Interesting to have the converse of South Africa 1999 as an IMO problem... Darij

Let $P$, $M$ and $N$ be the midpoints of $BD$, $CF$ and $CG$ respectively. Since $ABCD$ is a parallelogram, $AC$ also passes through $P$. Since $EC = EF$, $EM\perp CF$; similarly $EN\perp CG$. Consider the homothety centred at $C$ of ratio $\frac{1}2$. Clearly $A$, $F$ and $G$ are mapped to $P$, $M$ and $N$ respectively. Hence $P$, $M$ and $N$ are collinear. Let $P'$ be the foot of perpendicular from $E$ to $BD$. Since $E$ lies on the circumcircle of $\triangle BCD$, $P'$, $M$ and $N$ are collinear (Simson line). Therefore, $P$ and $P'$ coincide, or $E$ lies on the perpendiculr bisector of $BD$. Since $\angle BPE =\angle CME = 90^{\circ}$ and $\angle PBE =\angle MCE$ (angles in the same segment), $\triangle PEB$ is similar to $\triangle MEC$. Now $\angle MNC =\angle MEC =\angle PEB =\frac{1}2\angle DEB =\frac{1}2\angle DCB =\frac{1}2\ angle DAB$. We have proved that the line $l$ and the Simson line $PMN$ are related by a homothety, so they are parallel. Therefore the angle between $l$ and $AD$ is equal to the angle between $MN$ and $BC$, implying that $l$ is the bisector of $\angle DAB$.

Here is a ‘philosophical’ solution based on converse mentioned by Darij. Let us suppose the converse is proved. Lets move line l and look for moving of point E, the circumcenter of FGC. Its easy to show (or obvious for more experienced people:)) that E passes some quadratic curve which must be hyperbola since it has an asympthote and its unbounded. In our interval (while l intersect CD in ints interior) it is some connected part of hyperbola. Also it is all at the same side of line CD so there are at most two situations when E is on circle BCD. The first is our desired and the second is trivial E=F=G=C.

I only want to rewrite the problem and make solution of leepakhin clearly Problem:Given a parallelogramen $ABCD$ and a line $l$ is through $A$ intersects segment $CD$ at $F$ and $BC$ at $G$.If exist $E$ is a point on circumcircle of triangle $BCD$ such that $E$ is circumcenter of $CFG$. Prove that $l$ is bisector of $\angle DAB$. Solution:Let $P,M,N$ be the midpoints of $BD, CF, CG$ resp.$ABCD$ is a parallelogram, $AC$ also passes through $P$. Easily seen $P,M,N$ are colinear (by homothety or they are in the same line parallel with $l$) Since $EC=EF\Rightarrow EM\perp CF$ similarly $EN\perp CG$. Thus $MN$ is simson line of $E$ w.r.t $BCD$ but $MN\cap BD=\{P\}\Rightarrow EP\perp BD$ but $P$ is midpoint of $BD\Rightarrow EB=ED\Rightarrow \angle DBE=\angle BDE\ (1)$ But $\angle DBE=\angle DCE,\angle BDE=\angle ACG\ (2)$($ACED$ is cyclic). Combine $(1),(2)$ we get $\angle DCE=\angle ACG\Rightarrow \angle CEF=\angle CEG\Rightarrow EC$ is midline of $FG\Rightarrow$ triangle $CFG$ is isoceles $\Rightarrow\angle DAG=\angle AGC=\frac{180^{0}-\angle FCG}{2}=\frac{\angle BCD}{2}=\frac{\angle BAD}{2}$ Note: Invense problem: Given a parallelogramen $ABCD$ and a line $l$ is bisector of $BAD$, $l$ intersect lines $CD$ and $BC$ at $F,G$.Prove that circumcenter of triangle $CFG$ lie on circumcircle of triangle $BCD$.

Attachments:

i have solved this problem with some easy trigonometry(at home )

please post your solution.

I solved it by showing that BEG and DED are similar. What do you think was this problem on the shortlist? (I think some G3 )

Albanian Eagle wrote: I solved it by showing that BEG and DED are similar. You mean DEC Since $AD \parallel BG$, then $\angle DAG = \angle BGA$. So we have to prove $ABG$ is isosceles. $F, C, G$ all lie on the circle with center $E$ and radius $EF$. Let $K$ be the intersection between this circle and the circumcircle $\mathcal O$ of $BCD$. As proved early $E$ is the midpoint of arc $BD$ on $\mathcal O$. Then $BCKD$ is a trapezoid and $\angle ECK = \angle EBC$. Since $BD \parallel KC$, then $\angle EBD = \angle ECG = \angle EGC$ and by LAL $\Delta EDC \equiv \Delta EBG$. Finally $ABG$ is isosceles and the thesis follows.

