$M$ is the midpoint of $\overline{BC}$ and $H$ is the orthocenter of $\triangle ABC.$ Since the circle with diameter $\overline{BC}$ is orthogonal to the circle $\omega$ through $A,E,F,H,$ then $ME,MF$ are tangents of $\omega.$ Let $Q \equiv EF \cap BC.$ Since $(B,C,D,Q)=-1,$ we have $MB^2=ME^2=MF^2=MD \cdot MQ$ $\Longrightarrow$ $\odot(DEQ)$ and $\odot(DFQ)$ are tangent to $\omega$ at $E,F$ $\Longrightarrow$ $\odot(DEQ)$ and $\odot(DFQ)$ coincide with $\omega_1$ and $\omega_2$ $\Longrightarrow$ $P \equiv Q.$

Let $O$ be the centre of $\omega$, on which the orthocentre $H$ clearly lies diametrically opposite to $A$. An angle chase shows that $O,E,D,F$ are concyclic: $\angle EOF=2\angle EAF=2A$ and of course $\angle FDE=180^{\circ}-2A$. Invert with respect to $\omega$ (we now know $D'$ lies on $EF$). The problem becomes: In the triangle $AEF$ with circumcentre $O$ and circumcircle $\omega$, the diameter $AH$ meets $EF$ at $D'$. The circle tangent to $\omega$ at $E$ and passing through $D'$ and the circle tangent to $\omega$ at $F$ passing through $D'$ intersect at $D'$ and $P$. Prove that $OP\perp PD'$. Note that we can actually ignore points $B'$ and $C'$, who also lie on the circle with diameter $OD'$ along with $P$, so we've only really inverted $D,\omega_1,\omega_2$... Anyway: Consider the tangent $\ell$ to $\omega$ and $(OD'E)$ at $E$. By considering the angle $ED'$ makes with this tangent, we get, by the alternate segment theorem that this angle is both equal to $\angle D'PE$ and $\angle FAE$. Hence $\angle D'PE=\angle FAE$ and similarly we get that $\angle FPD'=\angle FAE$. Therefore $\angle FPE=\angle FPD'+\angle D'PE=2\angle FAE=\angle FOE$ so $EPOF$ is cyclic. Now consider the diameter $OO'$ of $(OEF)$. Since $OF=OE$, $O'$ is the midpoint of the arc $EF$ not containing $O$. Now, since $\angle FPD'=\angle EPD'$, we have that $PD'$ is the bisector of $\angle FPE$ and so $PD'$ passes through $O'$. Hence $\angle OPD'=\angle OPO'=90^{\circ}$ and the result follows.

First inverting the figure wrt $H$ with power $HA.HD$ we get the following problem In $\triangle ABC$, two circles $\omega_1, \omega_2$ pass through $A$ and touch the line $BC$ at $B,C$ respectively. Show that the intersection point of these two circles lies on $\odot AEF$. Now inverting this figure again wrt $A$ with power $AH.AD$ we get the following problem: In $\triangle ABC$, two tangents are drawn at the points $E,F$ to the circle $\odot AEF,$ then these two tangent lines intersect on $BC,$ and this is obvious. They meet on at the midpoint of $BC$.

Let $H$ be the orthocenter and $M$ be midpoint of $BC$, and second intersection point of $\omega_1,\omega_2$ be $Q$. $\angle{AEH}=\angle{AFH}=\frac{\pi}{2}$ implies $H$ is on $\omega$. Note that $B,F,E,C$ lie on the circle with diameter $BC$. Then $\angle{HEM}=\angle{BEM}=\angle{EBM}=\angle{EAD}=\angle{EAH}$, which means $ME$ tangents $\omega$. Similarly so does $MF$. Then we can conclude that radical axis of $(\omega,\omega_1),(\omega,\omega_2),(\omega_1,\omega_2)$ is $EM,FM,DQ$. thus $M,D,Q$ is collinear, which is equivalent to the claim.

The radical axes of the three circles are $DQ$ and the tangents to $w$ at $E,F$. But those two tangents meet at $M$, the midpoint of $BC$, because $M$ and the center of $w$ are antipodal on the nine-point circle.

