Just bash it Set $a=1~,~c=-1$ and so $|b|=1$ Let $M$ be the midpoint of $CD$ we see that it's collinear with the midpoint of both $AB$ and $AC$ So $\exists t \in \mathbb{R}$ such that$: m=t(b+1)$ Hence $d=2m-c=2tb+2t+1$ Now to evaluate $e$ we have: $$|ME|^2=|MA|^2\implies(e-m)(\overline{e}-\overline{m})=(m-1)(\overline{m}-1)\implies \overline{e}=\frac{(m-1)(m-b)+m(e-m)}{b(e-m)}\implies \overline{e}=\frac{(tb+t)e-(tb^2+2tb+t-b)}{b(e-(tb+t))}$$$$\angle AED=90^\circ\implies\frac{\overline{e}-\overline{a}}{\overline{d}-\overline{e}}=\frac{e-a}{e-d}\implies\frac{\overline{e}-\overline{a}}{\overline{d}-\overline{a}}=\frac{e-a}{2e-a-d}\implies\overline{e}=\frac{(\overline{d}-\overline{a})(e-a)+\overline{a}(2e-a-d)}{2e-a-d}\implies\overline{e}=\frac{(tb+t+b)e-(tb^2+2tb+t+b)}{b(e-(tb+t+1))}$$By equating we get: $$(e-(tb+t+1))((tb+t)e-(tb^2+2tb+t-b))=(e-(tb+t))((tb+t+b)e-(tb^2+2tb+t+b))$$Which is a quadratic accept $1$ as a root, the other can be obtained by vieta's formula: $$e=\frac{(tb+t)(tb^2+2tb+t+b)-(tb+t+1)(tb^2+2tb+t-b)}{b}=\frac{tb^2+b-t}{b}$$define $N$ as the midpoint of $BE$ to get: $$n=\frac{tb^2+b^2+b-t}{2b}$$Now to find $o$ we first notice that it's pure imaginary and then: $$|OD|^2=|OA|^2\implies (o-d)(-o-\overline{d})=(o-1)(-o-1)\implies o= \frac{d\overline{d}-1}{\overline{d}-d}\implies o=\frac{2tb+2t+b+1}{1-b}$$And finally we finish with collinearity: $$\frac{n-a}{a-o}=\frac{(b-1)^2}{4b}\in\mathbb{R}~~~~~~~~~\blacksquare$$

Just synthetic it [asy][asy] size(6cm); defaultpen(fontsize(10pt)); import olympiad; import geometry; pair A = (1.3,1.0); pair B = (0,0); pair E = (4,0); pair O1 = circumcenter(A,B,E); pair X = 2*O1-A; pair C = extension(B,X,A,2*O1-E); pair D = A+X-C; pair M = (C+D)/2; pair N = (B+E)/2; pair O = circumcenter(A,C,D); fill(A--C--X--cycle, mediumgray); fill(A--B--E--cycle, mediumgray); draw(C--X--D, linewidth(0.7)); draw(A--M, linewidth(0.7), StickIntervalMarker(1, 2, p=linewidth(0.7), size=0.2cm)); draw(M--X, linewidth(0.7), StickIntervalMarker(1, 2, p=linewidth(0.7), size=0.2cm)); draw(C--M, linewidth(0.7), StickIntervalMarker(1, 1, p=linewidth(0.7), size=0.2cm)); draw(M--D, linewidth(0.7), StickIntervalMarker(1, 1, p=linewidth(0.7), size=0.2cm)); draw(A--O, dashed); draw(rightanglemark(A,B,C), black); draw(rightanglemark(A,E,D), black); draw(A--B--C--D--E--cycle, linewidth(1)); draw(A--C--X--cycle, black); draw(A--B--E--cycle, black); dot("$A$", A, dir(98)); dot("$B$", B, dir(172)); dot("$E$", E, dir(25)); dot("$X$", X, dir(-82)); dot("$C$", C, dir(-143)); dot("$D$", D, dir(-20)); dot("$M$", M, dir(-136)); dot("$N$", N, dir(62)); dot("$O$", O, dir(-56)); [/asy][/asy] Let $X = BC\cap DE$, so by angle conditions, $X$ is the antipode of $A$ in $\odot(ABE)$. In particular, $ACXD$ is a parallelogram. Next, note that $$\left.\begin{array}{r} \angle CAX = \angle AXE = \angle ABE \\[2pt] \angle CXA = \angle BEA \end{array} \right\} \implies \triangle ABE\stackrel-\sim\triangle ACX.$$Thus, if $M$ and $N$ are midpoints of $CD$ and $BE$, then $$\measuredangle ACM = -\measuredangle BAN = 90^\circ - \measuredangle(BX,AN) = 90^\circ - \measuredangle DAN,$$implying the conclusion.

Lmao ratio lemma ...... Let $F = BC \cap ED$, $X = AO \cap BE$ and $Y$ be midpoint of $BE$ and $CD$. Observe that $\Box ABFE$ cyclic with diameter $AF$. Hence $Y$ lie on $AF$. Which also give us $\Box ACFD$ parallelogram. $$\measuredangle EAO = \measuredangle EAD + \measuredangle DAO =90 - \measuredangle ADE + 90 - \measuredangle ACD = \measuredangle DCF$$And similarly $\measuredangle OAB = \measuredangle FDC$. Note from $\measuredangle CFD = \measuredangle ADE = \measuredangle BCA$ we have $\triangle ADE \sim \triangle ACB$. Now using ratio lemma in $\triangle CFD$ and $\triangle AEB$ be have $$\frac{sin \measuredangle DCF}{sin \measuredangle FDC}=\frac{FD}{FC}=\frac{AC}{AD}=\frac{AB}{AE}$$Hence $\frac{EX}{XB} = \frac{sin \measuredangle EAO \cdot AE}{sin \measuredangle OAB \cdot AB} = 1 \Rightarrow X$ is midpoint of $EB$. $\blacksquare$

Let $A'$ be the reflection of $A$ over $M$, the midpoint of $CD$. Then $MA = MB = ME = MA'$, so $\angle ABA' = \angle AEA' = 90^{\circ}$. Therefore $C$ lies on $A'B$ and $D$ lies on $A'E$. Additionally, $ACA'D$ is a parallelogram because $M$ is the midpoint of both $CD$ and $AA'$. We'll show $\angle NAC + \angle CDA = 90^{\circ}$, which implies $N \in \overline{AO}$ where $N$ is the midpoint of $BE$ as $\angle CDA = 90^{\circ} - \angle OAC$. To do so, use complex numbers with unit circle the circumcircle of $\triangle ABE$. If $E'=-e$, then $C = AE'\cap BA'$ as $AC\parallel A'D$ and $D = -C$, so \begin{align*} c &= \frac{(a-e)(-ab)-(b-a)(-ae)}{-ae+ab} = \frac{2be-ae-ab}{b-e}\\ d &= \frac{-2be+ae+ab}{b-e}, \quad n = \frac{b+e}{2}. \end{align*}Finally, we compute: \begin{align*} \frac{n-a}{c-a}\cdot \frac{c-d}{a-d} &= \frac{\frac{b+e}{2}-a}{\frac{2be-ae-ab}{b-e}-a}\cdot \frac{2\left(\frac{2be-ae-ab}{b-e}\right)}{a+\frac{2be-ae-ab}{b-e}}\\ &=\frac{1}{4}(b+e-2a)(2/a-1/b-1/e)\cdot\frac{b-e}{4a(e-a)(b-a)} \end{align*}This can easily be checked to be purely imaginary, as desired.

