A interesting problem, but it is pretty difficult.

I proved that $\frac{ED}{FD}=\frac{sinC.cosC}{sinB.cosB}$ I also noticed that $O$, $H$ and $R$ are collinear but I coudn't prove that. If that's true it's easy to finish the problem. Notice that $OR$ is the angle bissector of $\angle PRQ$.

Jack

cadge_nottosh wrote:

Jack sorry if this sounds stupid put why does R lies on $QC_{1}$ and $PB_{1}$

Sorry, in typing it up I missed some steps. If $ QC_{1}$ intersects $ PB_{1}$ at $ T$ then one can show with a simple angle chase that OPQT is cyclic. Now by Pascal $ T$, $ B$ adn $ C$ are collinear, so $ T$ is the intersection of circle OPQ with BC, which is R. Hope that's clearer! Jack

Here's a sketch of my solution: Lemma: Suppose $ AHPQ$ is cyclic with $ Q,P$ on $ AB,AC$ respectively. There exists a point $ Z$ on $ BC$ such that $ BZ = QZ$ and $ CZ = PZ$. Proof: It suffices to prove that the perpendicular bisectors of $ BQ$ and $ CP$ meet on $ BC$, which is equivalent to $ \frac {BQ}{2 cos B} + \frac {CP}{2 cos C} = a$ (*). If $ \angle GHF = \angle PHE = x$, then using the Sin Rule repeatedly gives us $ BQ = BF - QF = (a cos B) \frac {tan a - tan x}{tan a}$ and $ CP = CE + EQ = (a cos C) \frac {tan a + tan x}{tan a}$, from which (*) follows immediately. Now to the problem: Consider the point Z on BC such that $ BZ = QZ$ and $ CZ = PZ$. A quick angle chase, gives us that $ \angle QZP = 180 - 2A$, so $ QZPO$ is concyclic. But since the circumcircle of $ QOP$ is tangent to $ BC$ at $ R$, it follows that $ R$ and $ Z$ coincide. Thus $ BQ = QR$ and $ CR = RP$. A quick angle chase also gives that $ QRP$ is similar to $ FDE$. Thus $ \frac {BR}{CR} = \frac {QR}{PR} = \frac {FD}{DE}$, as required.

Joao Guerreiro wrote: I proved that $\frac{ED}{FD}=\frac{sinC.cosC}{sinB.cosB}$ I also noticed that $O$, $H$ and $R$ are collinear but I coudn't prove that. If that's true it's easy to finish the problem. Notice that $OR$ is the angle bissector of $\angle PRQ$.

Let Circle of $\triangle BQH$ cut $BC$ at $R$. By Miquel theorem, $~$ $\square CPHR$ is cyclic. $\angle PRQ=\angle HRQ+\angle HRP=\angle HBQ+\angle HCP=180-2\angle A=180-\angle QOP$ Hence, $~$ $\square OPRQ$ is cyclic. By problem, this $R$ is that of the problem. $\angle HRP=\angle HRQ$ $~$ and then $R$, $H$, $O$ are collinear, since $~$ $OP=OQ$ By a little angle chase, $\triangle DEF$ is similar to $\triangle RPQ$ $~$ and $~$ $\angle RPC=\angle RCP$ Similarly, $~$ $\angle RQB=\angle RBQ$ $~$ $\therefore$ $ED/FD=PR/RQ=CR/RB$

Is this correct?

