$\Delta BAD \sim \Delta AEC \Rightarrow BC=CE$ angle chasing gives $ \angle OAE = 90-\angle ABC= \angle ADM$ where $M$ is $ \gamma \cap AO$ and, $\angle ACO= 90 - \angle ABC$ and we get $ DM \parallel CO$ so $DM$ is the midline and we are done.

My solution: Let $Q$ be the intersection of $BD$ and $AE$. Claim 1:$C$ is the midpoint of $BE$. As $AB=AC$ we get that $\angle ACB=\angle ABC...(1)$. Additionally $\angle QAD=\angle ABQ$ so $\angle ABC-\angle ABQ=\angle ACB-\angle QAD \implies \angle DBC=\angle AEB...(2)$. From $(1),(2)$ we get that $\triangle DBC \sim \triangle ABE$, so $\frac{DC}{AB}=\frac{DC}{AC}=\frac{1}{2}=\frac{BC}{BE} \implies C$ is the midpoint of $BE$. Claim 2:$ABOQ$ is cyclic: This follows directly from $\angle AQB=\angle QBE+\angle QEB=2\angle QEB-\angle AOB$ where $\angle QBE=\angle QEB$ follows from $\triangle DBC \sim \triangle ABE$. Note that $Q$ belongs to the bisector of $BC$ as $QB=QC$ so $Q, O, B$ are collinear. As a result $\angle ABQ=\angle AOQ=\angle DNC$ where $N=AO \cap \gamma$. We conclude that $ND\parallel CO$ and as $D$ is the midpoint of $AC$, $N$ is the midpoint of $AO$ Edit:Latex fixed. Thanks geometry6

Denote $ \angle ABD=x$ ,$\angle ABC=y$ .We will prove that $\angle AMD=x$ ,which finishes the proof . Since $AE $ is tangent to $(ABD)$ , $\angle EAC=x$. Obviosly $ \angle BAC=180^{\circ}-2y$ and $ \angle AOE=2y$.But $O$ is the circumcenter of $ \triangle ABE $.Therefore $\angle BOE =4y-2x$. Thus $\angle AOC=x$ ,which completes the proof ($MD$ is midline in $ \triangle AOC$).

Claim: $B$ is the midpoint of $CE$. Since $\angle DCB = \angle ABC$ and by the tangency condition $\angle CDB = \angle BAE$ so $\triangle DCB \sim \triangle ABE$. Hence $BC/BE = DC/AB = DC/AC = 1/2$. So our claim is proved. Now let $AO \cap (ADB) = X$. It suffices to show $XD \parallel OC$. But $\angle XDA = \angle OAE = 90 - \angle ABC = 90-\angle ACB = \angle OCA$.

Let $(ABD)$ intersects $AO$ at $M$. We prove that $<ADM=<ACO$, which is obviously sufficient. Let $<OAE=<OEA=<ADM=<ABM=x$. Then $<ABC=<ACB=90-x$. Thus $<BAC=2x$. Let $H$ be the projection of $A$ on $BC$. We have to prove that $<HAC=<ACO$, equivalently that $C$ is midpoint of $BE$. Now let $F$ be the midpoint of $AB$. Then $FDCB$ is cyclic, so $<FCD=<FBD=<CAE$, which implies that $CF$ is parralel to $AE$, hence the conclusion.

Who did proposed this problem so I can put the proposer at top of topic.

dangerousliri wrote: Who did proposed this problem so I can put the proposer at top of topic. If I recall correctly, Sam Bealing from the UK.

It's more natural to dilate from $A$ by factor 2. I'll change some labels too. Quote: In triangle $ABC$ with $BC=2AB$ tangent to circumcircle at $B$ meet median $AM$ at $E$. Prove that $O$ - circumcenter of $BME$ lays on $\odot ABC$. $AMC\cong EAB$ by ASA, so $A$ - midpoint of $ME$ and $AO\perp ME$. Then $\angle BAO=\angle B/2=\angle EBO$. $\blacksquare$

Let $M = AO \cup \gamma$. We will prove that $DM$ is middle line in triangle $AOC$. Part 1 - We prove $BC=CE$. Let $X\in (AC$ such that $BX||AE$. Then $\measuredangle AXB = \measuredangle CAE = \measuredangle ABD$, so triangles $ABX$ and $ADB$ are similar. Then $\frac{AB}{AD}=\frac{BX}{BD}$. Secondly, $\measuredangle DBC = \measuredangle B - \measuredangle ABD = \measuredangle C - \measuredangle CAE = \measuredangle AEB = \measuredangle CBX$, so $BC$ bisects $\measuredangle DBX$. Then $\frac{CX}{CD}=\frac{BX}{BD}$. Since $AD=DC$, we get $AB=CX$, so $AC=CX$. Therefor, $AEXB$ is a paralelogram, so $BC=CE$. Part 2 - We prove $MD||OC$. Since $C$ is the midpoint of $BE$ we have $OC \perp BC$. We will conclude with $MD \perp BC$ by showing that $\measuredangle BMD + \measuredangle MBC = \frac{\pi}{2}$. Indeed, $ \measuredangle BMD + \measuredangle MBC = \measuredangle BMD + \measuredangle MBD + \measuredangle DBC = \pi - \measuredangle BDM + \measuredangle DBC = \measuredangle OAB + \measuredangle DBC = \frac{\pi}{2} - \measuredangle AEB + \measuredangle DBC = \frac{\pi}{2}$.

