Let $AB\cap CD=G$, and $AD\cap BC=H$. Since $EM$ is the bisector of $DEM$ and $ME\perp GB$, with $Q=EM\cap DC$, then $(G,Q,C,D)=-1$. But $(G,F,C,D)=-1$, where $F=HM\cap CD$, since those 2 have 3 common points, we have $E=F$, so $H-M-E-Q$ collinear.So $(G,E,A,B)=-1$ Letting $BD'\perp AD$ and $AC'\perp CB$, $C'D'\cap AB=G'$, we have $C'D'\parallel CD$, $(G,E,A,B)=(G',E,A,B)=-1$, so $G=G'$, so $CD,C'D'$ coincide, so $BD,AC$ are altitudes, so $AB$ is the diameter.

Yep. That simple. Today's competition was horrible in terms of originality.

Do trigonometric ceva in $DEC$ and obtain $$\sin \angle DCA \cdot \sin (\angle DEA - \angle DCA) = sin \angle CDB \cdot \sin (\angle DEA - \angle CDB)$$$$\iff$$$$\frac{1}{2}(\cos \angle DEA - \cos (\angle DEA - 2 \angle DCA)) = \frac{1}{2}(\cos \angle DEA - \cos (\angle DEA - 2 \angle CDB))$$ $AB$ is not parallel to $CD$ so $\angle ACD + \angle BDC = \angle DEA \iff $ $M$ is incenter in $DEC \iff BCME$ is cyclic $\iff$ $AB$ diameter.

GGPiku wrote: Let $AB\cap CD=G$, and $AD\cap BC=H$. Since $EM$ is the bisector of $DEM$ and $ME\perp GB$, with $Q=EM\cap DC$, then $(G,Q,C,D)=-1$. But $(G,F,C,D)=-1$, where $F=HM\cap CD$, since those 2 have 3 common points, we have $E=F$, so $H-M-E-Q$ collinear.So $(G,E,A,B)=-1$. Letting $BD'\perp AD$ and $AC'\perp CB$, $C'D'\cap AB=G'$, we have $C'D'\parallel CD$, $(G,E,A,B)=(G',E,A,B)=-1$, so $G=G'$, so $CD,C'D'$ coincide, so $BD,AC$ are altitudes, so $AB$ is the diameter. FTFY

We can do this with areals. Let: $$E=(1,0,0) \, , \, C=(0,1,0) \, , \, D=(0,0,1) \quad \quad |CD|=a \, , \, |DE|=b \, , \, |CE|=c$$$M$ lies on the internal angle bisector of $\angle CED$ so as $ME \bot AB$ it follows $A,B$ lie on the internal angle bisector. Let: $$A=(\lambda,-b,c) \, , \, B=(\mu,-b,c) \quad \lambda,\mu \in \mathbb{R}$$From this we get: $$M=CA \cap BD=(\lambda \mu,-b \lambda,c \mu)$$And as $M$ is on the $E$ internal angle bisector: $$\frac{-b \lambda}{c \mu}=\frac{b}{c} \Leftrightarrow \mu=-\lambda$$. Now use the standard circle equation for $\odot ABCD$. As $C,D$ on the circle we get $v=w=0$. Solving for $u$ using $A$ gives: $$u=\frac{bc \left ((b-c) \lambda-a^2 \right)}{\lambda(\lambda-b+c}$$And using $B$ and the fact $ \mu=-\lambda$ we see: $$\frac{bc \left ((b-c) \lambda-a^2 \right)}{\lambda(\lambda-b+c}=\frac{bc \left((b-c) \lambda +a^2 \right)}{\lambda (c-b- \lambda)} \Leftrightarrow (b-c)(\lambda-a)(\lambda+a)=0$$But as $b-c \Leftrightarrow CE=ED$ which would mean $EM$ is the perpendicular bisector of $CD$ giving $AB \parallel CD$ contradiciting the problem statement we must have $\lambda=\pm a$. WLOG $\lambda=a$: $$A=(a,-b,c)\, , \, B=(-a,-b,c)$$But this gives $A,B$ are the $C,D$ excentres in $\triangle ECD$ and hence as $M$ is therefore the incentre of the triangle so: $$\angle BCA=\angle BDA=90^{\circ}$$And hence $\odot ABCD$ has diameter $AB$ as desired.

