APMO 2022/2 wrote: Let $ABC$ be a right triangle with $\angle B = 90^{\circ}$. Point $D$ lies on the line $CB$ such that $B$ is between $D$ and $C$. Let $E$ be the midpoint of $AD$ and let $F$ be the second intersection point of the circumcircle of $\triangle ACD$ and the circumcircle of $\triangle BDE$. Prove that as $D$ varies, the line $EF$ passes through a fixed point. We claim that the fixed point is the reflection of $C$ with respect to $B$. Let us define $G$ and $H$ to be the intersection of $EF$ with $BC$ and $(ADC)$ respectively. [asy][asy] import graph; size(9 cm); 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 = -10.22, xmax = 7.62, ymin = -5.46, ymax = 5.5; /* image dimensions */ pen ccqqqq = rgb(0.8,0,0); /* draw figures */ draw((-4.56,-0.88)--(-3.1,1.62)); draw((1.0412012644889352,-0.7984615384615372)--(-3.1,1.62)); draw((-4.56,-0.88)--(1.0412012644889352,-0.7984615384615372)); draw((-3.1,1.62)--(-4.1741590214067275,2.247308868501529)); draw(circle((-3.9346965626724275,1.4240361945777102), 0.8573914886285903)); draw(circle((-1.776916422566165,0.36408540456234306), 3.0484918718649974)); draw((-7.241201264488967,4.038461538461556)--(-1.4929577569177765,-2.6711526699600063), ccqqqq); draw((-4.1741590214067275,2.247308868501529)--(-7.241201264488967,4.038461538461556) ); draw((-4.56,-0.88)--(-4.1741590214067275,2.247308868501529)); /* dots and labels */ dot((-4.56,-0.88),dotstyle); label("$A$", (-4.95,-0.8), NE * labelscalefactor); dot((-3.1,1.62),dotstyle); label("$B$", (-3.12,1.72), NE * labelscalefactor); dot((1.0412012644889352,-0.7984615384615372),dotstyle); label("$C$", (1.12,-0.8), NE * labelscalefactor); dot((-4.1741590214067275,2.247308868501529),dotstyle); label("$D$", (-4.35,2.44), NE * labelscalefactor); dot((-4.3670795107033635,0.6836544342507644), dotstyle); label("$E$", (-4.28,0.84), NE * labelscalefactor); dot((-4.732651083553028,1.110366388286186), dotstyle); label("$F$", (-4.66,1.28), NE * labelscalefactor); dot((-7.241201264488967,4.038461538461556), dotstyle); label("$G$", (-7.46,4.2), NE * labelscalefactor); dot((-1.4929577569177765,-2.6711526699600063), dotstyle); label("$H$", (-1.42,-3), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] By Reim's Theorem, then $BE \parallel CH$. Then \[ \measuredangle ADC = \measuredangle EDB = \measuredangle DBE = \measuredangle DCH \]which implies $AH \parallel DC$. Now, note that as $E$ is the midpoint of $AD$ and $AH \parallel GD$, then $GDHA$ is a parallelogram, which implies $GE = EH$. As $EB \parallel HC$, we have $GB = BC$, which implies that $EF$ passes through the reflection of $C$ with respect to $B$.