Alternative proof of $C$ being the midpoint of $BE$: Let $(ABC)\cap AE=F$. Then, $\angle{FBC}=\angle{FAC}=\angle{ABD}$, so $BF$ is the $B$-symmedian of triangle $ABC$. Hence $(B,F;C,A)=-1$, and projecting through $A$ onto $BC$ gives $(B,E;C,P_{\infty})=-1$, so $C$ is the midpoint of $BE$.

Bashy, huh Solution : We claim that $C$ is nothing but the midpoint of $BC$. For this, using laws of sines in $\{\triangle ABD,\triangle CBD\},$ we have $$\frac{BC}{AB}=\frac{\sin\angle ABD}{\sin\angle CBD}=\frac{\sin\angle A}{\sin\angle C}$$and that in $\{\triangle ABC,\triangle AEC\},$ combined with above, gives, $$\frac{BC}{CE}=\frac{\sin \angle A}{\sin\angle B}\cdot \frac{\sin\angle CBD}{\sin\angle ABD}=\frac{\sin\angle A}{\sin\angle B}\cdot\frac{\sin \angle C}{\sin\angle A}=1$$proving the desired claim. Now it's easy to see that $\angle DBC=\angle AEC$. With this in hand, let $\overline{AO}\cap \odot ABD=X,$ then we have $$\angle BDX+\angle CBD=\angle BAX+\angle AEB=\angle BAO+\angle AEB=90^\circ-\angle AEB+\angle AEB=90^\circ$$So $\overline{DX} \perp \overline{BC}$ but also $\overline{OC} \perp \overline{BC},$ since $C$ is midpoint of $BE$ so $CO \parallel DX$ finishing the problem. $\blacksquare$

Mr.Mister wrote: $\Delta BAD \sim \Delta AEC \Rightarrow BC=CE$ angle chasing gives $ \angle OAE = 90-\angle ABC= \angle ADM$ where $M$ is $ \gamma \cap AO$ and, $\angle ACO= 90 - \angle ABC$ and we get $ DM \parallel CO$ so $DM$ is the midline and we are done. $\Delta BAD \sim \Delta AEC$ how？I think $\angle BDA, \angle CEA$ do not equal.

@above similarity does not require what you've said; for it, we need $\angle BDA=\angle ACE,\angle CEA=\angle DAB$, which is true.

Let $M$ be the midpoint of $AO$ and let lines $MD$ and $BC$ meet at $N$. First, observe that $MD\parallel OC\perp BE$, so $\angle BND=90^\circ$. We have \[\angle AEB = \angle ACB-\angle CAE = \angle ABC-\angle ABD=\angle CBD\]so that \[\angle BAM= \angle BAO = 90^\circ-\angle AEB = 90^\circ-\angle CBD = 90^\circ-\angle NBD = \angle BDN.\]Hence, $M, B, A, D$ are concyclic and we are done.

Let $F$ be the midpoint of $AB$ and $M = AO \cap \gamma$. $$\angle DBF = \angle DBA = \angle DAE = \angle DCF \implies FC \parallel AE$$Thus $C$ is the midpoint of $BE$ and $OC \perp BE$. $$\angle OCA = \angle 90^\circ - \angle ACB = 90^\circ - \frac{\angle BAC}{2}$$$$\angle MDA = \angle MBA = \angle MAC = \angle OAC = 90^\circ - \frac{\angle BAC}{2}$$Where the last angle equality occurs because $AO$ is isogonal to the $A- \text{Perpendicular}$ and since $\triangle ABC$ is isosceles, hence such angles. $$\implies \angle ADM = \angle ACO \implies DM \parallel CO$$Thus $M$ is the midpoint of $AO$.

As most of the solutions state we just prove first that C is the midpoint of $BE$.Afterwards we let $P=AO\cap\gamma$. Afterwards all it remains to prove that that $MP$ is parallel to $OC$ . We get $\angle PAE=\angle PMA=90-\angle B=90- \angle C=\angle OCA$ and hence we are done.

Nice. Posting for storage - seems lengthy yet is practically identical to the above solutions.