Lemma: Let $O$ just be any point on the perpendicular bisector of $AH$. Then, let $OH$ intersect $BC$ at $R'$. Then, $OPR'Q$ is cyclic. Proof: I will first show that $PH$ bisects angle $QPR'$. This is simple angle chasing: $\angle OPQ = 90^\circ - \frac{1}{2} \angle POQ = 90^\circ - A$, we have $\angle OPH = \angle 90^\circ - \frac{1}{2} \angle POH = 90^\circ - (90^\circ - B) = B$. Therefore, we have $\angle QPH = A+B - 90^\circ = 90^\circ - C$. But notice that $\angle HPR' = \angle HBR' = \angle EBC = 90^\circ - C$ (this follows from the fact that $PHR'B$ is cyclic because $\angle OHP = \angle OPH = B$) so we have $PH$ is the angle bisector of $\angle QPR'$. Similarly, we have $QH$ is the angle bisector of $\angle PQR'$ so therefore $H$ is the incenter of $PQR'$, meaning that $R'H$ is an angle bisector, and $O$ is the intersection of the angle bisector and the perpendicular bisector of $PQ$, so therefore $OPR'Q$ is cyclic, as desired. $\Box$ Now, I will take $O$ to be such that $OH \perp EF$. Let the intersection of $OH$ and $BC$ be $R''$ and I will show that $R = R''$ i.e. the circumcircle of $OPR''Q$ is tangent to $BC$. Let $X = BE \cap (APHQ)$ where $X \neq H$. From before, we know that $\angle POH = 180^\circ - 2B$, so that is the length of arc $PH$. But we also know that $\angle OHE = \angle FEA = \angle B$, so $\angle HOX = 180^\circ - 2B$, so arcs $PH$ and $HX$ are equal. Therefore, $QH$ is the angle bisector of $PQX$, implying that $QX$ passes through $R''$ (from all those observations in the lemma). Therefore, $\angle PQR'' = \angle PQX = \angle PHB = \angle PR''B$ so therefore $R'' = R$. This tells us that $OHR$ must be perpendicular to $EF$. Let $Y = HD \cap EF$. Because $\triangle FHE \sim \triangle BHC$, we can see that since $HR \perp EF$, $HR$ and $HY$ are isogonal w.r.t. $\triangle FHE$. That means that $HD$ and $HR$ are isogonal, which means $Y$ and $R$ are corresponding points between $HFE$ and $HBC$. Therefore, $BR/RC = FY/YE$. However, it is easy to see that $DH$ is the angle bisector of $\angle EDF$ so by the angle bisector theorem, we have that $FY/YE = FD/DE$ so $BR/RC = FD/DE$, as desired. $%Error. "blackbox" is a bad command. $

One of the key claims in this problem, which is showing the existence of a point $Z$ on $BC$ such that $BZ = QZ$ and $CZ = PZ$ is a good olympiad problem in itself: see December TST 2012/1 .

N.T.TUAN wrote: In acute triangle $ABC$ , segments $AD; BE$ , and $CF$ are its altitudes, and $H$ is its orthocenter. Circle $\omega$, centered at $O$, passes through $A$ and $H$ and intersects sides $AB$ and $AC$ again at $Q$ and $P$ (other than $A$), respectively. The circumcircle of triangle $OPQ$ is tangent to segment $BC$ at $R$. Prove that $\frac{CR}{BR}=\frac{ED}{FD}.$ Let $R' \overset{\text{def}}{:=} \odot(BHQ) \cap \odot(CHP)$ with $R' \ne H$ then Miquel's theorem asserts $R' \in \overline{BC}$. Notice $H$ is the spiral center $\triangle DEF \mapsto \triangle R'PQ$. Consequently, $\angle R'PQ=\angle EDF$ proving $R'$ lies on $\odot(PQR)$. Hence $R'=R$. Now observe $\angle BHQ=\angle RPQ=180^{\circ}-2\angle B$; hence $\angle (\overline{HA}, \overline{HR})=\angle (\overline{PQ}, \overline{EF})=\angle B-\angle C$. Let $X \overset{\text{def}}{:=} \overline{HA} \cap \overline{EF}$ and $R^{*}$ be isotomic conjugate of $R$ in $\overline{BC}$; then $\overline{AR^{*}}$ is a diameter of $\odot(ABC)$. Apply $\triangle AEF \sim \triangle ABC$ hence $\tfrac{BR}{RC}=\tfrac{CR^{*}}{BR^{*}}=\tfrac{FX}{EX}=\tfrac{DF}{DE}$ as desired. $\blacksquare$

This is definitely the longest angle chase I've ever done.