Jalil_Huseynov wrote: Let $ABC$ be a right triangle with $\angle B=90^{\circ}$. Point $D$ lies on the line $CB$ such that $B$ is between $D$ and $C$. Let $E$ be the midpoint of $AD$ and let $F$ be the seconf intersection point of the circumcircle of $\triangle ACD$ and the circumcircle of $\triangle BDE$. Prove that as $D$ varies, the line $EF$ passes through a fixed point. My sol during the test is kind of bold action as i had no ideas and i did like 6 diagrams before realicing this overkill works: Let $G$ the reflection of $C$ over $B$ and let $EF \cap BC=G'$, then for any such $D$ we prove that $G=G'$, so now we do mOvInG pOiNtS. Move $D$ along $BC$ then the map $f: D \to G$ is projective becuase $D \to CD \to CG \to G$ is projective. Now the map $g: D \to G'$ is projective becuase $D \to CD \to CB$ is projective and since spiral similarities are projective $CB \to EA$ is projective and now the map $D \to CD \to CB \to EA \to EF \to EG' \to G'$ is projective, so since both $f,g$ are projective maps it remains to show that $G=G'$ for 3 particular cases. Case 1: $D \equiv B$ Then $E$ is sent to the midpoint of $AB$ and $F$ would be sent to a point on $(ABC)$ such that $\angle EFB=90$, now let $EF \cap (ABC)=C'$ then $ABCC'$ is a square so that means $\angle GBE=90=\angle GCC'$ and $CC'=AB=2EB$ which means that there is an homothety mapping $E \to C'$ with scale of $2$ centered at $G$ which means $G,E,F,C'$ colinear which means $G=G'$. Case 2: $D \equiv G$ Then $EB$ is midbase of $\triangle AGC$ but that means $(GEB)$ tangent to $(GAC)$ which means that $D \equiv G \equiv F$ hence $G$ lies on $EF$ (coz its $F$ lol) which means $G=G'$. Case 3: $D \equiv \infty_{BC}$ (i.e. the point of infinity along $BC$) Then by the midbase $E$ is also a point of infinity but since the $A$-midbase of $\triangle ABC$ is parallel to $BC$ we have that $D \equiv \infty_{BC} \equiv E$, now that means $(BDE)$ is mapped the the line $BC$ and that means $F$ is mapped to $C$ which means that $G$ lies on $EF$ coz $EF$ was mapped to the line $BC$ so $G=G'$. Since we got $G=G'$ for 3 cases, we have it for any such $D$, hence we are done

I was disappointed when I finished the nongeo, because that means I had to work on the geo. -- Kaixin Wang, APMO 2022

Let $P=2B-C$, and let $F'=A+D-F$. Then $F$ is the spiral center taking $EA$ to $BC$, so it takes $DA$ to $PC$. $\measuredangle APD=\measuredangle DCA=\measuredangle DFA=\measuredangle AF'D$, so $A,F',P,D$ are cyclic. $\measuredangle F'PD=\measuredangle F'AD=\measuredangle FDA=\measuredangle FPC$, so $P-F-F'$. Since $F-E-F'$, $P$ is the desired fixed point.

Oh wow synthetic methods are sooooooo neat! I bashed it (meaning set up a Cartesian coordinate and calculate everything out using Algebra) and got a 3

Jalil_Huseynov wrote: Let $ABC$ be a right triangle with $\angle B=90^{\circ}$. Point $D$ lies on the line $CB$ such that $B$ is between $D$ and $C$. Let $E$ be the midpoint of $AD$ and let $F$ be the seconf intersection point of the circumcircle of $\triangle ACD$ and the circumcircle of $\triangle BDE$. Prove that as $D$ varies, the line $EF$ passes through a fixed point. A hint for the actual solution

Let $C'$ be the reflection of $C$ across $B$. I claim this is the fixed point. Redefine $F$ to be the $C'$ humpty point of $\triangle C'DA$. To show this is the $F$ in the problem, it suffices to show $F$ lies on $(ACD)$ and $(BDE)$. $\angle AFD = 180 - \angle AC'D = 180 - \angle ACD$ so $F \in (ACD)$. $\angle EFD = \angle FC'D + \angle FDC' = \angle FDE + \angle FDC' = \angle ADC' = 180 - \angle ADB = 180 - \angle EBD$ so $F \in (BDE)$, so we're done. $\blacksquare$