Retracted

SatisfiedMagma wrote: Are these angles directed here? There is no need to worry about configuration issues since the question states that ABC is acute

For some reason, I am really not able to follow anyone's solution. So, just for storage:

Let $M=\gamma \cap AO \ne A$ and $F$ is the midpoint of $AB$. First of all, we show $C$ as the midpoint of $BE$ by proving $\Delta DBC \sim \Delta AEB$ by simple angle chase. Now, $OC \perp BE$. It suffices to show that $MD \parallel OC$. As $F$ is midpoint, $OF \perp AB$ $\implies BOCF$ is cyclic. Again, $FDCB$ is cyclic as it is isosceles trapezium. Here, we go! An angle chase to finish it all: $$ \angle MDB= \angle MAB =\angle ABO = \theta $$Now, we nicely exploit the fact that $\angle FBO= \theta$ as $OA=OB$. Also, $\angle FOB= 90^\circ-\theta$. Using cyclic quadrilaterals, we discover: \[ \angle FDB= \angle FOB= 90^\circ-\theta\]This shows that $\angle MDF=90^\circ$ which shows that $OC \parallel DM$ as $FD \parallel BC$. This suffices as we have shown that $DM$ is mid-line of $\Delta ACO$. $\blacksquare$

By taking a homothety with scale factor $\tfrac{1}{2}$ we can rephrase the problem as follows: Rephrased Problem wrote: In $\triangle ABC$ with $\angle A$ acute, let $D$ be the midpoint of $\overline{AB}$. Suppose that $AD=AC$. Let $E$ be the intersection between $\overline{CD}$ and the tangent to $(ABC)$ at $A$. Prove that the center $O$ of $(ADE)$ lies on $(ABC)$ We can see by simple angle chasing that $\angle EAC=\angle ABC=\angle DBC$ and $$\angle ACE=180^\circ-\angle ACD=180^\circ-\angle ADC=\angle BDC.$$Further, as $BD=AD=AC$, it follows that $\triangle ACE \cong \triangle BDC$. Finally, we have $$\measuredangle AOC=\measuredangle AOD+\measuredangle DOC=2\measuredangle AED+\measuredangle DAE=2\measuredangle DCB+\measuredangle BAE=2\measuredangle DCB+\measuredangle BCA=2\measuredangle DCB+\measuredangle BCD+\measuredangle DCA=\measuredangle DCB+\measuredangle BDC=\measuredangle DBC=\measuredangle ABC,$$hence $ABCO$ is cyclic and $O$ lies on $(ABC)$ as desired. $\blacksquare$

Outline : By angel chasing, we can easily see that $\triangle{BAE} \sim \triangle{CDB}$. So we must have $\frac{AB}{CD}=2=\frac{EB}{CB} \implies C$ is the midpoint of $BE$. Note that we have that OC is perpendicular to BE. Now let $\angle{ABD}=\alpha, \angle{DBC}=\beta$. Note that we must have $\angle{AOC}=\angle{BOC}+\angle{AOB}=180-\alpha-2\beta+ 2 \beta = 180-\alpha = \angle{AXD}$ where $X$ is the intersection of $\gamma, AO$, so the result follows

Claim: $C$ is the midpoint of $\overline{BE}$ Proof: By an easy angle-chasing, We get $\triangle DBC\sim\triangle AEB$. Combine with $AD=DC=\frac{1}{2}AB$, We get $BC=CE$ as disired $\square$ Back to our problem, Note that the problem is equivalent to prove that $\overline{DM}\parallel\overline{OC}$ $(\dag)$ where $M$ is the midpoint of $\overline{AO}$ Consider $$\angle ADM=\angle OAE=90^\circ-\angle ABC=90^\circ-\angle ACB=\angle ACO$$So, $(\dag)$ is true, and we are done.

KICK OUT GEOGEBRA OUT OF YOUR LIFE, I MISSED A TRIVIAL SIMILARITY IN MY PREVIOUS PROBLEM JUST BECAUSE I CAN'T WRITE PROPERLY ON MY PC SCREEN(smh ignore this sentence i still use ggb :yawaymax:)

No figure cuz i stopped using ggb YAY!!

Let $F$ be midpoint of $AB$. $\angle FDC = \angle ADE$ and $\angle DCF = \angle DBF = \angle DAE$ ---> $CDF$ and $ACE$ are similar ---> $CE = BC$. we want to prove $ABDM$ is cyclic. $\angle MDA = \angle OCA = \angle 90 - \angle C = \angle 90 - \angle B = \angle OAE = \angle MBA$ ---> $ABDM$ is cyclic we're Done.

One can also show that $O$, $D$, $B$ and $C$ are concyclic and hence that $\angle ODB = 90^{\circ}$. (Done along the lines of the main solution, but actually not needed to solve the problem.)