Areas! Angle chase to find that $BC // AD$, $AC // DE$. Now let $A'$ be the antipode of $A$ on $(ABC)$. We now have that \begin{align*} [AA'B] = [AA'C] = [ADC] \end{align*}which is symmetric, so $AA'$ bisects $BE$ as desired.

Areas again https://dgrozev.wordpress.com/2024/07/17/an-easy-geometrical-problem-from-imo23-shortlist/

Let $X$ be the intersection of $BC$ and $DE.$ Then if $M$ is the circumcenter of $\triangle ABE,$ then $M$ is the midpoint of both $CD$ and $AX,$ so $ACXD$ is a parallelogram. Moreover, if we draw the line $\ell$ through $A$ parallel to $CD$, and let it intersect $DE$ and $BC$ at points $Y$ and $Z$, respectively, this implies that $ACD$ is the medial triangle of $\triangle XYZ.$ Now, let $F$ and $G$ be the feet of the altitudes from $Y, Z$ to $ZX, XY,$ respectively, so that $YE = EG$ and $ZB = BF.$ Then if $N$ is the midpoint of $FG,$ then since $AENB$ is a parallelogram, $AN$ bisects $BE.$ It suffices to show that $A,N,O$ are collinear. Indeed, since $AF = AG,$ they all lie on the perpendicular bisector of $FG,$ so we are done.

Let $BC \cap DE$ at $K$. If $M$ is center of $ABE$ and $K$ is on that circle, then $M$ is center of $ABKE$.Let $\angle EBM=\angle BEM=\alpha$, $\angle MEK=\angle MKE=\angle ABE=\beta$ so, $\angle MBE=\angle MKB=90-\alpha-\beta$.We know that $AM=MK$ and $CM=DM$ that means $ACKD$ is parallelogram.Thus,$AC,AD$ are parallel to $KD,KC$ respectively.Assume $AO \cap BC$ at $M'$.Let $\angle OAM=\gamma$.From parallelity $\angle KAC=\beta$ by cyclic condition $\implies$ $\angle BAC=\angle DAE=\alpha$ and $\angle OAD=90-\alpha-\beta-\gamma$.From center $O$ condition $\implies$ $\angle OCA=\beta+\gamma$, $\angle ODA=90-\alpha-\beta-\gamma$ and $\angle OCD=\angle ODC=\alpha$. From paralellity $\angle DCK=90-\beta-\gamma$ and $\angle CDK=\alpha+\beta+\gamma$. Finally, median sine to $\triangle CKD=\triangle ABE$ equation is true.

Just MMP it The crux of the problem is the following claim which we of course solve using MMP! $\textbf{CLAIM:}$ $ABA'D$ is a cyclic quadrilateral where $A'$ is the $A$-antipode. $E$ and $F$ are points on $BA'$ and $A'D$ respectively such that $AF \parallel A'D$ and $AE \parallel BA'$. Then $EF$, $AA$ and $BD$ are concurrent. $\textbf{PROOF:}$ [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. https://github.com/Bubu-Droid/dotfiles/blob/master/bubu-scripts/bubu-asy.py */ pair O = (11.71916,6.19787); pair A = (6.12566,25.60504); pair B = (-5.57187,16.63555); pair D = (28.92364,16.77764); pair Ap = (17.31267,-13.20928); pair E = (23.54528,2.88728); pair F = (-0.10694,9.50847); pair T = (-25.27638,16.55438); import graph; size(10cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); real xmin = -5, xmax = 5, ymin = -5, ymax = 5; pen ffxfqq = rgb(1.,0.49803,0.); draw(circle(O, 20.19715), linewidth(0.6)); draw(A--B, linewidth(0.6)); draw(B--D, linewidth(0.6)); draw(D--A, linewidth(0.6)); draw(A--T, linewidth(0.6)); draw(T--B, linewidth(0.6)); draw(T--E, linewidth(0.6)); draw(B--Ap, linewidth(0.6) + ffxfqq); draw(Ap--D, linewidth(0.6) + blue); draw(F--A, linewidth(0.6) + blue); draw(A--E, linewidth(0.6) + ffxfqq); draw(A--Ap, linewidth(0.6) + linetype("4 4") + red); dot("$O$", O, SW); dot("$A$", A, NW); dot("$B$", B, NW); dot("$D$", D, NE); dot("$A'$", Ap, dir(280)); dot("$E$", E, dir(-15)); dot("$F$", F, SW); dot("$T$", T, NW); [/asy][/asy] We fix the circle, $A$, $D$ and animate $B$ projectively on the circle. Clearly this fixes $A'$ and so the line $A'D$ is fixed. This further gives us that $A\infty_{A'D}$ is also fixed. Now we have that, \[ B \mapsto A'B \mapsto A'B \cap A\infty_{A'D} \equiv F \text{ via } \text{circle} \mapsto \mathcal{P}(A') \mapsto A\infty_{A'D} \]is projective. So the degree of $F$ is $1$. Also, we have that, \[ B \mapsto A'B \mapsto A'B \cap \ell_{\infty}\mapsto A\infty_{A'B} \mapsto A\infty_{A'B}\cap A'D \equiv E \text{ via } \text{ circle } \mapsto \mathcal{P}(A') \mapsto \ell_{\infty}\mapsto \mathcal{P}(A)\mapsto A'D \]is projective, where $\ell_{\infty}$ is the line at infinity. So the degree of $E$ is $1$. This gives us that the degree of line $EF$ is $=2$. Also note that degree of line $AA$ is $0$ as the line is essentially fixed. Now also note, \[ B\mapsto DB \text{ via }\text{ circle }\mapsto \mathcal{P}(D) \]is projective. This gives us that the degree of line $DB$ is $=1$. So the concurrency condition of $AA \cap BD \cap EF$ has degree $0 + 1 + 2 = 3$ (can be rephrased as the collinearity of $\overline{AA\cap BD - E - F}$). Thus it suffices to check that it is true for $3 + 1 = 4$ cases. $B = A$. In this case, note that $DB \equiv AD$, $F\equiv A$ and $E\equiv A'$ which clearly shows that $A$ is the common concurrency point and done. $B \equiv D$-antipode. In this case note that $F \equiv B$, $E \equiv D$, and so, $BD\equiv EF$. This clearly finishes. $B\equiv A'$. In this case, note that $E\equiv AA\cap BD$ which clearly finishes. $B\equiv D$. In this case, note that $F\equiv \infty_{A'D}$, $E\equiv \infty_{A'D}$ and so, we can remove this case using Zack's Lemma. [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. https://github.com/Bubu-Droid/dotfiles/blob/master/bubu-scripts/bubu-asy.py */ pair A = (6.90289,21.50125); pair B = (-3.09670,12.35254); pair E = (23.50200,17.08615); pair D = (20.67305,6.45038); pair C = (2.76527,5.94536); pair T = (-46.98268,4.54241); pair F = (-4.31783,5.74561); pair G = (27.75616,6.65014); pair S = (5.80296,60.50422); pair R = (7.21464,10.44674); import graph; size(10cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); real xmin = -5, xmax = 5, ymin = -5, ymax = 5; pen ffxfqq = rgb(1.,0.49803,0.); draw(circle((11.71916,6.19787), 16.04337), linewidth(0.6)); draw(A--B, linewidth(0.6)); draw(B--E, linewidth(0.6)); draw(E--A, linewidth(0.6)); draw(A--T, linewidth(0.6)); draw(T--B, linewidth(0.6)); draw(T--D, linewidth(0.6)); draw(C--A, linewidth(0.6) + blue); draw(A--D, linewidth(0.6) + ffxfqq); draw(D--G, linewidth(0.6)); draw(F--S, linewidth(0.6)); draw(S--G, linewidth(0.6)); draw(B--G, linewidth(0.6)); draw(E--F, linewidth(0.6)); draw(B--C, linewidth(0.6) + ffxfqq); draw(D--E, linewidth(0.6) + blue); draw(S--R, linewidth(0.6)); dot("$A$", A, NW); dot("$B$", B, NW); dot("$E$", E, NE); dot("$D$", D, dir(270)); dot("$C$", C, dir(270)); dot("$T$", T, NW); dot("$F$", F, NW); dot("$G$", G, NE); dot("$S$", S, NW); dot("$R$", R, dir(270)); [/asy][/asy] Now back to the original problem note that we have to prove $AO$ passes through the midpoint of $BE$. Let $M$ denote the midpoint of $BE$. But note that this is equivalent to proving that the $A$-altitude in $\triangle ACD$ and $AM$ are isogonal, i.e., the $A$-altitude is the symmedian of $\triangle ABE$. Let $CD$ meet the circle again at $F$ and $G$. Define $T=AA\cap BE\cap FG$, $R = BG\cap FE$ and $S = FB\cap GE$. Clearly $A$ lies on the polar of $T$. Now by Brokard's Theorem, we also have that $SR$ is the polar of $T$. This also gives us that $SR \perp FG$ since $FG$ is the diameter. This means that $AR\perp FG$. Now by Pascal on $BBGEEF$, we finally get that $AR$ is the $A$-altitude in $\triangle ACD$ which finishes.