Let $A'$ be the reflection of $A$ in the perpendicular bisector of $BC$, and let $H'$ be the reflection of $H$ in $BC$. It is well known that $A'H'$ is a diameter of $(ABC)$. Let $R=A'H'\cap BC$, and let $S=RH\cap\ell$ where $\ell$ is the perpendicular bisector of $AH$. Also let $X=RQ\cap BE$ and $Y=RP\cap CF$. We will first show that $(SPQR)$ is cyclic and tangent to $BC$, and then we will show that $BR/CR=DF/DE$. Combining these two facts will solve the problem. In fact the first claim is just angle chasing. Note that \[\angle PHR=\pi-\angle PHS=\pi-(\pi/2-\angle PSH/2)=\pi/2+\angle PAH=\pi-\angle PBR,\]so $PHRB$ is cyclic. We similarly learn that $QHRC$ is cyclic. Note then that \[\angle BPR = \angle BHR = \angle BH'R=\angle BH'A'=\pi-\angle BAA'=\angle PBR,\]so $RP=RB$, and similarly $RQ=RC$. Then, we have that \[\angle BXR = \angle QXE = \pi/2-\angle XQE = \pi/2-\angle QCR=\angle XBR,\]so in fact $RP=RB=RX$ and similarly $RQ=RC=RY$. Incidentally, since $\angle QXE=\pi/2-C=\angle QAH$, we have that $X\in\gamma$, and similarly $Y\in\gamma$. Now, we see that \[\angle SPQ = \pi/2-\angle QSP/2=\pi/2-\angle PAQ=\angle QCH=\angle QRH=\angle SRQ,\]so $(PQRS)$ ic cyclic. Finally, we show the tangency. Note that \[\angle PRB=\angle PHB=\pi-\angle PHX=\angle PQR,\]so the tangency holds. Now, all we have to show is that $BR/CR=DF/DE$. Let $N$ be the intersection of $AD$ and $EF$. By the angle bisector theorem, we have that $DF/DE=NF/NE$. Note that if $AEF$ is reflected in the angle bisector of $\angle A$, we would get a triangle that is a homothety at $A$ of $ABC$. Thus, letting $R'$ be the intersection of the isogonal conjugate of $AN$, namely $AO$ ($O$ is the center of $(ABC)$), with $BC$, we have that $CR'/BR'=NF/NE=DF/DE$. We see from the construction that $R$ is isotomic to $R'$, so $BR/CR=DF/DE$, as desired.

@above: I had exactly the same motivation too First off, we prove that $S,H,R$ are collinear. Let $SH\cap BC=R'$. It's straightforward to prove that $PHR'B$ is cyclic ($\angle SHP=90^\circ-\angle PAH=\angle ABR'$). Similarly, $QHR'C$ is cyclic. From these we get that $\angle SR'P=\angle SR'Q$. So $SR'$ bisects $\angle PR'Q \implies SPR'Q$ is cyclic (note that $SP=SQ$). Thus $R,R'$ both lie on $\odot(SPQ)$ and also lie on the same line $BC$. Well, we know that $\odot(SPQ)$ touches $BC$ at $R$, so we must have $R=R'$. So $S,H,R$ are collinear. $\qquad\blacksquare$ To see that $H$ is the incenter of $PQR$, note that it lies on the angle bisector of $\angle PRQ$ and also $\angle PHQ=90^\circ+\angle PRQ$ (easy angle chase). We have also shown above that $PBRH$ and $QCRH$ are cyclic. We will use these two to prove that $RP=RB$ and $RQ=RC$. We have $\angle QRC=\angle QPR=2\angle HPR=2\angle HBC=180^\circ-2\angle C\implies \angle RQC=\angle C\implies RQ=RC$. Similarly $RP=RB$. $\qquad\blacksquare$. Finally, note that $\Delta DEF\sim\Delta RQP$ (for e.g. see that $\angle PRQ=2\angle PRH=2\angle ABH=2\angle FDH=\angle FDE$. Similar for other angles). So $\frac{DF}{DE}=\frac{RP}{RQ}\implies \frac{DF}{DE}=\frac{RB}{RC}$.$\qquad\qquad\square$

This problem has a great gliding principle feel to it. Let $M$ be the midpoint of $AH$. Note that $FMEH$ is directly similar to $PSQH$. Let $SH$ meet $BC$ in $R'$. Then, $SH/HR' = MH/HD$, so $PSQR'$ is directly similar to $FMED$, which is cyclic. So $R' \equiv R$, and $H$ is the incenter of $PQR$. Then $\angle CRQ = \angle RPQ = 180 - 2\angle C, \angle BRP = \angle RQP = 180 - 2\angle B$, then $BR = RP, CR = RQ$, and $BR/CR = RP/RQ = DF/DE$, and we're done.