Let $AB$ and $CD$ intersect at $X$, $AD$ and $BC$ intersect at $Y$, and $ME$ and $CD$ intersect at $F$. Since $EF$ is the bisector of $\angle DEC$, and $\angle FEX=90^{\circ}$, we have that $(D, C; F, X)=-1$. Now, if we let $F'$ be the intersection of $YM$ and $DC$, we have that $(D, C; F', X)=-1$. Thus, $F=F'$, so $Y, F, M, E$ are collinear. Let $O$ be the center of $ABCD$. By Brokard, $O$ is the orthocenter of $\triangle XYM$, so $YM \perp OX$. But $YM \perp AB$, so $O$ lies on $AB$.

We can also do this with complex. Let $\odot ABCD$ be the unit circle with $A=a, B=b, \cdots$. Then it's easy to get: $$m=\frac{bd(a+c)-ac(b+d)}{bd-ac} \quad \quad \overline{m}=\frac{(b+d)-(a+c)}{bd-ac}$$Then dropping the perpendicular from $M$ to $AB$ gives: $$e=\frac{1}{2} \left (a+b+m-ab \overline{m} \right) \quad \quad \overline{e}=\frac{1}{2ab} \left (a+b+ab \overline{m}-m \right)$$The angle bisector condition is equivalent to: $$\frac{(e-m)^2}{(e-c)(d-e)} \in \mathbb{R} \Leftrightarrow \left (\frac{e-m}{\overline{(e-m)}} \right)^2 \left ( \frac{\overline{(e-c)}}{e-c} \right) \left ( \frac{\overline{(e-d)}}{e-d} \right)=1$$Now focussing on $(e-m)$ we see as $EM \bot AB$: $$\frac{e-m}{a-b}+\overline{\left (\frac{e-m}{a-b} \right)}=0 \Rightarrow \left (\frac{e-m}{\overline{(e-m)}} \right)^2=\left (\frac{a-b}{\overline{(a-b)}} \right)^2=\left(\frac{a-b}{\frac{b-a}{ab}} \right)^2=a^2 b^2$$ Now we look at $(e-c)$: $$2(e-c)=a+b+m-ab \overline{m}-2c=\frac{(b-c)(a^2+bd+ad-ab-2ac)}{bd-ac}$$$$2 \overline{(e-c)}=\frac{(b-c)(bcd+a^2c+abc-acd-2abd)}{(bd-ac)(abc)}$$Combining these we get: $$\left ( \frac{\overline{(e-c)}}{e-c} \right)=\frac{(bcd+a^2c+abc-acd-2abd)}{abc(a^2+bd+ad-ab-2ac)}$$And similarly for $d$ by interchanging $c \leftrightarrow d \, , \, a \leftrightarrow b$: $$\left ( \frac{\overline{(e-d)}}{e-d} \right)=\frac{(acd-bcd+b^2 d+abd-2abc)}{abd(b^2+ac+bc-ab-2bd)}$$ Hence our angle condition is equivalent to: $$1=a^2 b^2 \cdot \frac{(bcd+a^2c+abc-acd-2abd)}{abc(a^2+bd+ad-ab-2ac)} \cdot \frac{(acd-bcd+b^2 d+abd-2abc)}{abd(b^2+ac+bc-ab-2bd)}$$$$cd(a^2+bd+ad-ab-2ac)(b^2+ac+bc-ab-2bd)=(bcd+a^2c+abc-acd-2abd)(acd-bcd+b^2 d+abd-2abc)$$Which we can expand out cancel a lot of terms then factorise to give: $$2(a+b)(c-d)(ac-bd)(ab-cd)=0$$Clearly $c \neq d$. $$ac=bd \Leftrightarrow \frac{a}{b}=\frac{d}{c} \Leftrightarrow \angle BOA=\angle COD \Leftrightarrow AB=CD \Leftrightarrow AC \parallel BD$$Similarly: $$ab=cd \Leftrightarrow AB \parallel CD$$. The former is not true as the diagonals intersect and the latter is forbidden by the problem statement. Hence we must have: $$\boxed{a+b=0 \Leftrightarrow a=-b \Leftrightarrow AB \text{ diamter of } \odot ABCD}$$

