April wrote: In an acute triangle $ ABC$ segments $ BE$ and $ CF$ are altitudes. Two circles passing through the point $ A$ anf $ F$ and tangent to the line $ BC$ at the points $ P$ and $ Q$ so that $ B$ lies between $ C$ and $ Q$. Prove that lines $ PE$ and $ QF$ intersect on the circumcircle of triangle $ AEF$.

I have same soloutions as Mr.danh's. I have a question everybody : Isnt G4 a little easier than G3?

Thanks to Davood Vakili for a good problem.

Quote: In an acute triangle $ ABC$ segments $ BE$ and $ CF$ are altitudes. Two circles passing through the point $ A$ anf $ F$ and tangent to the line $ BC$ at the points $ P$ and $ Q$ so that $ B$ lies between $ C$ and $ Q$. Prove that lines $ PE$ and $ QF$ intersect on the circumcircle of triangle $ AEF$. Proposed by Davood Vakili, Iran Let $ H$ be the orthocenter of $ \triangle ABC$ and $ D$ is the projection of $ A$ onto $ BC$. Then consider the inversion through pole $ H$, power $ k = \overline {HA}.\overline {HD}$ $ = \overline {HE}.\overline {HB}$ $ = \overline {HC}.\overline {HF}$. We have $ \mathcal {I}(H,k)$ maps $ P$ into $ P'$ and $ Q$ is taken into $ Q'$, combine with the result that $ \mathcal {I}(H,k)$ maps $ BC$ into $ AEF$. Thus, $ P'$ and $ Q'$ both lie on $ (AEF)$. Therefore, obviously, $ H$ will be also the orthocenter of $ \triangle AQP$, and $ Q',$ $ P'$ respectively are the projections of $ P,$ $ Q$ onto $ AQ$ and $ AP$. Since $ AF$ is the radical axes of the two circles that pass through $ A,$ $ F$ and tangent to $ BC$ at $ P,$ $ Q$ respectively. Therefore, $ AF$ will goes to midpoint $ B$ of $ PQ$. In the other hand, we also have $ BP^2$ $ = BQ^2$ $ = \overline {BF}.\overline {BA}$ $ = \overline {BD}.\overline {BC}$. Then if $ C'$ is the intersection of $ P'Q'$ with $ PQ$, we have $ (QPDC')$ is harmonic. And by MacLaurin identity, $ BQ^2$ $ = BP^2$ $ = \overline {BD}.\overline {BC'}$, which implies $ C$ must coincide with $ C'$. Or we can say $ P',$ $ Q'$ and $ C$ are collinear. From $ (CDPQ)$ is harmonic, then $ \overline {CA}.\overline {CE}$ $ = \overline {CP}.\overline {CQ}$ $ = \overline {CD}.\overline {CB}$ $ \Longrightarrow$ $ E$ $ \in$ $ (AQP)$. The same argument holds for $ F$ $ \in$ $ (HPQ)$. Finally, we have $ \angle (FQ,EP)$ $ = \angle (FQ,QP) + \angle (QP,EP)$ $ = \angle (HF,HP') + \angle (AQ,AE)$ $ = \angle (AF,AP') + \angle (AP',AE)$ $ = \angle (AF,AE)$. Thus, if $ W$ is the intersection of $ EP$ with $ FQ$, then $ W$ $ \in$ $ (AEF)$. Our proof is completed then. $ \square$

Let $H$ be the orthocenter of $\triangle ABC$ and let $D$ the foot of the A-altitude. $B$ lying on the radical axis $AF$ of both circles has equal power with respect to them. Thus $BQ^2 = BP^2 = BF \cdot BA = BH \cdot BE = BD \cdot BC$ $\Longrightarrow$ circle $ \odot(PHE)$ is tangent to $ BC$ at $P,$ yielding $\angle EPQ = \angle BHP$ and also due to Newton's theorem, the cross ratio $(D,C,P,Q)$ is harmonic $\Longrightarrow$ $CP \cdot CQ = CD \cdot CB = CH \cdot CF,$ i.e. $FHPQ$ is cyclic $ \Longrightarrow$ $\angle FQP = \angle CHP.$ Therefore if $R\equiv QF \cap PE,$ we have $\angle QRP = \angle BAC$ $\Longrightarrow$ $ R \in \odot(AFE).$