My solution is more or less the same as the last couple; if you let $PB=PC$ on $BC$, showing that $P$ is the intersection of the tangents to the circle at $E$ and $F$ is straightforward, and then $PD$, the same line as $BC$, must be the radical axis so if the circles meet again they must meet on $BC$. This has one slight flaw, but that flaw is actually inherent in the problem statement. If $P=D$, then $AB=AC$ and the two circles $w_1$ and $w_2$ are tangent, so they don't meet on $BC$. If the problem were reworded to "If $w_1$ and $w_2$ meet at a point $P$ other than $D$, show that point is on $BC$" or if it was stated that the triangle was not isosceles, this would have been avoided.

let $AD$ be the diameter of circle $ABC$ and let $DH$ meet circle $ABC$ again at $R$. now let $AR$ meet $BC$ at $K$.since $\angle HRA=90=\angle HFA=\angle HEA$ $R$ is on $\omega$. since $\angle CRH=\angle CAD=90\angle CDA=90-\angle CBA=\angle BCH$ circle $CHR$ touches $BC$. Similarly circle $BHR$ touches $BC$. Consider inversion of pole $A$ and power $AE*AC$. it pictures $\omega$ into $BC$ so $R$ into $K$. it pictures $H,E,F$ into $D,C,B$ so $\omega$ touches circles $EDK$ and $FDK$ therefore $K=P$ so $P$ is on $BC$.

Nice and quick

Let $EF$ meet $BC$ at $X$. Now it is some easy angle chase that $X$ is the desired point.

First of all, it's easy to see that $AEHF$ is cyclic. Let $M$ be the pole of $EF$ wrt the circumcircle of $\triangle AEF$. Let $S$ be the point of intersection of the lines $EF$ and $AH$. Because $AEHF$ is cyclic, the polar of $S$ is the line $BC$, so $M$ lies on $BC$ because $S$ lies on the polar of $M$. Now if circle $\omega_1$ intersects $BC$ at $P$, and circle $\omega_2$ intersects $BC$ at $Q$, by power of point we have$$MD\cdot MP = MF^{2} = ME^{2} = MD\cdot MQ$$So $MP = MQ$, which means that $P = Q$;

After a inversion centered at $D$ with power (or radius or whatever it is called) $\sqrt{DB*DC}$, and reflecting the inverted figure in line $DA$, we see $\omega$ maps to itself, $F$ maps to $E$, $E$ maps to $F$, and the question becomes equivalent to proving whether the tangents to $\omega$ at $E$ and $F$ meet at $BC$ or not. But by a very well known property of nine-point circle, they meet at the midpoint of $BC$, hence QED.

Let $M$ be midpoint of $BC$ then $ME$, $MF$ are common tangents of ($\omega$; $\omega_1$), ($\omega$; $\omega_2$) then $M$ is radical center of $\omega$, $\omega_1$, $\omega_2$ But: $\omega_1$, $\omega_2$ have common point $D$ so $DM$ is radical axis of $\omega_1$, $\omega_2$ Hence: $\omega_1$, $\omega_2$ intersect again at a point on $BC$

Good problem although easy. math154 wrote: In acute triangle $ABC$, let $D,E,F$ denote the feet of the altitudes from $A,B,C$, respectively, and let $\omega$ be the circumcircle of $\triangle AEF$. Let $\omega_1$ and $\omega_2$ be the circles through $D$ tangent to $\omega$ at $E$ and $F$, respectively. Show that $\omega_1$ and $\omega_2$ meet at a point $P$ on $BC$ other than $D$. Ray Li. Let $X$ be the $A - $EX point of $\triangle ABC$. We claim that $X \equiv P$. $ $ Denote by $\Omega$ the circumcircle of $\triangle XFD$. Let $l$ be the tangent line to $\Omega$ at $F$. Then note that $$\angle (l , \overline{FH}) = \angle HFD - (\angle B - \angle C) = 90^{\circ} - \angle B = \angle FAH $$. So $l$ is also tangent to $\odot(AFHE) \implies {\Omega, \omega}$ are tangent. Therefore $\omega_2 = \Omega \implies X \in \omega_2$. Similarly $X \in \omega_1 \iff X \equiv P$.

Easy but Nice. ELMO Shortlist 2012 G2 wrote: In acute triangle $ABC$, let $D,E,F$ denote the feet of the altitudes from $A,B,C$, respectively, and let $\omega$ be the circumcircle of $\triangle AEF$. Let $\omega_1$ and $\omega_2$ be the circles through $D$ tangent to $\omega$ at $E$ and $F$, respectively. Show that $\omega_1$ and $\omega_2$ meet at a point $P$ on $BC$ other than $D$. Ray Li. Let the common tangent to $\omega$ and $\omega_1$ intersect at a point $M$ on $BC$. We will prove that $M$ is the midpoint of $BC$. Let $O$ be the midpoint of $AH$ which actually turns out to be the Circumcenter of $\odot(AFHE)$. So, $\angle OEM=\angle OEH+\angle BEM=\angle BHD+\angle BEM=90^\circ=\angle BHD+\angle HBM\implies BM=ME=MC\implies M $ is the midpoint of $BC$. Similarly we get that the Tangent to $\omega$ and $\omega_2$ intersect $BC$ at it's midpoint which is $M$. Hence, the Radical axis of $\omega_2$ and $\omega_1$ must concur at $M$. So, $DM$ must be it's Radical Axis, So as $D,M\in BC$, hence, $\omega_1$ and $\omega_2$ meet at a point $P$ on $BC$ other than $A$. $\blacksquare$