Let $EM$ intersect the perpendicular bisector of $CD$ at $P$. This is clearly the midpoint of arc $CD$ in $(EDC)$. However, notice that $MP$ is isogonal to the $M-$altitude in $\triangle MDC$, so $P$ is also the circumcenter of $(MDC)$. It follows that $M$ is the incenter of $\triangle EDC$. From this we easily conclude: $\angle CDM=\angle MDC=\angle MDC=\angle BAC=\angle BAM$ so $AEMD$ is cyclic $\implies \angle MDA=\frac{\pi}{2}\implies AB \text{is a diameter.} \Box $

microsoft_office_word wrote: A quadrilateral $ABCD$ is inscribed in a circle $k$ where $AB$ $>$ $CD$,and $AB$ is not paralel to $CD$.Point $M$ is the intersection of diagonals $AC$ and $BD$, and the perpendicular from $M$ to $AB$ intersects the segment $AB$ at a point $E$.If $EM$ bisects the angle $CED$ prove that $AB$ is diameter of $k$. Proposed by Emil Stoyanov,Bulgaria

By $Blanchet's$ $Theorem$ we get that $AD,BC$ and $EM$ concurr. So the problem is equivalent to this: [Let $ABC$ be a triangle and $AD$ an altitude. $BE$ and $CF$ are cevians concurrent with this altitude at $H$. Then $BCFE$ is cyclic if and only if $H$ is the orthocenter of $ABC$.] Let $ABC$ be the reference triangle and use barycentric coordinates. Let $H$ be a point on $AD$ and $BH$ intersect $AC$ at $E$ thus: $H= ( t : S_{AC} : S_{AB})$ and $E=( t : 0 : S_{AB})$. Let $(BCE)$ intersect $AB$ at $F$. We easily get $F=( c^{2}(S_{AB}+t)-b^{2}S_{AB} : b^{2}S_{AB} : 0 )$. Now $C,H,F$ are collinear if and only if $\frac {c^{2}(S_{AB}+t)-b^{2}S_{AB}}{b^{2}S_{AB}} = \frac {t}{S_{AC}}$ from which we find $t=S_{BC}$ which means that $H$ is the orthocenter of $\triangle{ABC}$ as needed $\blacksquare$ (Of course we get the result for t if amd only if b is not equal to c which is true from the problem statement)

InCtrl wrote: Let $EM$ intersect the perpendicular bisector of $CD$ at $P$. This is clearly the midpoint of arc $CD$ in $(EDC)$. However, notice that $MP$ is isogonal to the $M-$altitude in $\triangle MDC$, so $P$ is also the circumcenter of $(MDC)$. It follows that $M$ is the incenter of $\triangle EDC$. From this we easily conclude: $\angle CDM=\angle MDC=\angle MDC=\angle BAC=\angle BAM$ so $AEMD$ is cyclic $\implies \angle MDA=\frac{\pi}{2}\implies AB \text{is a diameter.} \Box $ MP is isogonal to the M - altitude? Please explain a little more! Thanks!

Let's assume that such a cyclic quadrilateral exists so that $AB$ is not a diameter. Let $A'$ and $B'$ be points on the diagonals $AC$ and $BD$ respectively such that $\angle A'DB = \angle B'CA = 90^{\circ}$. Since $A'B'CD$ is cyclic we get that $\angle B'A'C = \angle BDC = \angle BAC$ which further implies $AB \parallel A'B'$. That means $ME$ must be perpendicular to $A'B'$. Let $E'$ be the point of intersection of $ME$ and $A'B'$ . From the fact that $A'E'MD$ and $B'E'MC$ are cyclic by angle chasing we get that $\angle DE'M = \angle CE'M$. Since we assumed that $\angle DEM = \angle CEM$ triangles $\triangle CEE'$ and $\triangle DEE'$ must be equal, or in other words $CE = DE$. Because $\triangle CED$ is isosceles and since line $EM$ is its angle bisector it is also perpendicular to $CD$ which is not possible because $AB$ and $CD$ are not parallel. We got a contradiction! PS: If this post gets more than 40 likes im going to IMO

GGPiku wrote: Let $AB\cap CD=G$, and $AD\cap BC=H$. Since $EM$ is the bisector of $DEM$ and $ME\perp GB$, with $Q=EM\cap DC$, then $(G,Q,C,D)=-1$. But $(G,F,C,D)=-1$, where $F=HM\cap CD$, since those 2 have 3 common points, we have $E=F$, so $H-M-E-Q$ collinear.So $(G,E,A,B)=-1$ Letting $BD'\perp AD$ and $AC'\perp CB$, $C'D'\cap AB=G'$, we have $C'D'\parallel CD$, $(G,E,A,B)=(G',E,A,B)=-1$, so $G=G'$, so $CD,C'D'$ coincide, so $BD,AC$ are altitudes, so $AB$ is the diameter. $"(G,Q,C,D)=-1"$ What's the term for this in English? I think I know what it means but I want to be sure.