Here is my solution: First inversion with center A and radius AF. Now it is enough to prove that circumcircles of $AP_{1}E_1$ and $AQ_{1}F$ intersects at point R which belongs to the line $ E_{1}F$. This is equivalent to $\angle{AP_{1}E_{1}}=\angle{AQ_{1}F}$. We see $\angle{AP_{1}E_{1}}=\angle{P_{1}AF}+\angle{AFP_{1}}+\angle{P_{1}E_{1}F}+\angle{P_{1}FE_{1}}$ $=\angle{B_{1}P_{1}F}+\angle{B_{1}FP_{1}+\angle{LP_{1}F}+\angle{P_{1}FL}}$ (here we used that $FP$ is tangent to circumcircle of $AQB$ and then it is also tangent to circumcircle of $ELP_1$ ; $L$ belongs to $E_{1}F_{1}$ and $AL$ is perpendicular to ${E_{1}F_1}$) Because $\angle{AQ_{1}P_1}=\angle{B_{1}P_{1}F}+\angle{B_{1}FP_{1}}$. So we see that we need to show that ${P_{1}Q_{1}FL}$ is cyclic. That is true and we'll prove this if we take $S$ as midpoint of $AH$ (H is orthocenter of AEF). It holds that $\angle{SP_{1}F}=\angle{SQ_{1}F}=90=\angle{SLF}$. Done!

We have to prove that $ \widehat{BFQ}=\widehat{PEC} $.We have $ BQ^2=BA \cdot BF $,triangles $ BFQ,BQA $ are similar therefore $ \widehat{BFQ}=\widehat{BQA} $.It remains to prove that $ \widehat{PEC}=\widehat{BQA} $. Since $ BP=BQ $ and $ BP^2=BD \cdot BA $,$ D $ is inverse of $ C $ wrt circle $ (B,BP) $ so we have $ (C,D;P,Q)=-1 $.By a known theorem we have $ CP \cdot CQ=CD \cdot CB $,but $ EABD $ is cyclic so $ CE \cdot CA=CD \cdot CB=CP \cdot CQ $.Thus, $ PEAQ $ is cyclic,and $ \widehat{PEC}=\widehat{BQA} $.

Dear Mathlinkers, in order to have a more upper view for each problem, we can image that each proposed problem are built on a squeletton i.e. based on a well known situation. In this case, I have though to the Boutin's theorem which originated a chain of well known results and conduct to the result in question. This is only a personnal... to be developped... Sincerely Jean-Louis

My Solution : Consider $X$ such that $X\equiv(FQ)\cap(PE)$. If $X$ lies in $(AEF)$ then $\angle{XEA}=\angle{XFA}\Leftrightarrow \angle{CEP}=\angle{BFQ}$. Since $(AFQ)$ is tangent to $BC$ and $(AFP)$ si tangent to $BC$ then we have : $BQ^{2}=BF.BA=BP^{2}$. Otherwise, we have : $\angle{BFQ}=\angle{ABC}-\angle{FQC}=\angle{ABC}-\angle{BAQ}$ and $\angle{CEP}=\pi-\angle{ACB}-\angle{CPE}$. Hence we are left to prove that : $\angle{CAQ}=\angle{CPE}$ which means it suffices to prove that $\triangle{ACQ} \sim \triangle{CPE}$. From the fact that $\triangle{CFB} \sim \triangle{CDH}$ and $\triangle{BHD} \sim \triangle{BEC}$, we have : $\frac{BH}{BC}=\frac{BD}{BE}$ and $\frac{CH}{CB}=\frac{CD}{CF}$ which give us : $BH.BE+CH.CF=BC(CD+BD)=BC^{2}$. But we have : $BP^{2}=BF.BA=BH.BE$ and $CH.CF=CE.CA$, then we get : $CE.CA=BC^{2}-BP^{2}=(CB-BP)(CB+BQ)=CP.CQ$ and hence : $\frac{CA}{CQ}=\frac{CP}{CE}$ which means that $\triangle{CAQ} \sim \triangle{CPE}$ which give us the desired result.