Wait I think I finished differently than the above solutions... does the spiral similarity I used at the end work?

@above yes in my opinion it works

Let $R'=SH\cap BC$. We claim $HR'CQ$ and $HR'BP$ are cyclic. Indeed, since $NS\parallel BC$, \[ \angle HR'B=\angle HSN=\tfrac12 \angle HSA=\angle HPA=180-\angle HPB,\]proving $HR'CQ$ cyclic, and similarly $HR'BP$ cyclic. Now, note that $\angle HR'P=\angle HBP=\angle EBA=90-A$, and similarly $\angle HR'Q=90-A$. So $\angle PRQ=180-2A$, and furthermore $HR'$ bisects $\angle PRQ$. Now we want to use the real definition of $R$ to calculate $\angle PRQ$. We know $(PQRS)$ is cyclic, so \begin{align*} \angle PRQ & = 180-\angle PSQ = 180-\angle PSH-\angle QSH \\ &= 180-2\angle PAH -2\angle QAH = 180-2A. \end{align*}In conclusion, we know $R,R'\in BC$, and $\angle PRQ=\angle PR'Q$. This implies $R=R'$. So we now know the fundamental properties of the configuration: $S,H,R$ collinear, $HRBP$ cyclic, $HRCQ$ cyclic. from which it is easy to prove anything else. We know $HRBP$, $HRCQ$ are cyclic. Also, $HDBF$ is cyclic. This means $R$ is the center of spiral similarity $FD\to PR$ and $ED\to QR$. So $FD/PR = HD/HR=ED/QR$, i.e. $DF/DE = PR/QR$. So it remains to show that $PR/QR = BR/CR$. We claim that actually $PR=BR$ (and similarly $QR=CR$); this finishes by the above. Indeed, using the tangency, we get \begin{align*} \angle BRP &= \tfrac12 \widehat{PR} = \angle PSR = 2\angle PAD = 180-2B, \end{align*}and since $\angle PBR=B$, this implies $\triangle PRB$ is isosceles, finishing.

Claim 1: The quadrilaterals $BPHR$ and $CQHR$ are cyclic. Proof: By Miquel's Pivot Theorem, the circles $(BPH), (CQH)$ intersect again at a point $R'$ on $BC$. Compute $$\angle PRQ = 180^\circ -\angle PSQ = 180^\circ - 2 \angle A = \angle PBH +\angle QCH = \angle PR'H + \angle QR'H = \angle PR'Q$$implying $R=R'$ as desired. Since $\angle PRH = \angle PBH = 90^\circ-\angle A = \frac{1}{2} \angle PRQ$ we have that $H$ lies on the angle bisector of $\angle PRQ$; this implies by fact 5 that $H$ is the incenter of $PQR$. Claim 2: $\Delta DEF$ is similar to $\Delta RQP$. Proof: Note $\angle PQR = 2 \angle HQR = 2 \angle HCB = 180^\circ - 2 \angle B = \angle DEF$. Combined with the fact that $\angle PRQ = 180^\circ - 2 \angle A = \angle FDE$ the claim is proven. Since $\angle PRB = 180^\circ - 2 \angle B$ we get that $RB=RP$. Similarly $RC=RQ$ and thus by Claim 2 $$\frac{BR}{CR} = \frac{RP}{RQ} = \frac{DF}{DE}$$as desired.

Claim: $BPHR$ is cyclic, along with $CQHR.$ Proof: Suppose $(BPH)$ and $(CQH)$ intersect at point $R'.$ Since $APHQ$ is cyclic, $$\measuredangle HR'B + \measuredangle HR'C = \measuredangle HPA + \measuredangle HQA = 0,$$so $R' \in BC.$ Then, $$\measuredangle PR'B + \measuredangle CR'Q = \measuredangle PHB + \measuredangle CHQ = \measuredangle BHC + \measuredangle QHP = 2 \measuredangle CAB$$which means that $\measuredangle PR'Q = - 2 \measuredangle CAB.$ Since $S$ is the center of $(AH),$ $\measuredangle QSP = 2 \measuredangle CAB,$ so $R' \in (PSQ),$ meaning $R' \equiv R.$ Claim: $\triangle PQR \sim \triangle FED.$ Proof: This is another angle chase. We have $$\measuredangle PQR = \measuredangle PQH + \measuredangle HQR = \measuredangle FEH + \measuredangle HED = \measuredangle FED$$and $$\measuredangle QRP = \measuredangle QRH + \measuredangle HRP = \measuredangle EDH + \measuredangle HDF = \measuredangle EDF$$which readily gives the desired. To finish, remark that $$\measuredangle DEF = \measuredangle RQP = \measuredangle BRP = 180 - 2 \measuredangle ABC.$$This implies $$\measuredangle RPB = \measuredangle PBR = \measuredangle ABC$$so $BR = PR.$ Similarly $CR = QR.$ To finish, $\triangle PQR \sim \triangle DEF$ implies that $\frac{BR}{CR} = \frac{PR}{QR} =\frac{DF}{DE},$ as desired. $\blacksquare$