Pinko wrote: GGPiku wrote: Let $AB\cap CD=G$, and $AD\cap BC=H$. Since $EM$ is the bisector of $DEM$ and $ME\perp GB$, with $Q=EM\cap DC$, then $(G,Q,C,D)=-1$. But $(G,F,C,D)=-1$, where $F=HM\cap CD$, since those 2 have 3 common points, we have $E=F$, so $H-M-E-Q$ collinear.So $(G,E,A,B)=-1$ Letting $BD'\perp AD$ and $AC'\perp CB$, $C'D'\cap AB=G'$, we have $C'D'\parallel CD$, $(G,E,A,B)=(G',E,A,B)=-1$, so $G=G'$, so $CD,C'D'$ coincide, so $BD,AC$ are altitudes, so $AB$ is the diameter. $"(G,Q,C,D)=-1"$ What's the term for this in English? I think I know what it means but I want to be sure. It means that the four points are harmonic. It is called "cross-ratio".

This is a sketch, I'll write a full solution ASAP: let $l$ be the parallel to $AB$ through $M$; $l$ is tangent to the circumcircle of $DMC$, so $ME$ meets the axis of $DC$ in the centre $X$ of that circle, but such a point is, in the circumcircle of triangle $CDE$, the middle point of arc $CD$ not containing $E$ (because $ME$ is bisector, and it meets the axis of $DC$ in that point). But there is only one point on the bisector of $\widehat{DEC}$ lying on the circle centered in $X$ and passing through $D$ and $C$, namely the incentre, so $M$ is the incentre of $DEC$ from which it easily descents the thesis.

BronzaniCrnogorac wrote: PS: If this post gets more than 30 likes im going to IMO Why did you change it from 20 to 30? You really don’t want to go to IMO?

microsoft_office_word wrote: A quadrilateral $ABCD$ is inscribed in a circle $k$ where $AB$ $>$ $CD$,and $AB$ is not paralel to $CD$.Point $M$ is the intersection of diagonals $AC$ and $BD$, and the perpendicular from $M$ to $AB$ intersects the segment $AB$ at a point $E$.If $EM$ bisects the angle $CED$ prove that $AB$ is diameter of $k$. Proposed by Emil Stoyanov,Bulgaria A similar problem posted in the India forum: Quote: $\longrightarrow$ Let $\omega$ be a semicircle with diameter $CD$ and center $O$. Let $E,F$ be two arbitrary points on $\omega$ and let $EE \cap FF =Q$. Also let $ED \cap FC = P$. Prove that $PQ \perp CD$.

Isn’t the problem equivalent to proving that Blanchet’s theorem doesn’t have a converse?

MK4J wrote: Isn’t the problem equivalent to proving that Blanchet’s theorem doesn’t have a converse? No

Let $F=EM \cap CD,$ by Lemma 9.18, we have $(X, F; D, C)=-1.$ Now let $F'=YM \cap DC, E'=AB \cap YF'.$ From Lemma 9.11, $(X, E'; A, B)=-1,$ projecting through $Y,$ we have $(X, F'; D, C)=-1.$ This is enough to show that $F=F', E=E'.$ By Brocard, we have $XO \perp XM,$ but since $YM \perp AB,$ $O$ lies on $AB.$ $\blacksquare$

Let $AB \cap CD=X; AD \cap BC=Y$ $E,M,Y$ are collinear. By Brokard: $E$ is center of $k$. $\implies AB$ is diameter of $k$. $\blacksquare$

Dual w.r.t. a circle centered at $M$, with an arbitrary radius. We get the following equivalent problem: Quote: Let $ABC$ be a non-isosceles triangle. Then let $M$ be a point on the $BC$ bisector such that $M$ defines the same angles with the midpoints of $AB$ and $AC$ with $A$. Prove that $M$ is the circumcenter. Now, to prove this, let there be such triangle $ABC$ and $M$ not the circumcenter $O$. $P$ and $Q$ are the midpoints of $AB$ and $AC$. We have that $\angle MPA = \angle MQA$. By sine law in triangles $OMP$, $OMQ$ and $APQ$ (we use that $OPBR$ and $OQCR$ are cyclic, where $R$ is the midpoint of $BC$), we get that $MP \cdot AP = MQ \cdot AQ$. So the areas $AMP$ and $AMQ$ are equal and since their circumcircles are congruent, we have that $AMP \cong AMQ$. So $AP = AQ$, or the triangle $ABC$ is isosceles.