It's obvious that if we prove that AEPQ is cyclic then we will be done. Now take inversion in A which maps circle BCEF to itself and our thesis is equivalent to that $Q', P', C$ are collinear, but this is well-known result about polars.

Let us invert the figure wrt $A$ with power $AF.AC=AE.AB$.So $BC$ goes to $\odot AEF$.Now $\odot APF,\odot AQF$ goes to the tangents drawn from $C$ to $\odot AEF$.So the inverses of $P,Q$ (suppose $P',Q'$ ).So we just need to show that the circles $\odot AP'B, \odot AQ'C$ concur on $BC$.For this note that $P'Q'$ is the polar of $C$ wrt $\odot AEF$.So $B,P',Q'$ are collinear. Now the rest is just angle chasing.

Let M be the intersection point of QF and PE. We need to prove that ∠QMP = ∠BAC. Since ∠MQP = ∠QAB (QB is a tangent to the circle around △QFA), it is enough to prove that ∠QAB+∠BAC = ∠QMP+∠MQP, or equivalently, ∠QAE =∠EPC. Thereforewe need to prove that AQPE is a cyclic quadrilateral. From $BQ^2 = BF .BA = BP^2$ we get BP = BQ. Adding $BF .BA = BP^2$ to AF · AB = AE ·AC (which holds since BCEF is cyclic) we get $AB^2 = AE .AC+BP^2.$ From the Pythagorean theorem we have $AB^2 = AE^2+BE^2 = AE^2+BC^2-CE^2$, from which we get $BC^2-CE^2 = AE .EC+BP^2$ This implies that $BC^2-BP^2 = CE2+AE .EC$, or equivalently CE . (CE +AE) = (BC+BP)(BC−BP) =CQ·CP. Thus CE ·CA =CP·CQ and QPEA is cyclic.

I used trigonometry to prove that $QD:DH=AD:DP\Leftrightarrow QH\perp AP$, so $H$ is the orthocenter of $\triangle{APQ}$, thus $PH\perp AQ$. $BP^2=BF\cdot BA=BH\cdot BE$ implies $\angle{BPH}=\angle{BEP}$, and saecula saeculorum says $\angle{QFB}=\angle{AQB}$. Let the intersection be $X$, and get $\angle {XFH}=\frac{\pi}{2}-\angle {AQB}=\angle{HPB}=\angle{BEP}$. Thus $F,H,E,X$ cyclic.

let $D$ is foot of perpendicular from $A$ on $BC$. $BQ^{2}=BF.BA=BD.BA=BC^{2}-CD.BC\Rightarrow CE.CA=CD.BC=BC^{2}-BQ^{2}=BC^{2}-BP^{2}=CP.CQ$ so $AEPQ$ is cycle. $\widehat{QAB}=\widehat{FQB},\widehat{EPC}=\widehat{A}+\widehat{QAB\Rightarrow }$ lines PE and QF intersect on the circumcircle of triangle AEF. done

Define $H$ as orthocenter and $D$ such that $AD$ is altitude. Let $H'$ be reflection of $H$ across $BC$. Firstly notice that by power of point we have $BF \times BA = BD \times BC = DB \times (DC+DB) \Rightarrow BP^2-BD^2 = (BF \times BA) - BD^2 = DB \times DC = DH' \times DA$ but $BQ=BP$ because $BP^2=BF \times BA = BQ^2$. So $BP^2-BD^2 = PD \times DQ$ and so $PD \times DA = DA \times DH'$. From this we see that $H$ is the orthocenter of $APQ$. Notice that $BH \times BE = BF \times BA = BP^2$ and so the circumcircle of $PEH$ is tangent to $BC$. Therefore $\angle EPB = 180-\angle EHP$. The rest is just angle chasing. Also, we see $\angle BQF = \angle FAQ$. Since $PH \perp AQ$ and $BE \perp AC$ we see that $\angle EPB=180-\angle CAQ = 180 - \angle CAB - \angle FQB$ and so the lines $PE$ and $QF$ for an angle of $\angle CAB$, so we are done.