OK Another Solution. math154 wrote: In acute triangle $ABC$, let $D,E,F$ denote the feet of the altitudes from $A,B,C$, respectively, and let $\omega$ be the circumcircle of $\triangle AEF$. Let $\omega_1$ and $\omega_2$ be the circles through $D$ tangent to $\omega$ at $E$ and $F$, respectively. Show that $\omega_1$ and $\omega_2$ meet at a point $P$ on $BC$ other than $D$. Ray Li. Invert the figure around $A$ with radius $\sqrt{AH\cdot AD}$. The problem becomes Equivalent as follows. Inverted Problem wrote: $AF'E'$ is a triangle and $E'B',C'F',AH'$ are the altitudes with orthocentre $D'$. The circle passing through through $D',F'$ and tangent to $E'F'$ and a circle passing through $D',E'$ and tangent to $E'F'$ meets at a point $P'$. Then $P'\in\odot(AB'C')$. Notice that $D'$ is the $P'-\text{Humpty Point}$ of $\triangle F'P'E'$. Hence, $\angle F'AE'=\angle 180^\circ-\angle F'D'E'=\angle F'P'E'$. So, $AP'F'E'$ are concyclic. Also $P'D'$ bisects $F'E'$. Hence, $P'$ is the $A-\text{Queue Point}$ of $\triangle AF'E'$ or the Miquel Point of $F'B'C'E'$. Hence, $P'\in\odot(AB'C')$. Hence, Inverting back we get that $P\in BC$. $\blacksquare$ To know more about Queue Points (Name given by the User Math-Pi-Rate). Visit these two Blogposts of his. Part 1 Part 2

Let $H$ be the orthocenter of $\triangle ABC$, and let $N$ be the midpoint of $AH$. It is easy to see that $N$ is the center of circle $(AEHF)$. Consider the tangent to $\omega_1$ at $E$ and the tangent to $\omega_2$ at $F$. Call them lines $\ell_1$ and $\ell_2$. I claim they intersect on $BC$, namely the midpoint of $BC$. To see this, let $M$ be the midpoint of $BC$, and draw $ME, MF$. Look at $ME$. We have $$\angle BEM=\angle EBM=\angle HAE,$$(where the last step came from the fact that $\angle EBM$ and $\angle HAE$ are both complementary to $\angle BCA$) so $ME$ is tangent to $\omega_1$. In a similar fashion we can show that $MF$ is tangent to $\omega_2$. Since for each point on a circle there is only one tangent through that point, $ME$ and $MF$ are $\ell_1$ and $\ell_2$ respectively. This means that $\ell_1$ and $\ell_2$ intersect at $M$. Note that $\ell_1$ is the radical axis of $\omega$ and $\omega_1$ while $\ell_2$ is the radical axis of $\omega$ and $\omega_2$. Hence, $M$ is the radical center of the three circles. Since $D$ lies on $BC$ as well, $BC$ is the radical axis of $\omega_1$ and $\omega_2$, which means that $\omega_1$ and $\omega_2$ must intersect at another point on $BC$.