First Note that $E$ is not midpoint of $AB$. It's well-known that if $\angle CEM = \angle DEM$ then $ME$,$AD$ and $BC$ are concurrent at a point $P$. Claim1 : $PDMC$ is cyclic. Proof : Let $B'$ be reflection of $B$ across $E$ we have $\angle PB'M = \angle PBM = \angle CBD = \angle MAP$ so $PAB'M$ is cyclic and $\angle DCM = \angle DCA= \angle MBB' = \angle MB'B = \angle MPA = \angle MPD$ so $PDMC$ is cyclic. Now we have $\angle PCM = \angle 180 - \angle BCA = \angle 180 - \angle BDA = \angle PDM$ and $\angle PCM + \angle PDM = 180$ so $\angle PCM = \angle PDM = \angle 90$ so $\angle BCA = \angle BDA = \angle 90$ so $AB$ is the diameter as wanted. we're Done.

what in the world is blanchet [asy][asy] size(12cm); pair B = dir(0); pair C = dir(60); pair D = dir(110); pair A = dir(180); pair M = extension(A, C, B, D); pair E = foot(M, A, B); pair X = extension(A, C, D, E); pair Y = IP(Line(D, E, 10), unitcircle, 1); draw(unitcircle, blue); draw(A--B--C--D--cycle, blue); markscalefactor=0.03; draw(anglemark(C,E,M), cyan); draw(anglemark(M,E,D), cyan); draw(C--E--M, cyan+blue); draw(D--Y, cyan+blue); draw(A--C, cyan+blue); draw(A-(0, 0.5)--A+(0, 1.25), cyan+blue+dashed); draw(B-(0, 0.5)--B+(0, 1.25), cyan+blue+dashed); draw(Y--C+(0, 0.4), cyan+blue+dashed); draw(E--M+(0, 0.6), cyan+blue+dashed); dot("$A$", A, W); dot("$B$", B, E); dot("$C$", C, NE); dot("$D$", D, N); dot("$E$", E, SW); dot("$M$", M, NE); dot("$X$", X, NE); dot("$Y$", Y, SE); [/asy][/asy] Let $X = \overline{AC} \cap \overline{DE}$ and $Y = \overline{DE} \cap (ABC)$. By the right angle and bisectors lemma, we have $$-1 = (AM; XC) \overset{D}{=} (AB; YC).$$Thus, $ACBY$ is harmonic so $AA$, $BB$, and $YC$ concur at some point $T$ (possibly at infinity). By Pascal's on $ABBDYC$ we find that $T$ lies on line $EM$. This is enough to imply that $T$ must be the point at infinity along line $EM$, as otherwise $E$ must be the midpoint of $\overline{AB}$, which would mean that $\overline{AB} \parallel \overline{CD}$ by symmetry, contradiction. Therefore $\overline{AA} \parallel \overline{BB}$ so $\overline{AB}$ is indeed a diameter of $k$.

Denote $X = AD \cap BC$ and $Y - AB \cap BC$. Claim 1: $XE$ is an altitude of $\triangle XAB$, or $XME$ collinear. Suppose $N = XM \cap CY$ and $N' = ME \cap CY$. From the well known Ceva-Menalaus and Apollonian Circle Lemmas, we have \[(CD; NY) = (CD; N'Y) = -1,\] so $N = N'$. ${\color{blue} \Box}$ Claim 2: $M$ is the orthocenter of $\triangle XAB$. We want to show the orthocenter is the unique point on altitude $XE$ such that $\angle XAM = \angle XBM$. We can reflect $A$ over $E$ to $A'$ to form cyclic quadrilateral $A'BXM$. Then \[\angle MBA = \angle A'XM = \angle MXA = 90 - \angle XAB,\] which implies $BD$ is an altitude, so $M$ is the orthocenter. We finish by noting $\angle BDA = 90$, so $AB$ is a diameter. $\blacksquare$ [asy][asy] size(300); defaultpen(fontsize(10)); pair A, B, C, D, E, M, N, X, Y; A = dir(0); B = dir(180); C = dir(100); D = dir(50); M = extension(A, C, B, D); E = foot(M, A, B); N = extension(E, X, C, D); X = extension(A, D, B, C); Y = extension(A, B, C, D); draw(circumcircle(A, B, C)); draw(B--X--A--C--Y--B--D--E--C--X--E); dot("$A$", A, dir(315)); dot("$B$", B, dir(225)); dot("$C$", C, dir(120)); dot("$D$", D, dir(45)); dot("$E$", E, dir(270)); dot("$M$", M, dir(170)); dot(extension(E, X, C, D)); label("$N$", extension(E, X, C, D), dir(225)); dot("$X$", X, dir(90)); dot("$Y$", Y, dir(270)); [/asy][/asy]