Suppose that $QF$ intersects $(AEF)$ at $L$. We will prove $E,L,P$ are collinear. We have $BQ^2=BP^2=BF \cdot BA=BD \cdot BC$ so $BQ=BP$ and $BP^2=BD \cdot BC$. Let $K$ be the orthorcenter of triangle $ABC$ and $AD$ is the altitude. $AL \cap BC=R$ then $AR \perp KL$. Therefore $KDRL$ is cyclic. Hence $AL \cdot AR= AK \cdot AD= AF \cdot AB= AE \cdot AC$. It follows that $BFLR$ and $LECR$ are cyclic. Since $BFLR$ is cyclic then by power of point we get $QO^2-OK^2=QF \cdot QL= QB \cdot QR$. We have \[\begin{aligned} QO^2-OK^2 & =QO^2-(OD-DK)^2 \\ & =QD^2+DK \cdot DA \\ & = (QB+BD)^2+DB \cdot DC \\ & =QB^2+2QB \cdot BD + BD \cdot BC \\ & =2QB \cdot BD \end{aligned}\] Thus, $2BD=QR$ or $D$ is the midpoint of $QR$. Now we need to prove $\overline{PR} \cdot \overline{PC}= PO^2-OK^2$. Similarly, we obtain $PO^2-OK^2=PD^2-\overline{DB} \cdot \overline{DC}$. Hence, \[\begin{aligned} \overline{PR} \cdot \overline{PC}= PO^2-OK^2 & \iff \overline{PR} \cdot \overline{PC}=PD^2-\overline{DB} \cdot \overline{DC} \\ & \iff (\overline{PR}+ \overline{DP}) \cdot \overline{PC}= \overline{DP} \cdot ( \overline{DP}+ \overline{PC})- \overline{DB} \cdot \overline{DC} \\ & \iff \overline{DR} \cdot \overline{PC}= \overline{BP} \cdot \overline{DC} \\ & \iff \overline{QD} \cdot \overline{PC}= \overline{QB} \cdot \overline{DC} \\ & \iff (\overline{QB}+ \overline{BD}) \cdot \overline{PC}= \overline{QB} \cdot \overline{DC} \\ & \iff \overline{BD} \cdot \overline{PC}= \overline{QB} \cdot \overline{DP} \qquad (1) \end{aligned}\] Since $B,D$ are the mid point of $QP$ and $QR$ then $\overline{PR}=\overline{QR}- \overline{QP}=2 (\overline{DR}- \overline{BP})= 2 (\overline{PR}- \overline{BD})$. It follows that $\overline{PR}=2 \overline{BD}$. Thus, $(1) \iff \overline{PR} \cdot \overline{PC}= \overline{QP} \cdot \overline{DP}$. We have \[\begin{aligned} BP^2= \overline{BD} \cdot \overline{BC} & \implies 2BP^2= \overline{PR} \cdot \left( \overline{BP} + \overline{PC} \right) \\ & \iff 2 \overline{BP} (\overline{BP}- \overline{BD}) = \overline{PR} \cdot \overline{PC} \\ & \iff \overline{PR} \cdot \overline{PC}= \overline{QP} \cdot \overline{DP} \end{aligned}\] So $(1)$ is true or $\overline{PR} \cdot \overline{PC}= PO^2-OK^2$. $PE \cap (AEF)=L'$ then $\overline{PL'} \cdot \overline{PE}= \overline{PR} \cdot \overline{PC}=PO^2-OK^2$. It follows that $EL'RC$ is cyclic. Thus, $L' \equiv L$ or $E,L,P$ are collinear.