We have $F,M,G,H,I$ as the midpoints of $\overline{AC}, \overline{CD}, \overline{AD}, \overline{AB}, \overline{AE}$ respectively. Claim 1: $\overline{AC}$ and $ \overline{MH}$ intersect at $F$. $F$ is the circumcenter of $\Delta ABC$ since $\Delta ABC$ is a right triangle, so we get that $\overline{FH} \perp \overline{AB}$ but since $M$ is the circumcenter of $\Delta ABE$, we have that $\overline{MH} \perp \overline{AB}$ so we have the result. Claim 2: $\overline{AC}\parallel \overline{DE}$, and similarly $\overline{AD} \parallel \overline{BC}$. We have $$\angle BCA = 90^\circ -\angle BAC = \angle HFA = \angle CFM = \angle CAB$$so they are parallel. Let $OA$ intersect $BE$ at $N'$. We have that \[\frac{N'B}{N'E} = \frac{AB\sin(\angle OAB)}{AC\sin(\angle OAE)} = \frac{AC}{AD}\cdot \frac{\sin(\angle OAB)}{\sin(\angle OAE)} = \frac{AC}{AD}\cdot \frac{\sin(\angle ACD)}{\sin(\angle ADC)} = 1\]Therefore, $N' = N$, so we are done.

Solution using barycentric coordinates : Let $BC$ and $ED$ intersect at point $F$. Since $AF$ is a diameter, $M$ is the midpoint of $AF$ thus $ACFD$ is a parallelogram. Further $\angle{CAE} = \angle{CAF} + \angle{FAE} = \angle{AFE} + \angle{FAE} = 180^\circ - \angle{AEF} = 90^\circ$ and $\angle{BAD} = 90^\circ$ in the same manner. Let $A = (1,0,0), C = (0,1,0), D = (0,0,1)$ and let $CD = a, DA = c, AC = d$, further we define the Conway notation as $S_a = \frac{c^2 + d^2 - a^2}{2}$, $S_c$ and $S_d$ are defined in the same way. Now since $A + F = C + D$ (parallelogram law, where addition is done component wise), we have that $F = (-1,1,1)$ We now compute point $B = (x,y,z)$ where $x + y + z = 1$, since $B$ lies on $CF$ we have : $\begin{vmatrix} 0 & 1 & 0 \\ -1 & 1 & 1 \\ x & y & z \notag \end{vmatrix} = x + z = 0. $ Hence, implying that $y = 1$, and thus $B = (-z,1,z)$. Now using the fact that $\vec{AB} = (-z - 1, 1, z)$ is perpendicular to $\vec{AD} = (-1,0,1)$, by EFFT, we have : $0 = a^2 + c^2(-z - 1 - z) + d^2 \times - 1$, and thus $z = \frac{-S_a}{c^2}$ We thus get that $B = ( \frac{S_a}{c^2},1, \frac{-S_a}{c^2})$, and by symmetry that $E = ( \frac{S_a}{d^2},\frac{-S_a}{d^2}, 1)$ Hence the midpoint $N = (\frac{\frac{S_a}{c^2} + \frac{S_a}{d^2}}{2} = t, \frac{1 - \frac{S_a}{d^2}}{2}, \frac{1 - \frac{S_a}{c^2}}{2}) = (2t : \frac{d^2 - S_a}{d^2}:\frac{c^2 - S_a}{c^2}) = (2c^2d^2t = x : c^2S_c : d^2S_d)$ by the identities $c^2 - S_a = S_c$ and $d^2 - S_a = S_d$ But recall that $O = (a^2S_a : c^2S_c : d^2S_d)$, thus since $N = (x : c^2S_c : d^2S_d)$ it must lie on the cevian $AO$, as wanted.