This problem can be done using complex numbers, too.

let (ABD)=a (DBE)=b (EBC)=c and EC=sinb.k from sin laws DC=sin(b+c).k BC=sina.k FC=2.sinc.cosb.k GC=2.sinc.cos(a+c).k on the other hand we have FC/AB=GC/GB from similarity so writing the above results yields (sina+2.sinc.cos(a+c)).cosb=cos(a+c).sin(b+c) (sina+sin(2c+a)-sina).cosb=cos(a+c).sin(b+c) sin(2c+a+b)+sin(2c+a-b)=sin(2c+a+b)+sin(b-a) sin(b-a)=sin(2c+a-b) 1) b-a+2c+a-b=180 c=90 b=m(DEF)+c so m(ABC)>180 contradiction 2) b-a=2c+a-b so b=a+c hence FC=CG and AB=BG which proves our problem.

I solved this problem by coordinates

please show us how

Sepp wrote: Albanian Eagle wrote: I solved it by showing that BEG and DED are similar. You mean DEC Since $AD \parallel BG$, then $\angle DAG = \angle BGA$. So we have to prove $ABG$ is isosceles. $F, C, G$ all lie on the circle with center $E$ and radius $EF$. Let $K$ be the intersection between this circle and the circumcircle $\mathcal O$ of $BCD$. As proved early $E$ is the midpoint of arc $BD$ on $\mathcal O$. Then $BCKD$ is a trapezoid and $\angle ECK = \angle EBC$. Since $BD \parallel KC$, then $\angle EBD = \angle ECG = \angle EGC$ and by LAL $\Delta EDC \equiv \Delta EBG$. Finally $ABG$ is isosceles and the thesis follows. yeah! that's exactly my solution. Quite simple and straight foreward. You get the idea by seeing the proof of the other direction

Sepp wrote: Albanian Eagle wrote: I solved it by showing that BEG and DED are similar. You mean DEC Since $AD \parallel BG$, then $\angle DAG = \angle BGA$. So we have to prove $ABG$ is isosceles. $F, C, G$ all lie on the circle with center $E$ and radius $EF$. Let $K$ be the intersection between this circle and the circumcircle $\mathcal O$ of $BCD$. As proved early $E$ is the midpoint of arc $BD$ on $\mathcal O$. Then $BCKD$ is a trapezoid and $\angle ECK = \angle EBC$. Since $BD \parallel KC$, then $\angle EBD = \angle ECG = \angle EGC$ and by LAL $\Delta EDC \equiv \Delta EBG$. Finally $ABG$ is isosceles and the thesis follows. I want to ask you how do you get $E$ is the midpoint of arc $BD$ on $\mathcal O$

bolzano_1989 wrote: Sepp wrote: Albanian Eagle wrote: I solved it by showing that BEG and DED are similar. You mean DEC Since $AD \parallel BG$, then $\angle DAG = \angle BGA$. So we have to prove $ABG$ is isosceles. $F, C, G$ all lie on the circle with center $E$ and radius $EF$. Let $K$ be the intersection between this circle and the circumcircle $\mathcal O$ of $BCD$. As proved early $E$ is the midpoint of arc $BD$ on $\mathcal O$. Then $BCKD$ is a trapezoid and $\angle ECK = \angle EBC$. Since $BD \parallel KC$, then $\angle EBD = \angle ECG = \angle EGC$ and by LAL $\Delta EDC \equiv \Delta EBG$. Finally $ABG$ is isosceles and the thesis follows. I want to ask you how do you get $E$ is the midpoint of arc $BD$ on $\mathcal O$ Well using the notation of the first solution, it either follows from the Simson line argue, or even directly: P, M, N are collinear, CMEN is cyclic implies <PBE = <DCE = <MNE = <PNE, so PBNE is cyclic and <BPE is right.

easy geometry...

color wrote: please show us how I believe it is too robust... and long (I did this problem (and the fourth )using coordinates too)

Hm.. I solved both geometry problems with trigonometry.. Can somebody tell me how to use coordinates to solve geometry problems containing some circles ? It must be very hard..