Let $C'$ the reflection of $C$ over point $B$, and let $\overline{EF}\cap\overline{BC}=X$, we'll show $X=C'$ is true for any $D$. [asy][asy] size(7cm); import geometry; pair A,B,C,D,E,F,G; A=(-5,5); B=(-5,0); C=(-3.5,0); D=(-9,0); E=midpoint(segment(A,D)); G=(-6.5,0); F=intersectionpoints(circle(A,D,C),circle(B,D,E))[0]; draw(D--F); draw(A--B); draw(E--D); draw(D--B); draw(A--C); draw(E--B); draw(A--E, green); draw(F--E, green); draw(F--A, green); draw(F--B, red); draw(F--C, red); draw(B--C, red); draw(circle(B,D,E)); draw(circle(A,D,C)); dot(A^^B^^C^^D^^E^^F^^G); label("$A$",A,NE); label("$B$",B,SE); label("$C$",C,SE); label("$D$",D,W); label("$E$",E,N); label("$F$",F,S); label("$X=C'$",G,NW); [/asy][/asy] $F$ is Miquel's point of $AEBC\Rightarrow\triangle{FEA}\sim\triangle{FBC}$, so $$\frac{FE}{FB}=\frac{EA}{BC}=\frac{ED}{BC'}$$combining with $\angle{DEF}=\angle{DBF}$ we get $\triangle{DEF}=\triangle{C'BF}\Rightarrow \angle{DFE}=\angle{C'FB}$, but $\overline{DE}=\overline{BE}$ so finally $$\angle{EFB}=\angle{DFE}=\angle{C'FB}\Rightarrow X=C'$$as desired. $\blacksquare$

APMO 2019/3 vibes Let $T=\overline{FE} \cap (ADC)$ and $X=\overline{FE} \cap \overline{BC}$. I claim that $X$ is the desired fixed point. By Reim $TC \parallel EB \implies \measuredangle ADC =\measuredangle EDB = \measuredangle DBE = \measuredangle DCT \implies ATDC$ is an isosceles trapezium. $\angle XDE = \angle TAE$. $\angle XED = \angle TEA$. $ED = EA$ thus by SAS $\triangle EXD \cong \triangle ETA \implies ATDX$ is a parallelogram and because $AX = TD =AX$ it means $B$ is the midpoint of $\overline{CX}$ which finishes the problem.

Solution: We extend $EF$ to meet $CB$ at $C'$ and $(ADC)$ again at $G$. By Reim's theorem, $$BE//CG$$$$\angle GCD=\angle EBD=\angle EDB=D$$$$\implies \angle GCA=D-C$$$$\angle AGC=180^o-D$$So we get $$\angle GAC=C=\angle ACD$$which means $AG//CD \implies AG//C'D$ We have $E$ the midpoint of $AD$. Hence we clearly have $E$ the midpoint of $C'G$ By Converse of Midpoint theorem, $B$ is midpoint of $CC'$. Thus $C'$ is fixed and our proof is complete.

GianDR wrote: Jalil_Huseynov wrote: Let $ABC$ be a right triangle with $\angle B=90^{\circ}$. Point $D$ lies on the line $CB$ such that $B$ is between $D$ and $C$. Let $E$ be the midpoint of $AD$ and let $F$ be the seconf intersection point of the circumcircle of $\triangle ACD$ and the circumcircle of $\triangle BDE$. Prove that as $D$ varies, the line $EF$ passes through a fixed point. A hint for the actual solution

I have a pretty similar solution (1st time complex bashing (non egmo problems) coz my synthetic skills seem to have become garbage ) I tried cartesian for quite a while and the terms started to get really messy (same thing happens to me in complex usually but not this time) The synthetic solution is but I dont understand how I am supposed to guess the point, I wasted a lot of time thinking it was $EF \cap \odot(ABC)$ help Anyways the solution can be shortened considerably (calculation and neatness wise) by letting $C,D \in \mathbb{R}$

Let $FE$ intersect $BC,AB$ and the circumcircle of triangle $ADC$ at points $P,Q,R$ respectively. Claim 1: $AR \parallel BC$. Proof: This is easy: $$\angle EFC=\angle AFC-\angle AFE=\angle ADB-(\angle AFD-\angle EFD)=\angle ADB-(180^\circ-\angle C)+(180^\circ-\angle EBD)=\angle C,$$ as desired $\blacksquare$ Claim 2: $\frac{AQ}{QB}=\frac{AR}{BC}$. Proof: Note that $$\frac{AQ}{QB}=\frac{AE}{EB} \cdot \frac{\sin \angle AEQ}{\sin \angle QEB}=\frac{\sin \angle FED}{\sin \angle FDB}=\frac{\sin \angle FBD}{\sin \angle FDB}=\frac{FD}{FB},$$ and so it suffices to show $\frac{FD}{FB}=\frac{AR}{BC}$. However, since $\angle FAE=\angle FCB$ and $$\angle FEA=180^\circ-\angle FED=180^\circ-\angle FBD=\angle FBC,$$ triangles $FEA$ and $FBC$ are similar, hence $$\frac{FE}{EA}=\frac{FB}{BC},$$ and since triangles $FED$ and $EAR$ are similar too, $$\frac{FB}{BC}=\frac{FE}{EA}=\frac{FD}{AR},$$ and so $\frac{FB}{BC}=\frac{FD}{AR}$, which easily rewrites to what we wanted to prove $\blacksquare$ To the problem, by Claims 1 and 2 we have $$\frac{AR}{BP}=\frac{AQ}{QB}=\frac{AR}{BC},$$ and so $BP=BC$, thus $P$ is the desired fixed point we are looking for, and so we are done.