Lemma 1: $BQHR$ is cyclic. Proof: Draw $(BQH)$, and define $(BQH) \cap BC = R^\prime$. By Miquel, we see that $CR^\prime HP$ is cyclic. We then proceed with an angle chase: \begin{align*} \angle QR^\prime P &= \angle QR^\prime H + \angle HR^\prime P \\ &= \angle QBH + \angle HCP \\ &= (90^\circ - \angle A) + (90^\circ - \angle A) \\ \angle QR^\prime P &= 180^\circ - 2 \angle A \end{align*}By inscribed angle, $\angle QOP = 2 \angle A$. From here, it is obvious that $QOPR^\prime$ is cyclic, and that $R^\prime$ lies on $(QOP)$. This is only possible if $R^\prime = R$, which means that $BQHR$ is cyclic. $\Box$ Lemma 2: $\triangle PQR \sim \triangle EFD$. Proof: We can easily see that $BFHD$ and $CEHD$ are cyclic. This allows us to do the following angle chase: \begin{align*} \angle FDE &= \angle FDH + \angle HDE \\ &= \angle FBH + \angle HCE \\ &= 180^\circ - 2 \angle A \\ \angle FDE &= \angle QRP \end{align*}A similar process can be repeated for the other angles of $\triangle FDE$, from which we can verify the Claim. $\Box$ Lemma 3: $BR = QR$ and $PR = RC$. Proof: Because $R$ is a point of tangency, $$ \angle BRQ = \angle QPR = 180^\circ - 2 \angle B $$From here, it is easy to see that $BRQ$ is isosceles, with $BR=RQ$. A similar process can be repeated for the second half of the claim, giving us $PR = RC$. Now, we apply Lemma 2 and finish with Lemma 3, with $$\frac{ED}{FD} = \frac{PR}{QR} = \frac{CR}{BR}$$as desired. $\blacksquare$

Tricky but very rewarding. Relax the tangency condition and let $T \neq R = \overline{SH} \cap \overline{BC}$. Assume $T$ is closer to $B$ than $C$. Notice that $$\angle PHS = 90^\circ - \angle BAD = \angle B = \angle PBT$$so $BPHT$ is cyclic. By Miquel theorem $CQHT$ is also cyclic. Claim. [Main Claim] $RB=RP$ and $RQ=RC$ similarly. Proof. Long angle chase! \begin{align*} 180^\circ - \angle APR &= 180^\circ - \angle APQ - \angle QPR \\ &= 180^\circ - \angle APQ - (180^\circ - \angle QTB) \\ &= \angle QTB - \angle APQ \\ &= \angle QTS + \angle STB - \angle APQ \\ &= 90^\circ - A + \angle APH - \angle APQ \\ &= 90^\circ - A + \angle QPH \\ &= 180^\circ - A - C = B. \end{align*}This implies the result. $\blacksquare$ To finish, the problem statement condenses into $\triangle RPQ \sim \triangle DFE$ when $T = R$. But this is evident as $\angle PRQ = 180^\circ - 2\angle A = \angle FDE$ and cyclic permutations.