Here's a weird synthetic solution. Let $M,M_1,M_2$ be the midpoints of $\overline{CD},\overline{AB},$ and $\overline{AE}$ respectively. Since $\overline{CB}\perp\overline{AB}$ and $\overline{MM_1}\perp\overline{AB},$ we have that $\overline{DA}\perp\overline{AB}$ as well. Similarly, $\overline{CA}\perp\overline{AE}.$ Therefore, $\angle EAD=90^\circ-\angle DAC=\angle CAB,$ so $EAD$ and $BAC$ are similar right triangles. Next, let $A'=\overline{ED}\cap\overline{BC}$ and let $N$ be the midpoint of $\overline{BE}.$ Since $A$ is the antipode of $A'$ with respect to $(A'BE),$ we know that $\overline{AN}$ passes through the orthocenter of triangle $A'BE,$ which we'll call $H.$ Let $X$ and $Y$ denote the feet of $H$ onto $\overline{AC}$ and $\overline{AD}$ respectively. Since $\overline{BH}\perp\overline{A'E},$ we in fact have $\overline{BH}\perp\overline{AC},$ so $X$ is the foot of $B$ onto $\overline{AC}.$ Similarly, $Y$ is the foot of $E$ onto $\overline{AD}.$ Now by similar triangles, $\tfrac{AX}{AY}=\tfrac{AC}{AD}.$ This implies (say, by homothety at $A$) that $H$ lies on $\overline{AO},$ which finishes because we already showed that $H$ lies on $\overline{AN}.$

Define $T=BC\cap DE$, $M$ as the midpoint of $CD$, and $N$ as the midpoint of $BE$. Note that $AETB$ is cyclic with center $M$, and $T$ being the antipode of $A$. This implies $AM=MT$ and $CM=MD$ implying $ADTC$ is a parallelogram. Claim: $\angle TCD=\angle NAE$ Proof: Note that it suffices to show $\triangle TCA\sim\triangle EAB$. Clearly: \[\angle ATC=\angle ATB=\angle AEB\]\[\angle TAC=\angle ATD=\angle ATE=\angle ABE,\]as desired $\square$ We can then finish by saying: \begin{align*} \angle EAO&=\angle OAD+\angle DAE\\ &=90-\angle ACD+\angle DAE\\ &=180-\angle CDT-\angle EDA\\ &=\angle ADC\\ &=\angle TCD\\ &=\angle EAN\\ \end{align*}implying $A,O,N$ are collinear $\blacksquare$

complete the cyclic quadrilateral and prove AGE \sim CMF following from ABE \sim ACF. angle chasing finishes

what the trig Let $P=ED\cap BC$, this lies on $(ABE)$ from opposite right angles. First of all, $180^\circ$ rotation around $M$ gives $DP\parallel AC$. Use ratio lemma on $\triangle BAE$, we want \[\frac{\sin{\angle BAO}}{\sin\angle OAE}\times \frac{AB}{AE}=1\] Now, note that from circumcentre property, $\measuredangle ACD=90^\circ-\measuredangle DAO=\measuredangle OAB$ . Further, $\measuredangle EAO=90^\circ+\measuredangle CAO=\measuredangle CDA$ from a quick angle chase with angle at center, and due to parallel lines. Hence, \[\frac{\sin{\angle BAO}}{\sin\angle OAE}=\frac{\sin{\angle DCA}}{\sin\angle ADC}=\frac{AD}{AC}\] Now we want $\frac{AD\times AB}{AC\times AE}=1$ but this is trivial as $\triangle AED\sim \triangle ABC$ from corresponding angles and right angle.

Sol:- Let $Q$ be the circumcenter of $ABE$. Let $U=BC \cap ED$, then $U$ is $A-$ antipode in $(ABE)$ ,hence $Q$ is common midpoint of $CD,AU$.Thus $ACUD$ is a parallelogram.Thus $\angle ACB=\angle ADE \implies \angle BAC=\angle EAD \implies AC,AD $ isogonal wrt $\angle BAE$.Let $A'$ be reflection of $A$ across $CD$. Clearly $UA' \parallel CD$ and $A' \in (ABE)$. $-1=(C,D;Q, \infty_{CD})\stackrel{U}{=}(B,E;A,A') \implies AM,AA'$ are also isogonal lines wrt both $\angle BAE, \angle CAD$. Since $AA'$ is altitude in $\Delta ACD \implies A-M-O$ are collinear.

Let $A = (0, 0)$, $B = (1, 0)$, $C = (1, c)$ and $D = (0, d)$. Let $N$ be the midpoint of $CD$, $\left(\frac12, \frac{c+d}2\right)$. $E$ must lie on the intersection of $(AD)$ and the circle centered at $N$ through $A$, $B$. Thus $AE$ (the radical axis) is perpendicularly bisected by $LN$, where $L$ is the midpoint of $AD$. Thus $E$ is the reflection of $A$ in $LN$. Let $D'$ be the reflection of $D$ in $LN$, i.e. the foot from $D$ to $AC$ (by midpoint theorem) - lines $AC$ and $DD'$ have equations $y=cx$ and $y=\frac{-x}c + d$, so they intersect at $\left(\frac{cd}{c^2 + 1}, \frac{c^2d}{c^2 + 1}\right)$. As $E$ is the reflection of $D'$ in $L$ (because $AEDD'$ is a rectangle), it is $\left(-\frac{cd}{c^2 + 1}, \frac{d}{c^2 + 1}\right)$ and so if $M$ is the midpoint of $BE$ then $M = \left(\frac{c^2 - cd + 1}{2(c^2 + 1)}, \frac{d}{2(c^2 + 1)}\right)$. Finally, $\angle ACD = \arctan(c) - \arctan(c-d) = \arctan\left(\frac{d}{1 + c^2 - cd}\right)$, while $\angle MAB$ is just $\arctan \frac{y}x$ for $M$, which is identical to the expression for $\angle ACD$. Since $\angle BAD = 90^\circ$ (for $NA=NB$), $\angle DCA = 90^\circ - \angle OAD$ finishes.

Note that $K = BC \cap ED$ lies on $(ABE)$. Let $M$ be the midpoint of $CD$. Then, $\triangle AMD \cong \triangle KMC$ and $\triangle AMC \cong \triangle KMD$. Therefore, $AC \parallel EK$ and $AD \parallel BK$. So, $\angle BAD = \angle EAC = 90$. Let $J = AO \cap BE$. Note that \begin{align*} \angle ACM = \angle ACO + \angle OCD &= \angle CAO + \left(90 - \angle COM\right) \\ &= \angle CAO + \left(90 - \angle CAD\right) = \angle CAO + \angle CAB = \angle BAJ. \end{align*}Additionally, $\angle CAM = \angle AKE = \angle BAE$. Therefore, $\triangle BAJ \sim \triangle ACM$ and similarly $\triangle EAJ \sim \triangle ADM$. So, \begin{align*} \dfrac{BJ}{JE} = \dfrac{BJ : AJ}{EJ : AJ} = \dfrac{AM : CM}{AM : DM} = 1. \end{align*}Therefore, $BJ = JE$, so $AO$ passes through $J$ which is the midpoint of $BE$.

Let $ED$ and $BC$ intersect at $X$ and let $M$ and $N$ be the midpoints of $CD$ and $BE$. Let $A'$ be the antipode of $A$ in $(ACD)$ and let $B'$ and $E'$ be the foot of $B$ and $E$ onto $AC$ and $AD$. Let $BB'$ and $EE'$ intersect at $Y$. As $AEXB$ is cyclic with diameter $AX$ we have that $M$ is the midpoint of $AX$ and further that $ACXD$ is a parallelogram. Thus by similarity we have that $B'E'\parallel CD$. By homothety, $A$, $Y$, and $A'$ are collinear. As $O$ is the midpoint of $AA'$ and $N$ is the midpoint of $AY$, we are done.