Cute one. Shortest solution maybe ? Let $EF$ meet $BC$ at $S$. Note that $\angle SFD = \angle EBD = \angle EDB \implies \angle EDF = \angle DSF \implies ED^2 = EF.ES \implies EA^2 = EF.ES \implies \angle FSA = \angle FAE$ so Now Note that $\angle ASB = \angle ASF + \angle FSB = \angle FAE + \angle FDE = \angle 180 - \angle AFD = \angle ACB$ and Note that $AB \perp CS$ so $ASC$ is isosceles so $S$ is reflection of $C$ across $B$.

kinda cheated by using the hint in #7... Let $EF\cap (ABC)=Y$ and $EF\cap BC=X$. As $E$ is the midpoint of $AD$, we have $\angle{ADC}=\angle{EDB}=\angle{EBD}$. Reim's implies that $BE || CY$ and thus, $\angle{ADC}=\angle{YCD}$. So, $YD=AC$ and $AY || CD\implies AY || DX$. Using the congruence of $\triangle EDX$ and $\triangle EAY$, we get that $AYDX$ is a parallelogram. Thus, $AX=YD=AC$ which implies that $X$ is the reflection of $C$ wrt $B$.

I claim that $EF \cap BC=T$ is the required point. Let $EF \cap (ABC)=G$ Then from reims we know that $BE||CG$ As $E$ is the midpoint of $AD$ therefore $\angle ADC=\angle EDB=\angle EDC=\angle GCD=180-\angle GAD \implies AGCD \ \text{isosceles trapezium}$ From here we can easily get $\triangle TDE \cong \triangle AGE$ which implies $E$ is the midpoint of $TG$ and from mid-point theorem we get $B$ as the midpoint of $CT$ which finishes our problem.

My thoughts go to everyone who draw their right triangles with the 90 degrees at the top

