_1C_1AA_3 is cyclic. Since H is also on this circle, HA_3A=HB_1A=90 degrees=DA_3A; Then DA_3 passes through H and so do the others in the problem, as desired. $\blacksquare$

Claim: The three lines concur at the orthocenter $H$ of $\triangle{ABC}$. Proof: Note that $BCB_1C_1$ is cyclic with a center at $D$. By Brocard's Theorem, we can say that $D$ is the orthocenter of $\triangle{AHA_2}$, This means that $H$ lies on the perpendicular from $D$ to $AA_2$. Using a similar approach for the other lines, we can say that $H$ lies on the perpendicular from $E$ to $BB_2$, and $H$ lies on the perpendicular from $F$ to $CC_2$. This means that these three altitudes concur at $H$ as desired.

We claim the desired concurrency is the orthocenter $H$. We first show the foot from $D$ to $AA_2$ passes through $H$, as the other two follow symmetrically. Note $D$ is the circumcenter of cyclic quadrilateral $BCB_1C_1$, and \begin{align*} A &= BC_1 \cap B_1C \\ A_2 &= BC \cap B_1C_1 \\ H &= BB_1 \cap CC_1. \end{align*} Using Brocard's on $BCB_1C_1$, we find $D$ is the orthocenter of $\triangle AA_2H$; thus $DH \perp AA_2$. $\blacksquare$ [asy][asy] size(225); defaultpen(linewidth(0.4)+fontsize(8)); pair A, B, C, B1, C1, D, A2, H; A = dir(120); B = dir(210); C = dir(330); B1 = foot(B, C, A); C1 = foot(C, A, B); D = .5B + .5C; A2 = extension(B1, C1, B, C); H = orthocenter(A, B, C); filldraw(B--C--B1--C1--cycle, palegreen); draw(D--A--A2--B1--C--A2); draw(A--B--C--cycle); draw(B--B1^^C--C1); draw(A--H--A2--cycle, blue+linewidth(.55)); draw(D--foot(D, A, A2), dashed); draw(circumcircle(B, C, B1), blue); dot("$A$", A, N); dot("$B$", B, SW); dot("$C$", C, SE); dot("$B_1$", B1, NE); dot("$C_1$", C1, NW); dot("$A_2$", A2, S); dot("$D$", D, S); dot("$H$", H, S); [/asy][/asy]

I claim that these lines concur at the orthocenter $H$ of $ABC$. To prove this, it suffices to show $DH \perp AA_2$, and analogously apply a symmetrical argument for $E$ and $F$. Claim: We have that $DH \perp AA_2$. Proof: Notice that $D$ is the circumcenter of $(BC_1B_1C)$. Now by Brokard's theorem, we have $AA_2$ is the polar of $H$, so that indeed $DH \perp AA_2$, as desired. $\blacksquare$

Let $H$ be the orthocenter of $\triangle ABC$. Let $A_3$ be the foot of the perpendicular from $D$ to $AA_2$. We identify that $A_2$ is the $A$-Ex Point and $A_3$ is the $A-$Queue Point ($(ABC)\cap (AH) \neq A$). It is well known that $D-H-A_3$. Thus, the perpendicular from $D$ to $AA_2$ must pass through the orthocenter. Similarly, the other perpendicular must also pass through $H$ implying that the perpendicular from $D$ to $AA_2$ , the perpendicular from $E$ to $BB_2$ and the perpendicular from $F$ to $CC_2$ are indeed concurrent, at $H$.

hehe By Brokard (Brocard?) on $B_1CBC_1$, we have that $D$ is the orthocenter of $\triangle HAA_2$, which gives that $H$ is the orthocenter of $DAA_2,$ and thus $H$ lies on the perpendicular from $D$ to $AA_2.$ Similarly, it also lies on the other two lines, so we're done.

I claim that these lines concur at the orthocenter $H$ of $ABC$. Proof) Let $Q$ be intersection point of $(AB_1HC_1)$ and $(ABC)$. We know by Radical axis theorem $A_2 , Q , A$ are collinear We know that $ Q,H,D$ are collinear and $\angle HQA = 90$ Therefore We have that $DH \perp AA_2$ Therefore the perpendiculars concur at $H $

Note that $A_2$ is the $A$ expoint in $\Delta ABC$. It is well known that the orthocenter of $\Delta AA_2D$ is the orthocenter $H$ of $\Delta ABC$ so the perpendicular from $D$ to $AA_2$ passes through $H$ and so do the other two perpendiculars.

Note that by orthocenter properties, we have that $BC_1B_1C$ is cyclic with center $D$. Then, by Brocard's Theorem WRT the complete quadrilateral of $BC_1B_1C$, we have that $AA_2$ is the polar of $BB_1\cap CC_1$, the orthocenter of $\triangle ABC$, which we will call $H$. Then, by polar properties, this means that the line through $H$ perpendicular to $AA_2$ passes through the center of $BB_1\cap CC_1$, or $D$. Therefore the line through $D$ perpendicular to $AA_2$ passes through $H$. Similarly, we can find that the lines through $E$ and $F$, perpendicular to $BB_2$ and $CC_2$, respectively, also pass through $H$. Therefore all three lines concur at $H$, as desired. This completes our proof.