Let $A'$ be the reflection of $A$ across the midpoint $M$ of $CD$. $A', D, E$ and $A', C, B$ are collinear and $ACA'D$ is a parallelogram. If $A''$ is the reflection of $A$ across $O$, $A''D \perp AD \parallel CA'$ and $A''C \perp DA'$. Thus $A''$ is the orthocenter of $A'CD$. It is sufficient to prove that $AA''$ bisects $BE$ or $A$ is on the line connecting the orthocenters of $A'CD$ and $ABE$. Define $G = AB \cap A'E$ and $F = AE \cap A'B$. By spiral similarity, it is sufficient to prove that $\frac{BC}{BF} = \frac{ED}{EG}$ which is obvious due to $ACF \cup B \sim ADG \cup E$.

Let $M$ and $P$ be the midpoints of $CD$ and $AE$. Then $CB \parallel MP$, and since $M$ and $P$ are midpoints it follows that $CB \parallel MP \parallel DA$. Now let $AO$ intersect $BE$ at $F$. Then since $M$ is the circumcenter of $\triangle ABE$, $\angle AEF = \angle AMP = \angle MAD$, and $\angle FAE = 90^\circ - \angle OAC = \angle ADM$, so $\triangle AEF \sim \triangle DAM$. Similarly, we also have $\triangle ABF \sim \triangle CAM$. Therefore, \[ \frac{BF}{FE} = \frac{BF}{FA} \cdot \frac{FA}{FE} = \frac{AM}{MC} \cdot \frac{MD}{AM} = 1. \]

why just why are all G1s 5 minute solves let $M_1$ be the midpoint of $CD$, let $M_2$ be the intesection of $AO$ and $BE$, and let $P$ be the intersection of $BC$ and $ED$ since $M_1$ is the circumcenter of $AEB$ and $ABP=AEP=90^\circ$, we have that $AM_1P$ are collinear and $M_1$ is the midpoint of $AP$, and we can literally just angle chase to get that $AC$ is parallel to $DE$, and $AD$ is parallel to $BC$ we also angle chase for $OCD=ODC=BAC=EAD$ since $BAD=EAC=90^\circ$ then, $EAM_2=ADM_1$, and we also have that $M_1AD=APC=AEM_2$, so $EAM_2$ and $ADM_1$ are similar also, more angle chasing gets that $ADP$ is similar to $EAB$, and since $M_1$ is the midpoint of $AP$, similarity yields that $M_2$ is the midpoint of $BE$

[asy][asy] size(8cm); pair A = dir(150); pair B = dir(210); pair F = dir(330); pair E = dir(110); pair D = extension(A,A+F-B,E,F); pair C = -D; pair O = circumcenter(A,C,D); draw(unitcircle); draw(A--B--F--E--cycle); draw(A--D--C--cycle); draw(B--E); draw(A--F); draw(A--O); dot("$P$",(0,0),dir(0)); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); dot("$O$",O,dir(-O)); [/asy][/asy] Let $F$ be the intersection point of $ED$ and $BC$ extended so that $ABFE$ is cyclic (since $\angle ABF = \angle AEF = 90^\circ$), and let $P$ be the circumcenter of $(ABFE)$. Then note that since $AF$ and $CD$ bisect each other, we have $ADFC$ is a parallelogram. This means that \[\angle ADE = \angle CFD = \angle BCA\]and combining with the fact that $\angle AED = \angle ABC = 90^\circ$, we have $\triangle AED \sim \triangle ABC$. Also, \[\angle BAO = 90^\circ - \angle OAD = \angle ACD,\qquad \angle EAO = \angle ADC.\]Therefore, \[\frac{AB}{AE}\frac{\sin \angle BAO}{\sin \angle OAE} = \frac{AC}{AD}\frac{\sin \angle ACD}{\sin \angle ADC} = 1\]which implies that $AO$ passes through the midpoint of $BE$.

I think angle addition formula works best, lol $ABE$'s circumcentre being $CD$'s midpoint directly shows $AD\parallel BC$ and $AC\parallel ED$. Now let $\angle ADC=\alpha$, $\angle CAD=\beta$, and say $AO$ and $BE$ meets at point $M$. First we have $\frac{EM}{MB}=\frac{AE\sin\alpha}{AB\sin(\alpha+\beta)}$. Some trig shows $AE=AD\sin\beta$, $AD=AB(\cot\alpha+\cot\beta)$. Thus $AE\sin\alpha=AB(\sin\alpha\cos\beta+\sin\beta\cos\alpha)=AB\sin(\alpha+\beta)$. Now we have $EM=MB$ and the problem is done!

Let $X$ denote the intersection of lines $\overline{DE}$ and $\overline{BC}$ and $O_1$ denote the circumcenter of $\triangle ABE$. [asy][asy] import geometry; size(12cm); pen pri; pri=RGB(24, 105, 174); pen sec; sec=RGB(217, 165, 179); pen tri; tri=RGB(126, 123, 235); pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pair A = dir(150); pair B = dir(210); pair E = dir(100); pair O_1=circumcenter(A,B,E); pair X = 2O_1-A; pair C = extension(B,X,A,2*O_1-E); pair D = A+X-C; pair O =circumcenter(A,C,D); pair M= (B+E)/2; filldraw((path)(A--B--C--D--E--cycle), white+0.1*pri, pri); draw(D--X); draw(C--X); draw(A--O); draw(B--E,dotted); draw(A--D,dashed); draw(A--C,dashed); dot("$A$", A ,dir(172)); dot("$B$", B, dir(190)); dot("$E$", E, dir(82)); dot("$O_1$", O_1, dir(30)); dot("$X$", X, dir(0)); dot("$C$", C, dir(220)); dot("$D$", D, dir(82)); dot("$O$", O, dir(82)); dot("$M$", M, dir(10)); [/asy][/asy] We can start off with the proof of the following claim. Claim : Lines $\overline{DE}$ and $\overline{AC}$ and lines $\overline{AD}$ and $\overline{BC}$ are parallel. Proof : Since $\measuredangle XEA = 90 ^\circ = \measuredangle XBA$, the points $X$ , $B$ , $A$ and $E$ are concyclic. Now, this means $O_1$ is simply the center of $(ABXE)$ and thus, is the midpoint of $AX$. This then means that segments $CD$ and $AX$ bisect each other, so $ADXC$ must be a parallelogram. Thus, the desired claim follows. Now, note that \[\measuredangle ADE = \measuredangle DAC = \measuredangle BCA\]and $\measuredangle ABC = 90^\circ = \measuredangle DEA$, so $\triangle ABC \sim \triangle AED$. Further, since $AD$ and $BC$ are parallel, $AD \perp AB$. Similarly, $AC \perp AE$. Now, note that \[2 \measuredangle OAB = 2(90 + \measuredangle OAD) = 2\measuredangle OAD = \measuredangle AOD = 2\measuredangle ACD\]Thus, $\measuredangle OAB = \measuredangle ACD$. Similarly, it follows that $\measuredangle EAO = \measuredangle CDA$. Now, denote by $M$ the intersection of segments $AO$ and $BE$. We now invoke some trigonometry. By the Ratio Lemma on $\triangle ABE$ we have, \begin{align*} \frac{BM}{ME} &= \frac{ AB}{AE} \cdot \frac{\sin \angle OAB}{\sin \angle EAO}\\ &= \frac{AB}{AE} \cdot \frac{\sin \angle ACD}{\sin \angle CDA}\\ &= \frac{AB}{AE} \cdot \frac{AD}{AC}\\ &= 1 \end{align*}due to the previously established similarity. Thus, $BM=ME$, implying that $M$ is indeed the midpoint of segment $BE$ as desired.