A complex bash that is far too complicated: Set $(ACD)$ to be the unit circle. We have \[b=\frac{1}{2}\left(a+\frac{(c-d)\overline{a}+\overline{c}d-c\overline{d}}{\overline{c}-\overline{d}}\right)=\frac{1}{2}\left(a+c+d-\frac{cd}{a}\right).\]\[e=\frac{1}{2}(a+d).\]First we calculate $f$ by using the fact that $F$ lies on the unit circle. \[\frac{d-f}{f-e}\div\frac{\overline{d}-\overline{f}}{\overline{f}-\overline{e}}=\frac{d-b}{b-e}\div\frac{\overline{d}-\overline{b}}{\overline{b}-\overline{e}}.\]The left hand side can be computed as follows: \[\frac{d-f}{\overline{d}-\overline{f}}=\frac{d-f}{\frac{1}{d}-\frac{1}{f}}=-df.\]\[\frac{f-e}{\overline{f}-\overline{e}}=\frac{f-\frac{a+d}{2}}{\frac{1}{f}-\frac{\frac{1}{a}+\frac{1}{d}}{2}}=\frac{adf(2f-a-d)}{2ad-af-df}.\]Thus, \[LHS=\frac{2ad-(a+d)f}{a(a+d-2f)}.\]The right hand side can be computed as follows: \[\frac{d-b}{\overline{d}-\overline{b}}=\frac{d-c}{\overline{d}-\overline{c}}=-cd.\]\[\frac{b-e}{\overline{b}-\overline{e}}=\frac{\frac{c}{2}\left(1-\frac{d}{a}\right)}{\frac{1}{2c}\left(1-\frac{a}{d}\right)}=\frac{c^2\left(\frac{a-d}{a}\right)}{\frac{d-a}{d}}=-\frac{c^2 d}{a}.\]Thus, \[RHS=\frac{cd}{c^2 d/a}=\frac{a}{c}.\]Now we solve for $f$: \[\frac{2ad-(a+d)f}{a(c+d-2f)}=\frac{a}{c}.\]\[2acd-(ac+cd)f=a^2(a+d-2f).\]\[f=\frac{a(2cd-a^2-ad)}{ac+cd-2a^2}.\]We now claim that $EF$ passes through the reflection of $C$ across $B$. Define this point as $P$ with $p=2b-c=a+d-\frac{cd}{a}$. Then we wish to show that \[\frac{p-f}{\overline{p}-\overline{f}}=\frac{p-e}{\overline{p}-\overline{e}}.\]For the LHS, we have \begin{align*} p-f &= \frac{a^2+ad-cd}{a}-\frac{a(2cd-a^2-ad)}{ac+cd-2a^2} \\&= \frac{a^3c+2a^2cd-a^4+acd^2-ac^2d-a^3d-c^2d^2}{a(ac+cd-2a^2)}. \end{align*}Thus, \begin{align*} \frac{p-f}{\overline{p}-\overline{f}}&=\frac{a^3c+2a^2cd-a^4+acd^2-ac^2d-a^3d-c^2d^2}{a(ac+cd-2a^2)} \div \frac{\frac{1}{a^3c}+\frac{2}{a^2cd}-\frac{1}{a^4}+\frac{1}{acd^2}-\frac{1}{ac^2d}-\frac{1}{a^3d}-\frac{1}{c^2d^2}}{\frac{1}{a}\left(\frac{1}{ac}+\frac{1}{cd}-\frac{2}{a^2}\right)} \\&= \frac{a^3c+2a^2cd-a^4+acd^2-ac^2d-a^3d-c^2d^2}{a(ac+cd-2a^2)}\div\frac{acd^2+2a^2cd-c^2d^2+a^3c-a^3d-ac^2d-a^4}{acd(ad+a^2-2cd)} \\&= \frac{cd(ad+a^2-2cd)}{ac+cd-2a^2}. \end{align*}On the other hand, \[p-e=a+d-\frac{cd}{a}-\frac{1}{2}(a+d)=\frac{a^2+ad-2cd}{2a}.\]\[\frac{p-e}{\overline{p}-\overline{e}}=\frac{a+d-\frac{2cd}{a}}{\frac{1}{a}+\frac{1}{d}-\frac{2a}{cd}}=\frac{cd(a^2+ad-2cd)}{ac+cd-2a^2}.\]This shows that $P, E, F$ are collinear.

Let $X=EF\cap CD$ and $Y=EF\cap (ACD)$ Note that by power of point from $X$ to $(BDE)$ and $(ACD)$ $$XD\cdot XB=XF\cdot XE \hspace{0.5cm} XD\cdot XC=XF\cdot XY \Rightarrow \frac{XB}{XC}=\frac{XE}{XY} \Rightarrow BE\parallel CY$$Note that $\angle BDE=\angle EBD=\angle YCB$, therefore $AY\parallel DC$ Since $DE=EA$ we have that $AYDX$ is a parellelogram and $E$ midpoint of $XY$ therefore $B$ midpoint of $XC$ so $X$ is a fixed point on $EF$, as desired.

Let the line through $A$ parallel to $BC$ intersect $(ACD)$ again at $P$, and let $Q$ be the reflection of $C$ over $B$. We claim that $Q$ is the fixed point. We have $$\measuredangle PFD = \measuredangle PAD = \measuredangle BDA = \measuredangle EBD = \measuredangle EFD$$ so $P$ lies on line $EF$. But $APDQ$ is a parallelogram, so $Q$ lies on line $PE$, as desired.