math154 wrote: In acute triangle $ABC$, let $D,E,F$ denote the feet of the altitudes from $A,B,C$, respectively, and let $\omega$ be the circumcircle of $\triangle AEF$. Let $\omega_1$ and $\omega_2$ be the circles through $D$ tangent to $\omega$ at $E$ and $F$, respectively. Show that $\omega_1$ and $\omega_2$ meet at a point $P$ on $BC$ other than $D$. Ray Li. Nice problem. A new elegant solution which I do not think any one above has posted after a brief reading of all the beautiful solutions. Let line $\overline{EF} \cap \overline{BC} = X$, we claim that $X \equiv P$. We see that if $M$ is the midpoint of segment $BC$ and $K$ be the midpoint of segment $AH$, then $\angle MFK = 90^\circ$ since $M, F, K$ lie on the nine-point circle of $\triangle ABC$ and $M, K$ are antipodes of each other with respect to nine-point circle of $\triangle ABC$. Now, since circumcircles of $\triangle AEF$ and $\triangle DFP$ are tangent to each other at $F$, we must have that $\angle MFT = \angle MFK = 90^\circ \dots (1)$ where $T$ is the circumcenter of $\triangle DFP$. Therefore, if we perform an inversion $\Gamma$ about the circumcircle of $\square BCEF$, then since the circumcircles of $\square BCEF$ and $\triangle DFP$ are orthogonal to each other due to $(1)$, then circumcircle of $\triangle DFP$ must be preserved under $\Gamma$, but we see that $D$ inverts to $X$ under $\Gamma$ (this is because if $D$ inverts to $D'$ under $\Gamma$, then we see that $D' \in$ line $\overline{DM}$ $=$ line $\overline{BC}$ and since $D, E, F, M$ lie on the nine-point circle of $\triangle ABC$, this nine-point circle inverts to line $\overline{EF}$ and so $D' \in $ line $\overline{EF}$, implying that $D' =$ line $\overline{BC} \cap$ line $\overline{EF} = X$) and so $X \in$ circumcircle of $\triangle DFP$ and similarly circumcircle of $\triangle DEP$ is preserved under $\Gamma$, therefore in the same manner $X \in$ circumcircle of $\triangle DEP$, which means that $X = $ circumcircle of $\triangle DFP \cap$ circumcircle of $\triangle DEP \neq D$ and $X \in $ line $\overline{BC}$ and so $P = X =$ circumcircle of $\triangle DFP \cap$ circumcircle of $\triangle DEP \in$ line $\overline{BC}$ as desired.

The pair wise radical axes of the 3 circles are concurrent. Tangents at $E,F$ to $(AEF)$ meet at the midpoint of BC ($M$). Hence the other intersection point of omega1 and omega2 lie on $MD$ or $BC$.

Cute. Take points $P_1$ and $P_2$ on $BC$ such that $(DP_1E)$ and $(DP_2F)$ are indeed tangent to $\omega$. Verify that $EE\cap FF\cap BC=M$, the midpoint of $BC$, has equal power to the three circles. Hence $MD\cdot MP_1=MD\cdot MP_2$, i.e. $P_1\equiv P_2\equiv P$.

Let's take $M$ which is a midpoint of $BC$ From three-tangents lemma $MF=ME$ and $MF$ tangents to $\omega_2$ $ME$ tangents to $\omega_1$ thus $M$ on the radical axis line of $\omega_1$ and $\omega_2$ Let's take $Z$ as intersection point of $MD$ line and $\omega_1$ hence $MF^2=MD.MZ$ SO $ME^2=MF^2$ which implies $Z$ is also on the $\omega_2$ and $Z$ on the $BC$ which finish the problem

WLOG assume $AB < AC$. Let $H$ be the orthocenter and $X = BC \cap EF$. Then $EB$ bisects $\angle DEX$, so the $AE$ passes through midpoint of arc $DX$ not containing $E$. Hence there exists a homothety at $E$ sending $\omega$ to $(DEX)$, meaning the two are tangent, so $(DEX) = \omega_1$. $FA$ is the external angle bisector $\angle DFX$, so we use the same steps to get $(DFX) = \omega_2$. Hence $X$ is our desired point on $BC$. $\blacksquare$

Let $L$ be the mid-point of $\overline{AH}$ and $M$ be the mid-point of $\overline{BC}$. Then, $M,D,E,F,L$ lie on the nine-point circle. Moreover, $\angle MDL=90^\circ=\angle MEL=\angle MFL$. Note that $L$ is the center of circle $(AEF)$ as $H\in(AEF)$ and $\angle AEH=\angle AFH=90^\circ$. It follows that $ME$ and $MF$ are tangents to circle $(AEF)$. Since $\omega_1,\omega_2$ are tangent to $(AEF)$ as well, which means that $ME,MF$ is the radical axis of $(AEF),\omega_1$ and $(AEF),\omega_2$, respectively. Hence, $M$ is the radical centre of $(AEF),\omega_1,\omega_2$. So, $M$ lies on the radical axis of $\omega_1,\omega_2$. So, $MD$ is the radical axis of the two circles. Since $P\in MD$, and $M,D,B,C$ are collinear, it follows that $P\in BC$.