Let $\{M\} = QF \cap EP$. We observe that $\odot AFE = \odot FEH$, since $AFHE$ is cyclic. So, we need to prove that $FEMH$ is cyclic, namely that $\angle HEM \equiv \angle HFM$. Because $\odot FAP$ and $\odot FAQ$ are tangent to $BC$, we get $BQ^2=BP^2=BF \cdot BA \Rightarrow BP^2 = BH \cdot BE$ $\Rightarrow \Delta BHP$ ~ $\Delta BPE \Rightarrow \angle BEP \equiv \angle HPB$. So, we must have $ \angle HPQ \equiv \angle MFH$, namely that $QFHP$ is cyclic. This reduces to $CH \cdot CF =QC \cdot PC \Leftrightarrow BC \cdot AC \cos (\widehat{ACB}) = QC \cdot PC $, which is equivalent to $BC \cdot AC \cos (\widehat{ACB}) = (BC + BP)(BC - BP) = BC^2 - BF \cdot BA$ $\Leftrightarrow$ $BC \cdot AC \cos (\widehat{ACB}) = BC^2 - BC \cdot BA \cos (\widehat{ABC}) $, which is obvious (after reducing the factor $BC$).

First,observe that BQ and BP are tangents,so we have that <BQA=<BFQ.Now,we need to prove that <BFQ+<PEA=180,whic is equivalent showing that <BQA+PEA=<PBA+PEA=180,so we prove that QPEA is a cyclic.Now,let D be the foot of the perpendicular of A on to BC.CP*CQ=CB*CB-BP*BP=CB*CB-BF*BA=CB*CB-BD*CB=CB*CD=CE*CA,so we are finished.

First, we note that $BF\cdot BA = BP^2 = BQ^2$, so $B$ is the midpoint of $PQ$, so $\odot(PQ)$ and $\odot(AEF)$ are orthogonal. Thus, $P$ lies on the polar of $Q$ w.r.t. $\odot(AEF)$. Now, let $PE$ and $PF$ intersects $\odot(AEF)$ again at $E_1$ and $F_1$, and let $Q' = E_1F\cap F_1E$. By Pascal's theorem on $E_1EEF_1FF$, we have $P$, $Q'$, $EE\cap FF$ (which is midpoint of $BC$) are collinear, so $Q'\in BC$. Moreover, by Brokard, $Q'$ also lies on the polar of $P$, so $Q=Q'$, and we are done.