We will first prove a crucial claim, and this is in fact the main idea of the problem. \begin{claim} There exists a point $R'$ on $BC$ such that $R'B=R'Q$ and $R'C=R'P$. \end{claim} \begin{proof} This is saying that $Q$ and $P$ are the reflections of $B$ and $C$ across the feet of the altitudes from $R$ to $AB$ and $AC$. Call the feet $P_B$ and $P_C$. However, by similar triangles, we have $$\frac{BP_B}{BF}=\frac{BR}{BC}$$and $$\frac{CP_C}{CE}=\frac{CR}{BC},$$so it suffices to show that $$\frac{BP_B}{BF}+\frac{CP_C}{CE}=1$$since if this is true then one of the isosceles triangles implies the other. This is of course equivalent to $$\frac{BQ}{BF}+\frac{CP}{CE}=2.$$ Note that $P$ and $Q$ are related by $APHQ$ cyclic. To show this, we will barybash wrt $\triangle ABC$. Suppose $P=(p,0,1-p)$ and $Q=(q,1-q,0)$. Letting $k=S_AS_B+S_AS_C+S_BS_C$, The above condition that $$\frac{BQ}{BF}+\frac{CP}{CE}=2$$is equivalent to $$\frac{p}{\frac{S_C}{S_C+S_A}}+\frac{q}{\frac{S_B}{S_B+S_A}}=2$$or multiplying by $S_BS_C$, this is $$b^2pS_B+c^2qS_C=2S_BS_C.$$By plugging in $A,P,Q$, the equation of circle $(APHQ)$ is then $$(x+y+z)(c^2qy+b^2pz)=a^2yz+b^2xz+c^2xy.$$We know that $H=(S_BS_C,S_AS_C,S_AS_B)$ also lies on this circle, so $$k(c^2qS_AS_C+b^2pS_AS_B)=\sum_{cyc} a^2S_BS_C.$$Dividing by $kS_A$ gives the desired result. \end{proof} Now, the problem is not too difficult from here. \begin{claim} $R'$ as defined above lies on $(OPQ)$. \end{claim} \begin{proof} Let $\angle A=\alpha$ etc. Then, $$\angle QRB=180-2\beta,\angle PRC=180-2\gamma,$$which means that $$\angle QRP=180-(180-2\beta + 180-2\gamma)=2\beta+2\gamma-180=180-2\alpha$$and $$\angle QOP=2\angle QAP=2\alpha.$$\end{proof} This means that, in the given problem statement, $R$ is the same point as the point $R'$ we defined. Thus, it suffices to show that $$\frac{R'B}{R'C}=\frac{DF}{DE}.$$However, note that $$\angle QR'P=\angle FDE=180-2\alpha,$$and $$\angle QPR'=\angle QR'B=180-2\beta,$$so $\triangle DEF\sim \triangle R'PQ$ and we are done. remark: We can actually draw in $R$ accurately on the diagram immediately by using the length condition the problem asks us to prove. From here, we can get a rough estimate of where the circle $(APHQ)$ should be from the point $R$, and admittedly some trial and error/binary searching. From here it seems true that both $BR=RQ$ and $CR=RP$, so we then prove that such a point exists and then show it satisfies all of the desired properties. However, I noticed something weird about this problem. When doing the barybash, the coefficients of $p$ and $q$ in the condition that is equivalent to the existence of $R'$ are "in the same proportion" as the coefficients of $p$ and $q$ in the condition that $APQH$ is cyclic, allowing us to scale directly. This was "lucky" in a sense, as this means that the existence of $R'$ does not rely on the fact that $(OPQ)$ is tangent to $BC$, as we never used that condition in the bary bash, and only requires $APQH$ to be cyclic.