As earlier, $B,E$ are the reflections of $A$ in the $C$- and $D$-midlines of triangle $ACD$, so $\overline{BC}\parallel\overline{AD}$ and $\overline{AC}\parallel\overline{DE}$. Note also that $\overline{AB}$ and $\overline{AE}$ are isogonal in $\angle CAD$. We present an alternative finish: Define $T$ as the foot from $A$ to $\overline{CD}$. Claim: T is the A-Dumpty point of triangle $ABE$. Proof. We show that $\triangle TBA\overset +\sim TAE$. This follows because of cyclic quads $ABCT$ and $ATDE$: \[\measuredangle TBA=\measuredangle TCA=\measuredangle TDE=\measuredangle TAE \]and $\measuredangle TEA=\measuredangle TAB$ by symmetry, so we win.$\qquad\qquad\square$ Above claim implies that $A$-altitude of triangle $ACD$ is a symmedian in triangle $ABE$, so its isogonal $\overline{AO}$ is a median.

Rename $C\rightarrow B, D\rightarrow C, B\rightarrow P,E\rightarrow Q$. Note that $BP$ and $CQ$ meet on the circle at the antipode of $A$, call it $A'$. Thus $A',B,P$ collinear implies $A,C,P'$ collinear, where $P'$ is the antipode of $P$. Define $Q$ similarly. Now rename $P'\rightarrow Q, Q'\rightarrow P$. Then we have rewritten the problem as the following: In triangle $ABC$, side $BC$ has midpoint $M$. The circle with center $M$ passing through $A$ hits lines $AB$ and $AC$ again at $P$ and $Q$, respectively. Let $R$ be the midpoint of $PQ$. Prove that the reflection of $R$ over $M$ is collinear with $A$ and the circumcenter of $ABC$. Now we can finally complex bash. Let $(ABC)$ be the unit circle. Let $X$ be the foot from $M$ to $AB$. Then it is the midpoint of $B$ and the foot from $C$ to $AB$, so it is \[\frac12\left(b+\frac12\left(a+b+c-\frac{ab}{c}\right)\right),\]so \[p=b+\frac12\left(a+b+c-\frac{ab}{c}\right)-a.\]Similarly, \[q=c+\frac12\left(a+b+c-\frac{ac}{b}\right)-a.\]Now I will start being lazy and working mod $a\mathbb{R}$, since our goal is to eventually prove that the reflection thing is in $a\mathbb{R}$ and we will only be using addition and subtraction from now on. So basically, \[r=\frac12\left(b+c-2a+\frac12\left(2a+2b+2c-\frac{ab}{c}-\frac{ac}{b}\right)\right).\]We can ignore the $a$ terms, and also $\frac bc+\frac cb\in\mathbb{R}$. So \[r\sim\frac12\left(b+c+(b+c)\right)=b+c.\]So the reflection of $R$ over $M=\frac12(b+c)$ is indeed $\in a\mathbb{R}$, as desired. $\blacksquare$

Let $A' \equiv BC \cap DE,$ $O'$ be center of $(ABE),$ $M$ be midpoint of $BE$. We have $ACA'D$ is parallelogram. Then $AC \perp AE$. So $\angle{DAE} = 90^{\circ} - \angle{CAD} = 90^{\circ} - \angle{CA'D} = \angle{BAE} - 90^{\circ} = \angle{EO'M}$. Hence $\triangle ADE \sim \triangle EMO'$. From this, we have $\dfrac{AE}{AD} = \dfrac{EM}{EO'} = \dfrac{EM}{AO'}$. But $\angle{AEM} = \angle{AA'B} = \angle{O'AD}$ then $\triangle AEM \sim \triangle DAO'$. Therefore, $\angle{EAM} = \angle{ADO'} = 90^{\circ} + \angle{OAC} = \angle{EAO}$ or $A, M, O$ are collinear

a better complex bush here: Put $A(a), C(c), D(d)$. Let $BC$ and $ED$ intersect at $X$. Then $X(d+c-a)$. We easily compute $B$ and $D$ as just foots of perpendiculars from $A$ to $CX$ and $DX$, getting $B((a-d+ad/c+c)/2)), D((a-c+ac/d+d)/2)$. Now we are done

Let $AO \cap BE = \{N\}$. One can easily conclude that $BC || AD$ and $AC || ED$. Hence, $\angle BAC = x$, $\angle DAC = 90 - x$, $\angle DAE = x$. Also let $\angle ADO = y$. Hence, $\angle EAN = x+y$, $\angle NAB = 90-y$, $\angle ADC = x+y$, $\angle ACD = 90-y$. By similarity and sine theorem: $\frac{AE}{AB} = \frac{AD}{AC} = \frac{\sin(90-y)}{\sin(x)}$ Also by ratio lemma at $\triangle{AEB}$: $\frac{EN}{NB} = \frac{AE}{AB}.\frac{\sin(x+y)}{\sin(90-y)} = \frac{\sin(90-y)}{\sin(x+y)}.\frac{\sin(x+y)}{\sin(90-y)} = 1$ Hence, $EN = NB$, we are done. (Similar solution when $\triangle ADC$ is non-acute.) $\blacksquare$

Use reference triangle $ACD$. Let $M$, $P$, and $Q$ be the midpoints of segments $CD$, $AC$, and $AD$, respectively. Notice that $MA=MB$ and $PA=PB$, so $B$ is the reflection of $A$ over $\overline{MP}$. Similarly, $E$ is the reflection of $A$ over $\overline{MQ}$. Since $\overline{MP}$ and $\overline{MQ}$ are midlines, we have $\overline{CB} \parallel \overline{AD}$ and $\overline{DE} \parallel \overline{AC}$. Thus, we have \[[ABO]=[ABQ]=[ACQ]=\frac{[ACD]}{2}=[ADP]=[AEP]=[AEO].\]Thus, $\overline{AO}$ bisects $\overline{BE}$, as desired. $\blacksquare$