UwU This problem was soo gooooddd!!! Loved it!! But why do I not see Inversion anymore these days. %T__T [asy][asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ /* A few re-additions are done using bubu-asy.py. This adds the dps, xmin, linewidth, fontsize and directions. */ pair A = (-29.15078,120.53093); pair B = (-88.16206,-43.66321); pair C = (64.00328,-43.66321); pair E = (26.95112,21.64522); pair F = (-70.75554,4.76903); pair H = (-29.15078,-10.18369); pair Pp = (-1.97925,-4.26783); pair Qp = (-94.36779,50.89344); pair Q = (-182.92214,-43.66321); pair P = (6.59800,-43.66321); pair X = (36.06622,50.89344); pair S = (-120.21173,-132.83883); pair R = (-56.11240,45.51240); import graph; size(13.1cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); pen ffxfqq = rgb(1,0.49803,0); draw(A--B, linewidth(0.5)); draw(C--A, linewidth(0.5)); draw(circle((-12.07939,21.69409), 100.30030), linewidth(0.5)); draw(circle((-29.15078,55.17361), 65.35731), linewidth(0.5) + blue); draw(circle(B, 94.76007), linewidth(0.5) + blue); draw(A--Q, linewidth(0.5) + blue); draw(A--P, linewidth(0.5) + blue); draw(Q--C, linewidth(0.5)); draw(C--F, linewidth(0.5)); draw(B--E, linewidth(0.5)); draw(Pp--Q, linewidth(0.5) + ffxfqq); draw(Qp--P, linewidth(0.5) + ffxfqq); draw(Q--X, linewidth(0.5) + linetype("4 4")); draw(X--P, linewidth(0.5) + linetype("4 4")); draw(B--S, linewidth(0.5)); dot("$A$", A, N); dot("$B$", B, SW); dot("$C$", C, SE); dot("$E$", E, dir(0)); dot("$F$", F, 1.5*dir(180)); dot("$H$", H, dir(270)); dot("$P^*$", Pp, dir(0)); dot("$Q^*$", Qp, NW); dot("$Q$", Q, SW); dot("$P$", P, SE); dot("$X$", X, dir(0)); dot("$S$", S, SW); dot("$R$", R, NW); [/asy][/asy] Let $H$ denote the orthocenter of $\triangle ABC$. Now let $R$ and $S$ be the second intersections of $\odot(B,BP)$ with $AB$. Firstly, from PoP we get that $BF\cdot BA=BP^2=BQ^2$ which gives us that $\odot(B,BP)$ is orthogonal to $\odot(AEF)$. Now perform an Inversion $\mathbf{I}(A,\sqrt{AF\cdot AB})$. Note that the condition $BF\cdot BA=BP^2$ gives us that $A$ is the inverse of $F$ w.r.t. $\odot(B,BP)$. This by using EGMO lemma 9.17 gives us that $AF\cdot AB=AR\cdot AS$. So we get that $\mathbf{I}(A,\sqrt{AF\cdot AB})$ fixes $\odot(PQR)$. Now note that the Inversion swaps $\odot(AEF)\leftrightarrow BC$. This thus gives that the intersections of $\odot(AEF)\cap\odot(PQR)$ and that of $\odot(PQR)\cap BC$ get swapped under this Inversion. Let the second intersections of $AP$ and $AQ$ with $\odot(PQR)$ be $P^*$ and $Q^*$ respectively. Now as $PQ$ is the diameter of $\odot(PQR)$, we get that $\measuredangle PP^*Q=90^\circ$. Similarly as $AH$ is the diameter of $\odot(AEF)$, we get that $\measuredangle AP'H=90^\circ$. This thus finally gives that $\overline{P^*-H-Q}$ are collinear. Similarly we also get that $\overline{Q^*-H-P}$ are also collinear. To finally finish, we have,\[-1=(P,Q;B,\infty_{BC})\overset{A}{=}(P^*,Q^*;F,A)\overset{Q}{=}(H,A;QF\cap\odot(AH),Q').\]We also have,\[-1=(P,Q;B,\infty_{BC})\overset{H}{=}(Q',P';E,H)\overset{P}{=}(H,A;PE\cap\odot(AH),Q').\] This finishes as $PE\cap\odot(AH)\equiv QF\cap\odot(AH)$ and we are done. Remark: Wizard_32 wrote: Cute result! I wonder if there is a way to accurately construct the point $P.$ yes there indeed is! you can take the circle that is orthogonal to $\odot(AEF)$ and consider its intersections with $BC$ to get $P$ and $Q$ as I did in my diagram.

By well-known properties, it is clear that $F$ is the A-Humpty Point of $\triangle APQ$, from which it follows that $H$ is also the orthocenter of $\triangle APQ$. As such, $(QP; DC) = -1$, so $CE \cdot CA = CD \cdot CB = CP \cdot CP$, so $AEPQ$ is cyclic. Angle chasing using the fact that $\angle EPQ = 180^{\circ} - \angle EAQ$ and $\angle FQB = \angle BAQ$ gives that $\angle FAE = \angle FGE$, as desired.

Let $D$ be the foot of the $A$-altitude and $H$ be the orthocenter in $\triangle ABC$. Because $BQ^2 = BP^2 = BF \cdot BA$, we find that $F$ is the $A$-humpty point of $\triangle APQ$, and since $\overline{HF} \perp \overline{AF}$, we also see that $H$ is the orthocenter in $\triangle APQ$. Therefore, if we let $P_1$ be the foot from $P$ onto $\overline{AQ}$, then $AEHFP_1$ is cyclic. Thus, letting $X$ be the second intersection of $QF$ with $(AEHFP_1)$, we see that $$QF \cdot QX = QP_1 \cdot QA = QD \cdot QP,$$as $AP_1DP$ is cyclic, so $FXPD$ is cyclic too. Thus, $$\measuredangle FXP = \measuredangle FDP = \measuredangle FDC = \measuredangle FAC = \measuredangle FAE = \measuredangle FXE,$$which implies that $X$ lies on line $EP$, as needed.