Let $EF$ and $DC$ intersect each other at point $K$. We know that $ED=EB=EA$. By easy angle chasing we consider that $\angle EBF=\angle EKA$, which gives that $ED^2=EA^2=EB^2=EF*EK$, which gives that $\angle AKC=\angle AKE+\angle EKC=\angle FAE+\angle FDE=180-\angle AFD=\angle ACK$, implying that $ACK$ is isosceles and from $AB$ perp. to $KC$, we have $BC=BK$. Then $EF$ passes through symmetric of $C$ respect to $B$, which completes the solution.

Let $R$ be the reflection of $C$ over $B$, and let $R'$ be the reflection of $R$ over $E$. Note that $CDAR'$ is an isosceles trapezoid, now if $F'=RR'\cap (ACD)$ then by Reim's ($CR'\parallel BE$) it follows that $F'\in (BED)$, done.

attached sol

Let $G = EF \cap BC, K = EB \cap AC$. Then we'll prove that $BG = BC$, thus $G$ passes through a fixed point $G$, completing the proof. Note that $ED = EB$, thus $FE$ is the external angle bisector of $\angle BFD$. Therefore $\frac{GB-BD}{GB} = \frac{FD}{FB}$. Thus in order to prove $BG = BC$, we only need to prove that $\frac{FD}{FB} = \frac{BC-BD}{BC}$. Note that $F$ is the Miquel point of $AEBC$, thus $\triangle AFD \sim \triangle KFB$. Hence we get $\frac{FD}{FB} = \frac{AD}{KB}$. Let $L$ be the reflection of $D$ wrt $B$. Then we have $BD = BL$, thus $BK \parallel AL$. Therefore $\frac{BC-BD}{BC} = \frac{LC}{BC} = \frac{AL}{BK} = \frac{AD}{BK} = \frac{FD}{FB}$, so we're done. $\blacksquare$

Let $T=EF\cap BC$ and $K,F$ be the intersections of $\overline{TE}$ with $(ADC)$. A fast angle-chase: \[\measuredangle KAD=\measuredangle KFD=\measuredangle TFD=\measuredangle EFD=\measuredangle EBD=\measuredangle BDE=\measuredangle BDA.\]So, $AK\parallel DC$. Note that $E$ lies on mid-line of $\overline{AK}$ and $\overline{BC}$. Hence, we have \[\frac12=\frac{TE}{TK}=\frac{TB}{TC},\]where lengths are directed. So, $T$ is the desired ``fixed point" which is the reflection of $C$ about $B$. $\blacksquare$

Firstly, let line $EF$ meet $BC$ at point $G$. Let $EF$ meet $(\Delta ADC)$ again at $H$. By property of cyclic quadrilaterals, $\angle EBG= \angle GFD$, $\angle HCG= \angle GFD$ $\implies HC \parallel BE$ Let $\angle EBG= \angle HCG= \theta$. $\Delta ADB$ is a right angled triangle, with mid point $E \implies EB=ED$. Therefore, $\angle EDB=\theta$. As $\angle EDC= \angle HCD=\alpha$(say), $AHCD$ is a cyclic isosceles trapezoid $\implies AH \parallel CD \implies AH \parallel BG$. Now, by $AAS$ congruency, $\Delta GED \cong \Delta HEA \implies EH=EG \implies E$ is the mid point of $GH \implies AHDG$ is a parallelogram. Then, $\angle AHD= \angle AGD= \alpha$ and $AGC$ is isosceles. As $AB$ is the altitude on the unequal side of an isosceles triangle, it is also the median $\implies \boxed{GB=BC}$, and the point $G$ is always the reflection of $C$ about the line $AB$ and is independent of the choice of $D$.

This fixed point is the reflection of $C$ over $B$. Let this point be $X$; we will show that $X$, $E$ and $F$ are collinear. Extend $\overline{EF}$ to meet $(ADC)$ again at $G$. By Reim's theorem, $\overline{EB} \parallel \overline{GC}$, so it follows that $ADCG$ is an isosceles trapezoid with $\overline{AG} \parallel \overline{CD}$. Clearly, $AG = CB - DB = XB - DB = XD$, so it follows that $E$ is the midpoint of $\overline{XG}$; therefore, $X$, $F$, $E$ and $G$ are collinear.