Let $AD \cap BC = X$ and $CD \cap AB = Y$. By right angles and bisectors we get that $(A, M; G, C) = -1$ and $(A, M; G, C) \overset{D}= (A, B; E, Y) = -1$ which implies that $EM$, $AD$ and $BC$ concur by Ceva-Menelaus. Then by Brokard's we get that the center of $k$ is the orthocenter $\triangle XMY$ so $YE$ passes through $O \implies AB$ is a diameter as desired.

Really too easy for words. I first saw this problem as a G4 and nearly got a heart attack, Problem 1 makes a lot more sense. We let $P = \overline{AD} \cap \overline{BC}$ and $X= \overline{AB}\cap \overline{CD}$. Now, we show the following key result. Claim : Points $E$ , $M$ and $P$ are collinear. Proof : Let $R_1$ and $R_2$ be the intersections of $\overline{CD}$ with $\overline{EM}$ and $\overline{PM}$ respectively. Then, due to the Right Angles/Bisectors picture, we know that \[(DC;R_1X)=-1\]Further, from the Ceva/Menelaus picture we also have that \[(DC;R_2X)=-1\]Thus, $R_1=R_2$ and indeed points $E$ , $M$ and $P$ are collinear as claimed. Now, let $O$ be the center of $\Gamma$. Then, by Brokard's Theorem we know that $PM \perp XO$. But, due to the above observation we also have that $\overline{PM} \perp \overline{AB}$. Thus, the center $O$ must lie on line $\overline{AB}$ which implies that $AB$ is a diameter of $\Gamma$ as desired.

fakesolve $\implies$ hinted on pascal Assume otherwise, then let $L = \overline{ED} \cap k$, and $G= \overline{AC} \cap \overline{ED}$. By Right Angles and Bisectors $-1=(AM;GC) \stackrel{D} = (AB ; LC)$. let $\overline{A_{\perp}}$, and $\overline{B_{\perp}}$ be the tangent lines to $k$ at $A$, and $B$ respectively. Note that $\overline{A_{\perp}}$, $\overline{B_{\perp}}$, and $\overline{LC}$ must concur at a point call it $X$, as $ABLC$ is harmonic. Pascal on $ABBDLC$ gives us $E$, $M$, $X$ collinear. However $X$ lies on the perpendicular bisector of $\overline{AB}$, so $E$ must be the midpoint of $\overline{AB}$, which implies $\overline{CD} \parallel \overline{AB}$ contradiction.

Let $AD \cap BC=N$ , $ AB \cap CD=P$ and $EM \cap CD=K$. It is easy to see that $(D,C;K,P)=-1$ due to the famous right angle-angle bisector-harmonic lemma. This implies $K$ lies on the polar of P with respect to $(ABCD)$ hence $K$ and consequently $E$ lie on $NM$ due to brokard's. $$(D,C;K,P) \stackrel{N}{=} (A,B;E,P)=-1$$which implies $P$ is the $N$ expoint in $\Delta NAB$ which coupled with the given conditions implies that $C$ and $D$ are feet of altitudes which is what we needed to show.

Harmonic ratioes kill this problem.

just projective geometry finishes this. let \(DB \cap EC\) = \(K\) and \(AC \cap ED\) = \(L\) from the fact that \(EM\) bisects \(\angle CED\) and \(\angle EDB = 90^\circ\) we obtain \((D, K, M , B) = -1)\) then project from \(C\) the line \(DB\) onto \(ME\) to get \((BC \cap ME, M, DC \cap ME, E) = -1\) then do the same for the line \(CA\) to get \(AD, ME, CB\) concurrent the rest just finishes by brocard on \(ADCB\)

Nice solution using Pappus and Desargues theorem. Let $EM$ intersects $CD$ at $G$. $GA$ and $DE$ intersects at $J$ and $GB$ and $EC$ intersects at $I$. Then by Pappus theorem $J-M-I$ collinear, since $(D,C;G,F)=-1$ we know that $F$ lies on the line $J-M-I$. If we use Desargues theorem on triangles $JAD$ and $CBI$, we get that $AD, BC$ and $GE$ concurrent, let that concurrency point be $N$. By the Brocards theorem $NM$ is the polar of $F$, since $AB$ is perpendicular to $EN$, we get that $AB$ is a diameter.