Well, when the circle is circumcircle (around the triangle), it isn't difficult at all, although it might look ugly (or complicated) to some-you know, when you are looking for the equation of a circle which goes through three points, you have two ways to do it: first one is to solve the system of three circle equations, and the anotherone (the way I used) is to find the equations of two symetrals of the triangle's sides, and then find their crossing point-that's the circle's centre, and the radius is simplz calculated using the distance formula (Pythagoras...). On the other hand, whe the circle is inscribed... that's a bit more complicated...

Let $X$ be the second intersection of $AB$ and $BCED$, and $S$ be the second intersection of $(FCG)$ with $(BCED)$. It is easy to see that $DX=DA$. $\angle XSF=\angle XSC-\angle FSC=\angle BAD-\angle FGC=\angle XAD- \angle FAD=\angle XAF\Rightarrow XASF$ is cyclic. $\angle DES=\angle DCS=\frac{\angle FES}{2}\Rightarrow DE$ bisects $\angle FES$, and since $FE=ES$ and $DE=DE$, we have that triangles $DEF$ and $DES$ are congruent, hence $DS=DF$. By the fact that $AXFS$ is cyclic and $DA=DX$ and $DF=DS$ implies that $D$ is the center of $AXFS$. By the incenter lemma we have that $F$ is the incenter of triangle $CSX$. Now, $\angle XAF=\angle XSF=\angle FSC=\angle FGC=\angle FAD$ and we are done.

We first prove the converse, i.e. if $\ell$ is the bisector, then the center of $(CFG)$ lies on $(BCD):=\Gamma$. In this case, I claim that our $E$ is the arc midpoint of $BD$. Let $H$ be the point on $\Gamma$ such that $BDHC$ is an isosceles trapezoid, which by symmetry gives $EC=EH$. Equivalently, $H$ is the reflection of $A$ over $\overline{BD}$. Angle chasing gives $$\angle CGF=\angle FAD=\angle BAF=\angle DFA=\angle CFG,$$ hence $CF=CG$. Further, $$\angle GCE=180^\circ-\angle BCE=\angle BDE=\angle DBE=\angle FCE,$$ hence $\overline{CE}$ bisects $\angle FCG$. Combining these two facts, it follows that $E$ lies on the perpendicular bisector of $\overline{FG}$, i.e. $EF=EG$. Now, note that $DH=DA=DF$ (where the second equality comes from $\angle FAD=\angle DFA$). Further, $\overline{DE}$ is the bisector of $\angle FDH=\angle CDH$, as $E$ is the midpoint of minor arc $CH$. Hence $F$ and $H$ are reflections of each other over $\overline{ED}$, so $EF=EH$. Since $EC=EH$ and $EF=EG$, it follows that $EF=EG=EC$, so $E$ is indeed the circumcenter. We now use the converse (!!) to prove that there cannot exist two choices of $\ell$ such that the center of $(CFG)$ lies on $\Gamma$. This can probably be done with hyperbolas but we don't know anything about hyperbolas so we use a synthetic geometry proof instead. Suppose we had some $\ell'$ intersecting $\overline{CD}$ at $F'$ and $\overline{BC}$ at $G'$, such that the center $E'$ of $(CF'G')$ was on $\Gamma$ as well. Note that $E'$ lies the perpendicular bisector of $F'G'$, which intersects $\overline{CE}$ on $(CF'G')$—precisely, on the midpoint of arc $F'G'$ not containing $C$. Further, as $F'$ moves from $C$ to $D$, the length of $CE'$ strictly increases. If $F'$ lies on segment $\overline{CF}$, then $E'$ should lie on minor arc $CE$ of $\Gamma$, as it is closer to $C$ than $E$ is. On the other hand, said minor arc lies on the opposite side of $\overline{CE}$ as the perpendicular bisector of $F'G'$ does. This is because $CG'<CF' \iff DA=DF<DF'$ (from $\triangle F'CG' \sim \triangle F'DA$) which is true, so the angle bisector theorem implies that the midpoint of $\overline{F'G'}$ lies on the opposite side of $\overline{CE}$ as the minor arc does, and then our previous intersection finding implies that the perpendicular bisector will not "switch sides" within $(CF'G')$. Likewise, if $F'$ lies on segment $\overline{DF}$, then $E'$ should lie on major arc $CE$ of $\Gamma$, as it is farther from $c$ than $E$ is. On the other hand, this time we have $CG'>CF'$, so the midpoint of $\overline{F'G'}$ again lies on the opposite side of $\overline{CE}$ as the major arc does. These two cases capture all possibilities, so $\ell'$ cannot exist, as desired. $\blacksquare$