Let $F'$ be the reflection of $F$ wrt $B$. Note that \[ \angle CBF' = \angle DBF = \angle DEF. \]On the other hand, \[ \frac{FE}{ED} = \frac{FE}{EA} = \frac{FB}{BC} = \frac{F'B}{BC}. \]This implies triangles $FED$ and $F'BC$ are similar by Side-Angle-Side. In turn \[ \angle BF'C = \angle EFD = 180^\circ - \angle DBF = 180^\circ - \angle EDB = 180^\circ - \angle EFB. \]This implies $FE$ and $F'C$ are parallel. Since $B$ is the midpoint of $FF'$, this implies $B$ is the midpoint of $C$ and the intersection of $FE$ with $BC$. Conclusion follows

Pretty easy but quite nice. Let $M$ denote the intersection of lines $\overline{EF}$ and $\overline{BC}$. Then, we can notice the following important result. Claim : Triangles $\triangle EMA$ and $\triangle EFA$ are similar. Proof : Since $E$ is the midpoint of the hypotenuse $AD$ of right-angled triangle $\triangle ABD$, it follows that $E$ is the center of $(ADB)$. In particular, then $\triangle EDB$ is isoceles and thus, \[\measuredangle EFB = \measuredangle EDB = \measuredangle DBE = \measuredangle MBE \]from which it follows that $\triangle EMB \sim \triangle EFB$. But now, \[\frac{EM}{EB} = \frac{ EB}{EF}\]Thus, we have that \[EM \cdot EF = EB^2 = EA^2\]which implies that $\triangle EMA \sim \triangle EFA$, as we set out to show. Now, we are left with a simple angle chase. Note that, \begin{align*} \measuredangle DAB + \measuredangle BAM &= \measuredangle DAM\\ &= \measuredangle EAM \\ &= \measuredangle AFE \\ &= \measuredangle AFD + \measuredangle DFE\\ &= \measuredangle ACD + \measuredangle DBE\\ &= \measuredangle ACD + \measuredangle EDB \\ &= (90 + \measuredangle CAB) + (90 + \measuredangle DAB)\\ &= \measuredangle CAB + \measuredangle DAB \end{align*}Thus, it follows that $\measuredangle BAM = \measuredangle CAB$. But this implies that point $M$, which is on line $\overline{EF}$, is the reflection of $C$ across line $\overline{AB}$, which is indeed a fixed point as $D$ varies.