Let $(ABE)$ have center $M$, $(AC)$ have center $X$, $(AD)$ have center $Y$. Since $M,X$ both lie on the perpendicular bisector of $AB$, we know $MX \perp AB$, since $MX$ is parallel to $AD$ (midline), we have $AD \perp AB$, so $\angle CAB = 90 - \angle DAC$. Likewise, $\angle EAD = 90 - \angle DAC$, so $ABC$ and $AED$ are similar. Let the foot from $A$ to $CD$ be $K$. Observe $K$ lies on $(ABC)$ and $(ADE)$. Note that $AO, AK$ are isogonal in $\angle DAC$, and since $\angle EAD = \angle CAB$, they are also isogonal in $\angle EAB$. Thus it suffices to prove $AK$ is the $A$-symmedian of $\triangle BAE$. Since we know $AK$ is the radical axis of $(AC)$ and $(AD)$, it suffices to find two points on the $A$-symmedian that have equal power with respect to those two circles. One of these points is $A$, obviously. We then claim that the symmedian foot on $BE$ has equal power with respect to both circles. To do this, we use linearity of power of a point. Consider the function $f$ from points to reals that outputs the power of the point with respect to $(AC)$ minus the power of the point with respect to $(AD)$. This function is known to be linear. We calculate this $f(B), f(E)$. We know the power of $B$ with respect to $(AC)$ is just $0$, then $AB \perp AY$, so $AB$ is a tangent to $(AD)$, so the power of $B$ with respect to $AD$ is $AB^2$, so $f(B) = -AB^2$. Likewise, $f(E) = AE^2$. Thus, $f$ evaluated at the symmedian foot is just $\frac{f(B)AE^2 + f(E)AB^2}{AE^2 + AB^2} = 0$, as desired. We are done.

Sketch: Insert $A'$ = A-antipode, $ACA'D$ parallelogram, do some angle chase and then sine rule.

Why are G1's so hard... Took me three days to solve this one. Why lengths? Where is angle chasing? Anyways here is a solution with help from proxima1681 to help me find the length chase. Solution: Let $X$, $T$ and $M$ be the midpoints of $\overline{BE}$, $\overline{MT}$ and $\overline{CD}$ respectively. Set $\triangle ACD$ to be the reference triangle. The main claim is as follows. [asy][asy] import olympiad; import geometry; size(10cm); pair A = (-0.64,0.77); pair C = (0.93, -0.36); pair D = (-0.59, -0.81); pair O1 = (A+D)/2; pair O2 = (A+C)/2; pair E = intersectionpoints(line(D,(D+C-A)), circle((A+D)/2, length(O1--A)))[0]; pair B = intersectionpoints(line(C,(C+D-A)), circle(O2, length(O2--C)))[0]; pair M = (C+D)/2; pair O = circumcenter(A,C,D); pair X = extension(A,O,E,B); //(B+E)/2; sorry, diagram not accurate since coords of A,B,C, only up to 2 decimal points pair T = (A+E)/2; draw(A--B--C--D--E--A, blue); draw(M--E, red); draw(M--A, red); draw(M--B, red); draw(O--A, dashed+pink); draw(E--B); draw(A--D, blue); draw(A--C, blue); draw(M--T, deepgreen); dot("$A$", A, dir(A)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$B$", B, dir(B)); dot("$M$", M, dir(M)); dot("$X$", X, dir(X)); dot("$T$", T, dir(T)); dot("$O$", O, dir(O)); [/asy][/asy] Claim: $ED \parallel AC$ and $BC \parallel AD$. Proof: By symmetry, it is enough to show $ED \parallel AC$. Start by noticing that $MT \parallel DE$ since $\measuredangle MTE = \measuredangle DET = 90^\circ$. From here, instead of using midpoint theorem, we'll use complex numbers to finish. Note that $d - e \parallel (c+d)/2 - (e+a)/2$. But, this means \[d - e \parallel \left(\frac{c-a}{2} + \frac{d-e}{2}\right) \parallel c-a\]which is as desired. $\square$ Now, from the above set of parallel lines, we can conclude that $\measuredangle(ED, AD) = \measuredangle(BC,AC)$. Since $\measuredangle DEA = \measuredangle ABC = 90^\circ$, we can easily show that $\triangle EDA \sim \triangle BCA$. Finally, let $\angle EAX = x$ and $\angle XAB = y$. It is not hard to show that $x + y = \angle ADC + \angle ACD = D +C$. From Ratio Lemma and $\triangle EDA \sim \triangle BCA$, we get that \begin{align*} \frac{\sin x}{\sin y} = \frac{\overline{AB}}{\overline{AE}} &= \frac{\overline{AC}}{\overline{AD}} = \frac{\sin C}{\sin D} \\ \implies \sin x \sin D &= \sin C \sin y. \end{align*}Apply product to sum formula. Cancel $\cos(x-C)$ and $\cos(y-D)$ to get \[\cos(x+C) = \cos(y+D).\]Since $0 < x+C, y+D \le 2\pi$, we either have $x + C = y + D$, or $x + C + y +D = 2\pi$. If the second condition holds, then we can conclude that $C+D = \pi/2$. But then observe that in this case, $E \equiv D$ and $B \equiv C$. The result is obviously true for the degenerate case. Taking the next case, we can directly find out that $x = D$ and $y = C$. Here, we can conclude easily. Note that \begin{align*} \measuredangle XAD &= \measuredangle XAE - \measuredangle DAE \\ &= \measuredangle ADC - (90^\circ - \measuredangle EDA) \\ &= \measuredangle ADC + \measuredangle CAD + 90^\circ \\ &= 90^\circ - \measuredangle DCA = \measuredangle OAD \end{align*}which shows $O-X-A$ are collinear as desired. $\blacksquare$

This is a hard G1 but a satisfying one. Let $F =\overline{BC} \cap \overline{DE}$. Then $P$ is the center of $(ADFE)$, i.e. it is the midpoint of $\overline{AF}$, so $ACFD$ is a parallelogram. Let $M$ be the midpoint of $\overline{BE}$ and $G = \overline{BE} \cap \overline{CD}$. Claim: $AMPG$ is cyclic. Proof: Clearly $\angle PMG = 90^\circ$. Now let $H = \overline{AC} \cap (ADFE)$. By Pascal on $AAHEBF$, $\overline{AA} \cap \overline{BE}$ lies on $\overline{CP}$, hence $\angle GAP = 90^\circ$, which proves the result. $\blacksquare$ Then \begin{align*} \measuredangle OAP &= \measuredangle OAD + \measuredangle DAP - 90^\circ - \measuredangle DCA + \measuredangle DAP \\ &= -\measuredangle\left(\overline{FD}, \overline{BE}\right) - \measuredangle\left(\overline{DC}, \overline{CA}\right) \\ &= -\measuredangle\left(\overline{DC}, \overline{BE}\right) = \measuredangle MGP = \measuredangle MAP. \end{align*}So $M$ lies on $\overline{OA}$.