Let $\Gamma = (BCED)$, $P = DA \cap \Gamma$, $Q = BA \cap \Gamma$. We have $F = \lambda \cap CD$, and define $X = QF \cap \Gamma$. By Pascal on $PXQBCD$, we have $A = BQ \cap DP$, $F = XQ \cap CD$, and $BC \cap XP$ collinear. Hence they are collinear on $\lambda$, and since $G = \lambda \cap BC$ then $G$ lies on $XP$. Claim 1: $X$ is the $C$-dumpty point of $\triangle CGF$. Proof: $\angle XGC = \angle PGC = \angle GPD = \angle XPD = \angle XCD = \angle XCF$. Similarly, $\angle GCX = \angle BPX = \angle BQX = \angle CFX$. Hence, by AA we have $\triangle XCG \sim \triangle XFC$, and $X$ is the $C$-dumpty point. $\square$ It's well-known that $CX \perp EX$. Hence, since $C, X, E$ lie on $\Gamma$, either $X = E$ or $E$ is the $C$-antipode in $\Gamma$. Case 1: $X = E \Rightarrow \angle ECG = \angle EFC = \angle FCE \Rightarrow CE$ is the angle bisector of $\angle FCG$. Hence $\triangle CGF$ is isosceles. Since $\triangle FDA \sim \triangle FCG$ by AA we have $\angle FAB = \angle AFD = \angle DAF$, and $\lambda$ is the bisector of $\angle DAB$. Case 2: $E$ is the $C$-antipode. Then $(CGF)$ is the image of $\Gamma$ under homothety scale factor 2 at $C$, and so no other point on $(CGF)$ lies inside $\Gamma$. But $F \in (CGF)$ lies on segment $CD$ within $\Gamma$, contradiction.

Note that the Simson line of $E$ wrt $\triangle BCD$ is homothetic to $\ell$ wrt $C$ with factor $1/2,$ and therefore passes through the center of $ABCD$. This yields $|EB|=|ED|\implies \angle FCE=\angle DBE=\angle BDE=\angle GCE,$ so it's trivial that $|CF|=|CG|$ and hence $\angle BAF=\angle CFG=\angle CGF=\angle DAF,$ as desired.

The key is to construct the Simson Line of $E$ with respect to triangle $BCD$. By taking a homothety at $C$ with ratio $2$, it follows that this line is parallel to $\ell$. On the other hand, it is well-known that the the Simson Line is parallel to $\overline{CJ}$, where $J$ is the point at which the altitude at $E$ to $\overline{BD}$ meets $(BCD)$. Thus, it suffices to show $J$ is the arc midpoint. But the aforementioned homothety implies that the Simson Line bisects $\overline{AC}$, so it bisects $\overline{BD}$ too. This common midpoint $O$ is thus both the midpoint and foot from $E$ in triangle $EAB$, implying that it is isosceles. Thus, $\overline{EJ}$ bisects $\widehat{BD}$, which is enough.

Let $M,N,O$ be midpoints of $CG$, $CF$, and $BD$, respectively. Note that $EN\perp CD$ and $EM\perp BC$. Note that the perpendicular from $E$ onto $BD$ must lie on $MN$. Since $O,M,N$ are midpoints of $AC$, $FC$, and $GC$ respectively, $O$ lies on $MN$. Therefore, $EO\perp BD$. Now, $EB=ED$, $\angle EMB=\angle END$ and $\angle NDE=\angle EBM$ so $\triangle BME\cong \triangle DNE$. Therefore, $EM=EN$ which implies $CG=CF$. Thus, $$\angle FAD=\angle CGF=\angle CFG=\angle BAF$$ as desired.

Let $H_{b}$ and $H_{d}$ be the midpoints of $CG$ and $CF$, respectively. Notice that $\triangle ADF \sim \triangle GCF$ and $\triangle EH_{d}D \sim \triangle EH_{b}B$. The first follows from $ABCD$ being a parallelogram and the second follows from $\angle EDC = \angle EBC$ combined with $\overline{EH_{d}} \perp \overline{FC}$ and $\overline{EH_{b}} \perp \overline{CG}$. These give: $$\frac{\frac{1}{2}CF\cdot\tan\angle EFC}{\frac{1}{2}CG\cdot\tan\angle ECG} = \frac{EH_{d}}{EH_{b}} = \frac{DH_{d}}{BH_{b}} = \frac{DF + FH_{d}}{BC + CH_{b}} = \frac{DF + \frac{1}{2}CF}{BC + \frac{1}{2}CG} = \frac{CF}{CG}\cdot \frac{\frac{DF}{CF}+\frac{1}{2}}{\frac{AD}{CG}+\frac{1}{2}} = \frac{CF}{CG}$$ where the last equality follows from $\triangle ADF \sim \triangle GCF$. Now $\tan\angle EFC = \tan\angle ECG$ and the angles are acute, hence $\angle EFC = \angle ECG$. This implies that $\triangle EFC \cong \triangle EGC$, so $CF = CG$ and therefore $AD = DF$, implying that $\ell$ is the angle bisector of $\angle BAC$ as desired.