Let $EF\cap BC=C'$ and $EF\cap \odot(ADC)=G$. By Reim's theorem $BE \parallel CG$ and $AG \parallel DC$ because $\measuredangle BDA =\measuredangle EBD =\measuredangle GFD =\measuredangle GAD$, this gives us $AGCD$ is an isosceles trapezoid. Now, $\angle ADC =\angle EBD= \angle GCD\implies GD=AC$ and as $E$ is the midpoint of $AD$ and $AG \parallel C'D$, $AGC'D$ is a parallelogram. From this we get $AC'=DG=AC$ which implies that $C'$ is the reflection of $C$ w.r.t $B$, which is indeed our fixed point. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.5) + fontsize(9); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -6.665055418509338, xmax = 7.756561693440308, ymin = -5.836388202244625, ymax = 6.694138866518905; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); pen ccqqqq = rgb(0.8,0,0); draw((-1,5)--(-1,-2)--(3,-2)--cycle, linewidth(0.7) + zzttqq); /* draw figures */ draw((-1,5)--(-3.54,-2), linewidth(0.7) + ccqqqq); draw(circle((-2.27,-0.4804142857142859), 1.9804142857142855), linewidth(0.7)); draw(circle((-0.27,0.7742857142857141), 4.288305169235254), linewidth(0.7)); draw((-1,5)--(-1,-2), linewidth(0.7)); draw((-1,5)--(3,-2), linewidth(0.7) + ccqqqq); draw((-1,-2)--(-2.27,1.5), linewidth(0.7)); draw((0.46,5)--(3,-2), linewidth(0.7)); draw((-1,5)--(0.46,5), linewidth(0.7) + ccqqqq); draw((0.46,5)--(-3.54,-2), linewidth(0.7) + ccqqqq); draw((-1,5)--(-5,-2), linewidth(0.7) + ccqqqq); draw((-4.190819625537356,-0.9625892635094345)--(0.46,5), linewidth(0.7) + ccqqqq); draw((-5,-2)--(-4.190819625537356,-0.9625892635094345), linewidth(0.7) + dashed + ccqqqq); draw((-5,-2)--(-3.54,-2), linewidth(0.7) + dashed + ccqqqq); draw((-3.54,-2)--(3,-2), linewidth(0.7)); /* dots and labels */ dot((-1,5),dotstyle); label("$A$", (-0.9260082439706534,5.164822570724931), NE * labelscalefactor); dot((-1,-2),dotstyle); label("$B$", (-0.8108984152549779,-1.8733212421763692), NE * labelscalefactor); dot((3,-2),dotstyle); label("$C$", (3.0699472385878,-1.8404327196861763), NE * labelscalefactor); dot((-3.54,-2),dotstyle); label("$D$", (-3.4748687369606137,-1.8404327196861763), NE * labelscalefactor); dot((-2.27,1.5),linewidth(4pt) + dotstyle); label("$E$", (-2.570434368480305,1.6293064030291846), NE * labelscalefactor); dot((-4.190819625537356,-0.9625892635094345),linewidth(4pt) + dotstyle); label("$F$", (-4.741076852833046,-0.9359983512058689), NE * labelscalefactor); dot((-5,-2),linewidth(4pt) + dotstyle); label("$C'$", (-4.987740771509493,-2.333760557039071), NE * labelscalefactor); dot((0.46,5),linewidth(4pt) + dotstyle); label("$G$", (0.5210867455978401,5.131934048234738), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

First extend FE to meet (ADC),call it T ,it is easy to see that $BE \parallel CT$(simple angle chasing), and then another angle chasing gives that $AH\parallel DC$ so AHCD is an isoceles trapezoid and it is well known that HE meets DC at the reflection of C with respect to B and hence we are done. Q.E.D

I officially proclaim now that my synthetic skills are now complete bogus Construct $Q=\overline{BE}\cap\overline{AC}$; note that $F$ is the Miquel point of complete quadrilateral $DBCAEQ$ so if $P=\overline{BC}\cap\overline{EF}$ then $FDPB\sim FACQ$ for the following reason; by definition $\triangle FDB\sim FAQ$, and further as $EB=ED$, $\measuredangle EDP=\measuredangle EDB=\measuredangle EFB=\measuredangle DFE$ so by Alternate Segment $\overline{DE}$ tangent to $(FDP)$, from there by similar switch we must have $\triangle FDP\sim\triangle FAC$, which gets it. Now all we have to show is that $BP$ is constant, i.e. $BD\cdot\frac{CQ}{AQ}$ is constant. Note by Menelaus on $\triangle ADC$ with transversal $\overline{EBQ}$ that $\frac{AE}{ED}\frac{DB}{BC}\frac{CQ}{QA}=-1$ (lengths are directed), i.e. $\frac{DB\cdot CQ}{QA}=-BC$ as desired, so $BP=-BC$, i.e. $P=2B-C$ as vectors, which is very very clearly fixed

Notice that the spiral similarity with center $F$ sends $E$ to $A$ and $B$ to $C$(1). Let $EF$ intersect $BC$ at $X$. By very very easy angle chasing get that $F$ is the center of the spiral similarity sending $X$ to $C$ and $D$ to $A$. Notice that $E$ is the midpoint of $AD$ so by (1) $B$ is the midpoint of $XC$

Fast solution: Clearly $F - E - A'$, where $A'$ is the point such that $CDAA'$ is an isoceles trapezoid ($AA' \parallel CD$). Now, fix $ABC$ and move $D$ along the line $BC$. Clearly $deg D = deg E = deg A' = 1$ and $E = A' \iff D = \infty_{BC}$, so by Zack´s lemma: $deg EA' = deg E + deg A' - 1 = 1$, so $deg EF = 1$, as desired.

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