Nice question. Spent a lot of time on this

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8 cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.1) + fontsize(8); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -2.079116415157009, xmax = 21.981098681428076, ymin = -7.1854532813250165, ymax = 4.4581131063994714; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0.); draw((7.1,2.38)--(5.396656465371547,-4.856426824148215)--(12.469419555800402,-4.891440502813705)--cycle, linewidth(0.9) + zzttqq); /* draw figures */ draw((7.1,2.38)--(5.396656465371547,-4.856426824148215), linewidth(0.9) + zzttqq); draw((5.396656465371547,-4.856426824148215)--(12.469419555800402,-4.891440502813705), linewidth(0.9) + zzttqq); draw((12.469419555800402,-4.891440502813705)--(7.1,2.38), linewidth(0.9) + zzttqq); draw(circle((5.896667983403489,-0.6144011737757799), 3.2271421306902477), linewidth(0.9)); draw(circle((7.663095632463902,-2.375230415615365), 2.49415040811144), linewidth(0.9)); draw((5.396656465371547,-4.856426824148215)--(9.95695704784056,-1.4889830393352765), linewidth(0.9)); draw((7.1,2.38)--(7.064135239259586,-4.8646816695635025), linewidth(0.9)); draw((5.760148117239355,-3.3121812351314825)--(12.469419555800402,-4.891440502813705), linewidth(0.9)); /* dots and labels */ dot((7.1,2.38),dotstyle); label("$A$", (7.166317058022571,2.5332958986726597), NE * labelscalefactor); dot((5.396656465371547,-4.856426824148215),dotstyle); label("$B$", (5.462380513477529,-4.6926572254165215), NE * labelscalefactor); dot((12.469419555800402,-4.891440502813705),dotstyle); label("$C$", (12.530561735294,-4.739988796098328), NE * labelscalefactor); dot((7.064135239259586,-4.8646816695635025),linewidth(4.pt) + dotstyle); label("$D$", (7.134762677568033,-4.739988796098328), NE * labelscalefactor); dot((9.95695704784056,-1.4889830393352765),linewidth(4.pt) + dotstyle); label("$E$", (10.021988489158243,-1.3636700874627719), NE * labelscalefactor); dot((5.760148117239355,-3.3121812351314825),linewidth(4.pt) + dotstyle); label("$F$", (5.825255888704714,-3.1938241538259704), NE * labelscalefactor); dot((7.070294210556583,-3.620569467570122),linewidth(4.pt) + dotstyle); label("$H$", (7.134762677568033,-3.4935907681440805), NE * labelscalefactor); dot((5.896667983403489,-0.6144011737757799),dotstyle); label("$O$", (5.967250600750134,-0.46437024450844155), NE * labelscalefactor); dot((5.637988719646843,-3.8311590711280683),linewidth(4.pt) + dotstyle); label("$P$", (5.699038366886563,-3.698694241098577), NE * labelscalefactor); dot((9.112588532445418,-0.34551206294038433),linewidth(4.pt) + dotstyle); label("$Q$", (9.170020216885723,-0.21193520087213827), NE * labelscalefactor); dot((7.74455888871129,-4.868050103471679),linewidth(4.pt) + dotstyle); label("$R$", (7.813181857340596,-4.739988796098328), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] CLAIM-$O,H,R$ are collinear. PROOF- Let $OH \cap BC \equiv R'$ $\angle PHR'= 180 -\angle OHP= 90 + \angle PAH =90 + \angle BAD=180 - \angle PBR'$ Thus,$PHR'B$ is cyclic. Similarly $HQCR'$ is cyclic too. $\angle FDH= \angle HDE$ as $H$ is the incenter of $DEF$ and also $FHDB$ and $HDCE$ are cyclic. Therefore $\angle HDF= \angle HBF =\angle HBP =\angle HR'P$. Similarly, $\angle HDE= HCE$. This implies that $R'H$ is the bisector of $\angle PR'Q$. $\angle OPQ= \angle OQP=\angle ORP= \angle ORQ$ .Thus $R'$ lies on $(OPQ)$ and as it also lies on $BC$ implies $R \equiv R'$. CLAIM- $RP=RB$ and $RQ=RC$. PROOF- $\angle PQH=\angle PQR /2 = \angle PRB /2$ but $\angle RHP= 90+\angle PRB /2= 90+ \angle PQH= 90+ \angle PRB/2 = 180- \angle PBR$. Thus $\angle PBR= 90-\angle PRB/2$ implying $RP=RB$. A similar and symmetric angle chase for $RQ=RC$ suffices. It's easy to see that $\triangle PQR \sim \triangle FED$ as $\angle FDE= 2\angle FDH= 2\angle FBH =180- 2\angle BAC= \angle QRP$ (and similar angle chases for other equalities) We finally conclude, $\frac{ED}{FD} =\frac{QR}{PR}= \frac{CR}{BR}$.