Note that $M = \overline{EE} \cap \overline{FF}$, the midpoint of $\overline{BC}$ by three tangent lemma, is the radical center of $\omega_1, \omega_2$, and $(AEF)$. Hence $\overline{DM} \equiv \overline{BC}$ is the radical axis of $\omega_1$ and $\omega_2$, and the result follows.

They meet at the $A$-ex point; if we let this point be $X$, let $H$ be the orthocenter and let $N$ be the midpoint of $\overline{AH}$, then $\angle FXD = \angle DNM$ (it's well known that $\overline{NM} \perp \overline{FE}$). Because of the nine point circle, $\angle DNM = \angle DFM$, so $(XFD)$ is tangent to $\overline{FM}$. The three tangents lemma gives us that $\overline{MF}$ is tangent to $(AFE)$, so indeed $(XFD)$ is tangent to $(AFE)$. Likewise, $(XED)$ is tangent to $(AFE)$.

We claim that the point $P$ is just the intersection of $EF$ and $BC.$ To do so, we let $X = EF \cap BC,$ then we will show that $(DEX)$ and $\omega$ are tangent; this will imply by symmetry that $(DFX)$ and $\omega$ are tangent, which would finish. Let $M$ be the midpoint of $BC.$ By the Three Tangents Lemma, $EM$ is tangent to $\omega.$ We wish to show that $ME$ is also tangent to $(DEX).$ Note that $(X,D;B,C) = -1,$ so by the midpoint of harmonic bundles lemma, we see that $MB^2 = ME^2 = MD \cdot MX,$ which finishes.

Let $M$ be the midpoint of $BC$. It is well known that $B$, $F$, $E$, and $C$ lie on a circle centered at $M$ and lines $ME$ and $MF$ are tangent to $\omega$. Thus, line $ME$ is tangent to $w_1$ and line $MF$ is tangent to $w_2$, so $M$ has the same power with respect to $w_1$ and $w_2$. $D$ also has the same power $0$ with respect to both circles, so $\overline{DM} = \overline{BC}$ must be the radical axis of the two circles, making the result obvious.

It is well known that the tangents at $E$ and $F$ at $(AEF)$ concur at the midpoint of $BC$, which we will call $M$. So it follows that $M$ is the radical center of $\omega_1$, $\omega_2$ and $(AEF)$, and since $\omega_1$ and $\omega_2$ pass through $D$, the radical axis of the two circles is $\overline{DM} = \overline{BC}$, so the two circles intersect again on $BC$.

Let $M$ be the midpoint of $BC$. Observe that $AB \ne AC$, else $\omega_1, \omega_2$ cannot exist. Therefore $D \ne M$. Claim 1: $\omega_1$ and $\omega_2$ cannot be tangent to each other. Proof sketch: Note that from the three tangents lemma, $MD = ME = MF$. But the key idea is at least one of $\angle MDE, \angle MDF$ is $\ge 90^\circ$. Therefore, $M$ cannot be the circumcentre of $\Delta DEF$. (in the case that $AB = AC$, we get a contradiction anyways as $M = D$.) Therefore, $\omega_1$ and $\omega_2$ meet at $P \ne D$. Claim: The radical axis of the above two circles is line $BC$. (Note that this implies the statement) Proof: Again from the Three Tangents Lemma, $M$ is the radical centre of $\omega, \omega_1, \omega_2$, so it lies on the radical axis of these circles. But so does $D$, so the radical axis is line $MD$ which is also line $BC$. $\square$

Inversion with radius $\sqrt{AF.AB}$ switches $(F,B),(H,D),(E,C),(\omega, BC)$. Hence $\omega_1^*$ and $\omega_2^*$ intersect at $H$ humpty point of $HBC$ which lies on $(EHF)$ (well known). QED I am only capable of inverting and finding 2 line solutions. Dying of skill issue.

Observe that the radical center of $\omega, \omega_1, \omega_2$ is $M$, where $M$ is the midpoint of $\overline{BC}$ (by three tangents lemma). Now as $\omega_1, \omega_2$ meet at $D$ on $\overline{BC}$, it follows that the radical axis of $\omega_1, \omega_2$ is $\overline{BC}$, so that their second point of intersection (not equal to $D$) likewise lies on $\overline{BC}$, as desired.

$ $

Let $M$ be the midpoint of $BC$. It is well known that $ME,MF$ are tangent to $\omega$ and $DMEF$ is cyclic. We claim that $P=EF\cap BC$. It suffices to prove that $P$ lies on $\omega_1$, so it suffices to prove that $(MP)(MD)=ME^2$. But $ME^2=(MB)(MC)$ so this turns into the well known harmonic config since $(BC;DP)=-1$.