Got a hint to construct midpoints, but I think the finish is pretty unique! Let J,K,L be the midpoints of GC,FC,BD. Notice that by a homothety of 1/2 ratio at C we get AFG is a line implies LKJ is a line; Since EJ perp. BC, EK perp. CD, by converse of Simson line we get that EL perp. BD. Now note by pop we get $$\frac{DF}{FC}=\frac{AD}{CG}=\frac{BC}{CG}\implies\frac{DF}{BC}=\frac{FC}{CG}=\frac{AB}{BG}=\frac{CD}{BG}\implies BC\cdot CD=DF\cdot BG\implies^{BC\cdot BG=DF\cdot CD}CD=BG,BC=DF\implies BAG=BGA=GAD,$$ as desired. $\blacksquare$

Let $M$ be the midpoint of $\overline{BD}$, so by a homothety at $C$, $M$ is collinear with the feet of the altitudes from $E$ to $\overline{CB}$ and $\overline{CD}$. On the other hand, by Simson lines, these two feet are also collinear with the foot from $E$ to $\overline{BD}$, hence this foot is just $M$, so $E$ is the midpoint of arc $BCD$. Then $\angle FCE=\angle GCE$, so $CF=CG$ and thus $\angle GFC=\angle FGC \implies \angle GAB=\angle FAD$, which finishes the problem. $\blacksquare$ EDIT: bro ive done this problem before too whatttttttttttt

had to get hint ;-; my brain Just draw Simson line of $E$ to $BCD$, which passes through midpoint of $BD$. So $E$ is arc midpoint, finish since $\triangle ECF\sim\triangle EBD\sim \triangle EGC$

I have absolutely no clue how I even conjured this solution. We start with a key claim: Claim: Let $XYZ$ be a triangle, and let $P$ and $Q$ be on lines $XY$ and $XZ$ such that $PQ \parallel YZ.$ Let $R$ be on $YZ$ such that $YPQR$ is a parallelogram. Then regardless of the position of points $P$ and $Q$, the circumcircle of $\triangle YPR$ will pass through a fixed point other than $Y.$ Proof: It suffices to show that all circles of this type share a radical axis. To do so, we will use barycentric coordinates with $\triangle XYZ$ as the reference triangle. Let $X = (1,0,0), Y = (0,1,0),$ and $Z = (0,0,1).$ Also let $P = (1 - \lambda, \lambda, 0)$ and $Q = (1 - \lambda, 0, \lambda),$ where $\lambda$ is an arbitrary real number, so that $P - Q = \lambda(0,1,-1) = \lambda \cdot (Y-Z)$ and therefore $PQ \parallel YZ.$ Then $R = Q + Y - P = (0,1-\lambda, \lambda).$ Now we find the circumcircle of $\triangle YPR.$ A point $(x,y,z)$ on this circle must satisfy an equation of the form $$a^2 yz + b^2 zx + c^2 xy = ux + vy + wz,$$ where $a = YZ, b = ZX,$ and $c = XY.$ Plugging in $Y = (0,1,0)$ gives $v = 0,$ so $$a^2 yz + b^2 zx + c^2 xy = ux + wz.$$ Plugging in $P = (1 - \lambda, \lambda, 0)$ gives $u(1-\lambda) = c^2 \lambda(1-\lambda),$ so $u = c^2 \lambda.$ Similarly, plugging in $R = (0, 1-\lambda, \lambda)$ gives $w\lambda = a^2 \lambda(1-\lambda),$ so $w = a^2 (1-\lambda).$ Therefore, our circumcircle is $$\boxed{a^2 yz + b^2 zx + c^2 xy = c^2 \lambda x + a^2 (1-\lambda) z}.$$ If we let $\lambda_1$ and $\lambda_2$ be two distinct real values of $\lambda,$ then the radical axis of the two corresponding circles is given by $$(c^2 \lambda_1 - c^2 \lambda_2) x + (a^2 (1 - \lambda_1) - a^2 (1-\lambda_2))z = 0.$$ Dividing both sides by $\lambda_1 - \lambda_2$ gives $c^2 x - a^2 z = 0,$ which is clearly fixed. This concludes our (lengthy) proof of the claim. Now apply the claim with $(X,Y,Z,P,Q,R) = (G,C,F,B,A,D).$ Since $E$ is the circumcenter of $GCF,$ it has no relevance to $B,A,$ or $D,$ so $E$ is the fixed point as shown to exist by the claim. Taking $\lambda$ to be arbitrarily close to $0,$ we discover that $(FEC)$ is tangent to $CG.$ Hence $\measuredangle GCF = \measuredangle CEF = 2 \measuredangle CGF.$ However, by sum of angles in a triangle, $\measuredangle GCF = \measuredangle CGF + \measuredangle GFC,$ so $\measuredangle CGF = \measuredangle GFC.$ Since $CF \parallel AB$ and $DA \parallel CG,$ we get $\measuredangle CGF = \measuredangle DAF$ and $\measuredangle GFC = \measuredangle FAB.$ Thus $\measuredangle DAF = \measuredangle FAB,$ and we conclude. Ok reading the solutions above there is a Simson line??????