Let $D$ be the foot from $A$. Note that by Power of a Point, $$BP^2=BQ^2=BF\cdot BA.$$Notably, $BP=BQ.$ Claim 1: $(QP;DC)=-1$. We have $$BD\cdot BC=BF\cdot BA=BP^2,$$since $AFDC$ is cyclic, hence shown since $B$ is the midpoint of $PQ$. Claim 2: $AEPQ$ is cyclic. Note that we have $$CE\cdot CA=CD\cdot CB=CB(CB-BD)=CB^2-CB\cdot BD$$and $$CP\cdot CQ=CB^2-BP^2.$$Thus, it suffices to show that $$CB^2-CB\cdot BD=CB^2-BP^2,$$which is just $$BP^2=CB\cdot BD$$which is clearly true since $(QP;DC)=-1$. Now, since $AEPQ$ is cyclic, we have $$\angle CET=\angle AQP=\angle QFB=\angle AFT,$$so $AETF$ is cyclic and we are done. Note that $\angle AQP=\angle QFB$ is due to $(AFQ)$ being tangent to $BC$.

Solved with OronSH and GrantStar: Let $X$ be the intersection of lines $PE$ and $QF$. Claim: $QFHP$ is cyclic.

Since $BF \cdot BA = BP^2$ and $BF \cdot BA = BH \cdot BE$, we have $BH \cdot BE = BP^2$, hence $(BHE)$ is tangent to $\overline{BC}$. Now, we have that \begin{align*} = ~ & \angle XFH + \angle HEX \\ = ~ & (180^{\circ} - \angle QFH) + (180^{\circ} - \angle HEP) \\ = ~ & (180^{\circ} - \angle QFH) + (180^{\circ} - \angle HPQ) \\ = ~ & 180^{\circ}, \end{align*}hence $X$ lies on $(EHF)$ as desired.

First we obviously have \[BP^2=BF \cdot BA = BQ^2\]and thus, $BP=BQ$. Now, since $B$ is the midpoint of $PQ$ and \[BP^2=BF \cdot BA = BD \cdot BC\]we must have $(CD;PQ)=-1$. But this means, \[CP \cdot CQ = CD \cdot CB = CE \cdot CA\]and thus, $AEPQ$ must be cyclic. Now, let $R = \overline{QF} \cap \overline{PE}$. This gives us that, \begin{align*} \measuredangle AFR &= \measuredangle AQF + \measuredangle FAQ \\ &= \measuredangle AQF + \measuredangle FQP\\ &= \measuredangle AQP\\ &= \measuredangle AEP \\ &= \measuredangle AER \end{align*}and thus, $R$ must lie on the circumcircle of $\triangle AEF$ as required.

Checking all 90+ solutions above, I think this method is new (albeit rather assymmetrical). The motivation is trying to draw $P$ and $Q$ with just ruler and compass. Note that $F$ is the $A$-humpty point of $APQ$, so to construct $P$ and $Q$ we can proceed as follows; Let $G$ is reflection of $F$ across $B$, and $J$ is the reflection of $F$ across $BC$, then $P$ and $Q$ are merely the intersections of $(AGJ)$ with $BC$. Claim: $AEPQGJ$ cyclic. Proof: It suffice to prove $AEGJ$ cyclic. The main idea is that $DF$ and $DE$ are reflections across $BC$, so $J, D, E$ are colinear. Hence $\angle EJF=\angle FDA=90-\angle A$, so $\angle EJG=180-\angle A=\angle GAE$, and this proves the claim. By the claim, we get $\angle FQP+\angle EPQ=\angle QAF+180-\angle QAE=180-\angle A$, so $QF$ meet $PE$ at $(AEF)$.