popop614 wrote: Let $ABCDE$ be a convex pentagon such that $\angle ABC = \angle AED = 90^\circ$. Suppose that the midpoint of $CD$ is the circumcenter of triangle $ABE$. Let $O$ be the circumcenter of triangle $ACD$. Prove that line $AO$ passes through the midpoint of segment $BE$. Let $M,N$ be the midpoint of $\overline{CD},\overline{BE}$,respectively . let $P$ be a point on the plane such that $\measuredangle PBA=\measuredangle ABE$ and $\measuredangle PEA=\measuredangle EAB$ Then $A$ is the incenter of $\triangle PBE$, $M$ is the midpoint of arc $BE$, and $\odot (ABE)$ is the chicken-feet circle. then $AM,\odot(ABE),BC,DE$ meet at $P$-excenter (Denote by $Q$) Since $\overline{AM}=\overline{MQ}$ and $\overline{CM}=\overline{MD}$ we know $ACQD$ is a parallelogram.($\therefore AD\perp AB$ and $AC\perp AE$) $\therefore\measuredangle ACB=\measuredangle DQC=\measuredangle EDA$, let this value be $\theta$, so $\begin{cases} \frac{AB}{AE}=\frac{AC\sin\theta}{AD\sin\theta}=\frac{AC}{AD}=\frac{AC}{CQ}\\ \measuredangle ACQ\overset{ACQD}{=}\measuredangle DQC=\measuredangle EQB\overset{\odot{ABCE}}{=}\measuredangle EAB \end{cases}$ $\implies\triangle EAB\overset{-}{\sim}\triangle QCA$ and since $M,N$ are the midpoints of $AQ,BE$ so $\triangle EAB\cup N\overset{-}{\sim}\triangle QCA\cup M$ $\implies \measuredangle BAN=-\measuredangle ACM=\measuredangle MCA=\measuredangle BCA+\measuredangle MCQ=(90^{\circ}-\measuredangle CAB)+\measuredangle MDA$ $\overset{90^{\circ}-\measuredangle ADM=\measuredangle CAO}{=}\measuredangle BAC+\measuredangle CAO=\measuredangle BAO$ $\implies A-N-O$

Claim: Line $\color{blue}{AO}$ passes through the midpoint of segment $\color{blue}{BE}$. Proof: Define $N$ as the midpoint of $\overline{BE}$ and $X= BC \cap DE$ then clearly $A,B,X,E$ lie on a circle with diameter $\overline{AX}$. Also $A,M,X$ are collinear where $M$ is the midpoint of $CD$ as well as the center of $(ABE)$. Now toss the figure on the complex plane with $(ABE)$ as the unit circle. Let $a,b,c,d,e,o,x,m,n$ denote the complex numbers representing $A,B,C,D,E,O,X,M,N$ respectively. Then, $\boxed{m=0}$;$\boxed{n=\frac{b+e}{2}}$;$\boxed{x=-a}$ and $\boxed{d=-c}$. Now we have, $\boxed{c+(-ab)\overline{c}=-a+b}$ since $B-C-X$ and $\boxed{-c+(-ae)(-\overline{c})=-a+e}$ since $E-D-X$. On adding both the equations we get, $\boxed{\overline{c}=\frac{b+e-2a}{a(e-b)}}$. On conjugating we get, $\boxed{c=\frac{a(b+e)-2be}{(b-e)}}$. Now on applying the circumcenter formula one can get, $\boxed{o=\frac{ac(1-c\overline{c})}{(c-a^{2}\overline{c})}}$. Now, $$o-a=\frac{ac(1-c\overline{c})}{(c-a^{2}\overline{c})}-a=\frac{a\overline{c}(a^2-c^2)}{(c-a^{2}\overline{c})}$$Note that $c=a+\frac{2e(a-b)}{(b-e)}$ and $c-a^{2}\overline{c}=\frac{2(a-b)(e-a)}{(b-e)}$. On plugging in and after some obvious calculations we get, $$(o-a)=\frac{2be(2a-(b+e))}{(b-e)^2} \implies \frac{(o-a)}{(2a-(b+e))}=\frac{2be}{(b-e)^2}$$Notice that $\alpha=\frac{2be}{(b-e)^2} \implies \overline{\alpha}=\frac{2/(be)}{(b-e)^2/(be)^2}=\frac{2be}{(b-e)^2}=\alpha$ implying $\alpha$ is real. Thus, $\frac{(o-a)}{(2a-(b+e))}$ is real. Now, $A,N,O$ are collinear $\iff \frac{o-a}{\frac{b+e}{2}-a} \in \mathbb{R} \iff \frac{(o-a)}{(2a-(b+e))} \in\mathbb{R}$ which is indeed true. $\blacksquare$ ($\mathcal{QED}$)

The circumcenter of $\triangle ABE$ is the midpoint of $A$ and $P:=BC\cap DE$, by Thales; and so $ACPD$ is a parallelogram. In other words, $AC\parallel DE$ and $AD\parallel BC$, giving the following problem statement: Quote: Reference $\triangle ACD$ with feet $C'$ and $D'$ and orthocentre $H$. Construct rectangles $ABCC'$ and $AEDD'$ and parallelogram $ABA'E$. Show that $A'$ and $H$ are isogonal in $\angle A$ i.e. $\angle BAE$. But $ACD\sim BAA'\cong EA'A$ by SAS, easy from there. $\blacksquare$ Lesson Learned. Keep it simple, stupid

Let $A'$ be the intersection of $BC$ and $DE$. Then $A'$ is the antipode of $A$ on $(ABE)$. Let $M$ be the midpoint of $CD$, then $M$ also is the midpoint of $AA'$, so $ACA'D$ is a parallelogram. Now $\angle ABE=\angle AA'D$ and $\angle BEA=\angle BA'A=\angle A'AD$, so $\triangle ABE\sim \triangle DA'A$. Let $N$ be the midpoint of $BE$, then from the previous similarity we get that $\triangle ANE\sim \triangle DMA$. Now we angle chase to finish. \begin{align*} \angle EAN& =\angle ADM \\ & =\angle DCA' \\ & =\angle A'CA-\angle ACD \\ & =\angle A'DA-\angle ACD \\ & =180^\circ-\angle ADE-\angle ACD \\ & =\angle EAD+\angle DAO \\ & =\angle EAO \end{align*}This implies that $A, N, O$ are collinear which finishes the proof.

Helped by SilverBlaze_SY and Fibonacci_math. Let $BC \cap ED = R$. Claim: $ABRE$ is cyclic. Proof: As $\angle ABR + \angle AER = 180^{\circ}$ thus we are done. $\blacksquare$ Claim: $ACRD$ is a parallelogram. Proof: As $R$ is the antipode of $A$ and $CM = MD$, thus our claim is true. $\blacksquare$ Let $S$ be $AR \cap (ACD)$. Claim: $\triangle ABE \sim \triangle SDE$. Proof: $\angle SDC = \angle SAC$ (by angles in the same segment) $= \angle SRD$ (by alternate angles) $= \angle ABE$. Hence we go the same for $\angle AEB$ and by $A-A$ criterion we are done! Let $Q$ be the midpoint of $BE$. Claim: $\angle EAQ = \angle CSG$. Proof: From c.p.c.t by similar triangles. Claim: $\angle EAO = \angle CSG$. Proof: $\angle EAO = \angle EAD + \angle DAO = \angle EAD + \angle ODA$ $\angle ACG = 90 - \angle DAO$ Say $\angle DAO = x$ Thus, $\angle CSA = 180^{\circ} - (\angle SCA + \angle CAS) = \angle BAE + x - 90^{\circ}$. $\angle DAB = \angle ABC = 90^{\circ}$ due to parallel lines. $\angle CSA = \angle BAE - \angle DAB + x = \angle EAO$. As both of the angles lie on the same side and are equal, hence $A,Q,O$ are collinear.