We will solve this problem through a series of claims: Claim 1: Quadrilaterals $BPHR$ and $CQHR$ are cyclic. Proof: We have $\angle APH = \angle CQH$, so $H$ is the Miquel point of $\triangle ABC$. This implies that $(BFH)$ and $(CQH)$ intersect at a point $R'$ on $\overline{BC}$. Hence, it is sufficient to prove that $R=R'$. A short angle chase will suffice: \begin{align*} \angle PRQ &= 180^\circ-\angle PSQ \\ &= 180^\circ - 2 \angle PAQ \\ &= (90-\angle A) + (90 - \angle A) \\ &= \angle PBH + \angle QCH \\ &= \angle PR'H+ \angle QR'H = \angle PR'Q. \ \square \end{align*} The calculations from the proof of Claim 1 shows us that $\angle PRH = \frac{1}{2} \angle PRQ$, so $\overline{RH}$ is the angle bisector of $\angle PRQ$. Moreover, we have $SP=SQ=SH$, so $H$ is the incenter of $\triangle PQR$ by Fact 5. Claim 2: $\triangle DEF \sim \triangle RQP$ Proof: Note that \begin{align*} \angle PQR &= 2 \angle HQR = 2 \angle HCR \\ &= \angle HCD+\angle HAF = \angle HEF+\angle HED \\ &= \angle DEF. \end{align*} Also, from the proof of Claim 1, we have \[\angle PRQ = \angle PBH + \angle QCH = \angle FDH + \angle EDH = \angle FDE.\] The desired claim follows from AA similarity. $\square$ Claim 3: $RB=RP$, and symmetrically $RC=RQ$ Proof: Observe that \[\angle PRB = \angle PQR = 2 \angle HCR = 180^\circ-2\angle B. \ \square \] Remark: Second equality from the proof of Claim 2. Thus, \[\frac{BR}{CR} = \frac{RP}{RQ} = \frac{DF}{DE}. \ \square\]

Reinterpret the problem condition as letting $O$ vary and change the conclusion to be conditional on the tangency fact, and drop the definition of $R$ provided. In general, let $\overline{OH}$ intersect $(OPQ)$ again at $R$, so by incenter-excenter $H$ is the incenter of $\triangle RPQ$. Moreover, as $O$ varies the angles of $\triangle HPQ$ don't change, so neither do the angles of $\triangle RPQ$, which is in fact always similar to $\triangle DEF$, and thus the center of the spiral similarity between any two choices of $\triangle RPQ$ (corresponding to two choices of $O$) is $H$. Since $P$ and $Q$ move on fixed lines as $O$ varies, $R$ should as well. Moreover, when $O$ is the midpoint of $\overline{AH}$, $\triangle RPQ$ is just the orthic triangle, and when $O$ is chosen so that $\overline{OA} \perp \overline{AC}$ then $P=A$. In this case, one can check that having $R$ as the reflection of $B$ over $D$ and defining $Q$ to be the point on $\overline{AB}$ such that $\angle AQR=180^\circ-2\angle C$ will yield $\triangle RPQ \sim \triangle DEF$, hence $R \in \overline{BC}$ always. (Note: this entire section after obtaining $\triangle DEF \sim \triangle RPQ$ can more or less be replaced with noting that $(BRHQ)$ and $(CRHP)$ are cyclic and citing the converse of Miquel, but I'm synthetically challenged). Now when $(OPQ)$ is tangent to $\overline{BC}$, our definition of $R$ coincides with that of the problem. Moreover, $\triangle BRQ$ and $\triangle CRP$ are isosceles (with vertices $B$ and $C$ respectively), so $\frac{CR}{BR}=\frac{PR}{QR}=\frac{ED}{FD}$. $\blacksquare$

Denote $R' = SH \cap BC$ and $M$ as the midpoint of $AH$. We can chase spiral similarities to get \[\triangle HDR' \sim \triangle HEQ \sim \triangle HFP \sim \triangle HMS,\] which implies that $H$ is the Miquel point of both $\triangle PQR'$ and $\triangle DEF$ with respect to $ABC$. Hence the converse of the Spiral Triangle Lemma tells us the two are similar, so \[\angle PR'Q = 180 - 2A = 180 - \angle PSQ \implies R' = R.\] The newly added tangencies induces isosceles triangles $\triangle BPR$ and $\triangle CQR$, so \[\frac{BR}{CR} = \frac{PR}{QR} = \frac{DF}{DE}. \quad \blacksquare\]