Observe HFQP cyclic Then by power of point observe AEPQ cyclic Then angle chase We are done

From $\angle QAF = \angle QFB$ and $\angle PAF = \angle PFB$ we get $F$ is Humpty point of $\triangle APQ$ As $FH \perp AB$ and $AH \perp PQ$ we get $H$ is orthocenter of $\triangle APQ \implies Q,F,H,P$ is cyclic. by power of point $$CP.CQ=CH.CF=CE.CA$$ we get $A,E,P,Q$ cyclic. $$\angle FQP + \angle EPQ = \angle QAF + 180 - \angle QAE = \angle BAC$$$\blacksquare$

First PoP gives $BP^2=BF\cdot BA=BD\cdot BC$ where $D$ is the foot from $A$ to $BC$ so $CP\cdot CQ=BC^2-BP^2=CD\cdot BC=CA\cdot CE$ so $A,E,P,Q$ are concyclic. Now if $K$ is the reflection of $F$ over $B$ then $BP\cdot BQ=BP^2=BF\cdot BA=BK\cdot BA$ so $A,E,P,Q,K$ are concyclic. Since $PQ,FK$ share a midpoint we get $QF\parallel PK$ so $\measuredangle (QF,PE)=\measuredangle KPE=\measuredangle KAE=\measuredangle FAE,$ done.

Denote AD is an altitude and $PE \cap QF = X$. BP is tangent to (AFP) $\Rightarrow$ $BP^2 = BA.BF$. BQ is tangent to (AFQ) $\Rightarrow$ $BQ^2 = BA.BF$. So $BP^2 = BQ^2$ $\Rightarrow$ BP = BQ. $BP^2 = BC.BD$ $\Rightarrow$ from problem 1 (C, P; D, Q) = -1 and CP.CQ = CB.CD. From ABDE cyclic, CD.CB = CE.CA $\Rightarrow$ CP.CQ = CE.CA $\Rightarrow$ AEPQ is cyclic. Now $\angle CAB + \angle BAQ = \angle CPE$, $\angle BAQ = \angle BQF$, $\angle CPE - \angle PQX = \angle EXF$ $\Rightarrow$ $\angle EXF = \angle CAB = \angle EAF$ $\Rightarrow$ AFEX is cyclic and we are ready.

We invert the problem with center $A$ and radius $\sqrt{AF.AB}$. Now the new problem is- Quote: $P$ and $Q$ are two tangents from $B$ to $(AH)$ where $H$ is the orthocenter of the main triangle. Show that $(APC),(AQB),BC$ concur. We first angle chase to try and get rid of the circles. We get that the required condition is equivalent to $\angle APC= \angle AQB$ which by some more backward working is just equal to $C$ lying on $PQ$. Note that $PQ$ is the polar of $B$ with respect to $(AH)$ so it suffices to prove that $B$ lies on the polar of $C$ with respect to the same circle which just follows by brokard's theorem on cyclic quadrilateral $AEHF$.

xoinks Note that we just wish to show that $\angle PQF+\angle EPQ+\angle BAC=180^\circ$ by cyclics and triangles. Note that $\angle PQF=\angle QAB$ by alternate segment theorem, thus we desired that $\angle QAC+\angle EPQ$. Thus, it suffices to prove that $QAEP$ is cyclic. By power of a point at $C$, it suffices to show that $CP\cdot CQ=CE\cdot CA=CD\cdot CB$. It is obvious as $BF\cdot BA=BP^2=BQ^2$ that $BP=BQ$. Note that $(P,Q;D,C)=-1$ by Polars on $(PQ)$ (as $D$ inverts to $C$ and vice versa). Now, inverting at $C$ with radius $\sqrt{BD\cdot BC}$, we note that $(P',Q';B,P_\infty)=-1$. Thus, $B$ is the midpoint of $P'Q'$. Yet, this implies that if $CP=p$, and $CQ=q$, then $\frac{r^2}{p}-a=a-\frac{r^2}{q}$, or equivalently $2a=r^2\left(\frac{1}{p}+\frac{1}{q}\right)$. However, initially, we have $q-a=a-p$, thus $p+q=2a$. This means that \[p+q=2a=r^2\frac{p+q}{pq}\]Which means $pq=r^2$, as desired.