Different and short solution Assume that,the angle bisector of $\angle DAB \cap CD=G$ and $\angle DAB \cap BC=F$ call triangle $CGF$'s center $E$ we will prove $BCED$ cyclic.Let,$\angle DAG=\angle GAB=\angle AGD=\angle CGF=\angle CFG=\alpha$ from Ptolemy's sine lemma that is equivalent to show $CE\sin(2\alpha)+CB\sin(90-\alpha)=CD\sin(90+\alpha)$.We know $AD=DG=BC=x$ and $GC=CF=y$ ,we also know $AB=BF$.Sine Thrm to $\triangle CEG$ $\implies$ $CE\sin(2\alpha)=y\sin(90-\alpha)$.So, $\sin(90-\alpha)(x+y)=CD\sin(90+\alpha)=(x+y)\sin(90+\alpha)$ which is indeed true.

Magical solution, needed a pointer to look at Simson Lines. Let $M$ and $N$ denote the midpoints of $GC$ and $FC$. Now, let $P = \overline{AC} \cap \overline{MN}$. By the Midpoint Theorem, it follows that $P$ is the midpoint of $AC$. Since diagonals of a parallelogram bisect each other, it also follows that $P$ is the midpoint of $BD$. But now, consider the Simson Line of $E$ with respect to $\triangle PCD$. Clearly the feet of the perpendicular from $E$ to $GC$ and $FC$ and $M$ and $N$ and thus, $EP\perp BD$. But note that this means $\overline{EP}$ is in fact the perpendicular bisector of segment $BD$, which implies that $EB=ED$. Thus, $$\measuredangle BAD = \measuredangle DCB = \measuredangle DEB = 2\measuredangle DBE = 2 \measuredangle FCE = \measuredangle FEC = 2 \measuredangle FGC = 2\measuredangle FAD$$ which indeed implies that $\ell$ is the internal $\angle BAD-$bisector.

denote the perpendiculars from $E$ to $BC$, $CD$, and $BD$ as $P_1$, $P_2$, and $P_3$ respectively, and they are collinear by simson line it is easy to tell that $P_1$ is the midpoint of $CG$ and $P_2$ is the midpoint of $CF$, so $P_1P_2P_3$ is parallel to $AFG$ so $P_3$ is the midpoint of $BD$ and $AC$ $DBE=DCE$ since $BCDE$ is cyclic, and $GCE=BDE$ also since $BCDE$ is cyclic thus $FCE=GCE$ and we can prove $CE$ is perpendicular to $FG$ and $P_1P_2$ thus, $CP_1=CP_2$, and angle chasing gives $BAG=DAG=CP_1P_2$ thus, $\ell$ is the bisector of angle $DAB$

Let $\ell'$ be the image of $\ell$ under a homothety of factor $\tfrac{1}{2}$ centered at $C$. Simson's theorem on $E$ WRT $(BCD)$ tells us that the foot from $E$ to $\overline{BD}$ lies on $\ell'$. But the intersection of $\ell'$ with $\overline{BD}$ is always the midpoint of $\overline{BD}$, so $E$ is one of the arc midpoints. This determines $\ell$, as desired.