Let $F$ be the reflection of $A$ over $B$. By a 2x homothety at $A$, it suffices to show that $O$ lies on $(ACF)$. Note that $AE$ is also tangent to $(ACF)$. Let $\overset{\Huge\frown}{AC} = 2\alpha$ and $\overset{\Huge\frown}{CF} = 2\beta$. Let the perpendicular bisectors of $AB$ and $BF$ intersect $(ACF)$ at $X$ and $Y$, respectively, on arc $\overset{\Huge\frown}{ACF}$. Then $XY = \frac{1}{2}AF = AC$, so $\overset{\Huge\frown}{XY} = 2\alpha$. In particular, $\overset{\Huge\frown}{XF} = 2\alpha + \beta$. It follows that $\angle AXB = 180^\circ - 2\angle XAB = 180^\circ - 2\alpha - \beta$. But $\angle AEB = 180^\circ - \angle EAF - \angle CBA = 90^\circ - \alpha - \frac{1}{2}\beta$. Now we know $\angle AEB$ is acute, $X$ and $E$ lie on the same side of $AB$, $\angle AXB = 2\angle AEB$, and $XA = AB$, which is enough to conclude that $X$ is the circumcenter of $\triangle ABE$, as desired.

Let $ABC$ be an acute triangle with $AB=AC$, let $D$ be the midpoint of the side $AC$, and let $\gamma$ be the circumcircle of the triangle $ABD$. The tangent of $\gamma$ at $A$ crosses the line $BC$ at $E$. Let $O$ be the circumcenter of the triangle $ABE$. Prove that midpoint of the segment $AO$ lies on $\gamma$.($M$ is midpoint of $AO$) Claim: $C$ is midpoint of segment $BE$ Let $A,B,D$ be the reference triangle hence $A(1,0,0),B(0,1,0),D(0,0,1)$ where $c=2b$ (1). Since $D$ is midpoint of $BC$, we have $C(-1,0,2)$. It is well known that the tangent line crosses A has equation: $$b^2z+c^2y=0 \rightarrow^{(1)} z=-4y \\ (2) $$And line $BC$ has equation $z=-2x$ (3) from line formula. Then the intersection will be $x=2y=-\frac{z}{2}$ meaning $E(-2,-1,4)$. Observe that $\frac{E+B}{2}=C$. Hence C is midpoint of segment $BE$. $\square $ Now we will calculate the length of $AE$.Let $(ABC)$ be the reference triangle, Hence $A(1,0,0),B(0,1,0),C(0,1,0)$ Since $C$ is midpoint of $BE$, $E(0,-1,2)$. Then the displacement vector $\overrightarrow{AE}(1,1,-2)$.Putting coordinates to formula: \begin{align}\sum_{cyc} -a^2yz: |AE|= \sqrt{2a^2+c^2} \end{align}where $a=BC,b=AC,c=AB$. Now, let $ABE$ be the reference triangle hence $A(1,0,0),B(0,1,0),E(0,1,0)$.Now, lets look at $AB,BE,AE$: \begin{align*} & AB:c= c' \\ & BE: BE=2BC=2a=a' \\ &AE:\sqrt{2a^2+c^2}=b' \end{align*}Where $a,b,c$ is sides of triangle $ABC$. Since we have proven $C$ is midpoint, $C(0,\frac{1}{2},\frac{1}{2})$. And $D$ is midpoint of $AC$, $D(\frac{1}{2},\frac{1}{4},\frac{1}{4})$. It is well known that $O(a'^2S_A:b'^2S_B:c'^2S_C)$. where $a',b',c'$ are lengths of reference triangle. Using the length relationship we have shown above, we can simplify: $$O(4a^2(c^2-a^2):2a^2+c^2:3c^2)$$For calculational purposes, WLOG $a=1$. Hence $O(4(c^2-1):2+c^2:3c^2)$. Then $M(12c^2-6:2+c^2:3c^2)$ From the circle formula the circle $(ABD)$ has equation $$u=v=0, w=-2-c^2$$The last thing we have to show is $M$ satisfishes the circle equation, and we are done: \begin{align*}-4(2+c^2)(3c^2)-(2+c^2)(12c^2-6)(3c^2)-c^2(12c^2-6)(2+c^2)-(-2-c^2)(3c^2)(16c^2-4) \\ -12c^2-(12c^2-6)(3c^2)-c^2(12c^2-6)+3c^2(16c^2-4) \\ -12-3(12c^2-6)-(12c^2-6)+3(16c^2-4)\\ 0=0\\ \blacksquare \end{align*}[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(30.301792464859595cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -20.070770331125214, xmax = -7.76897786626562, ymin = 3.2091503332731603, ymax = 9.202166044283887; /* image dimensions */ pen ttttff = rgb(0.2,0.2,1.); draw((-15.695823578133023,6.058332269947024)--(-14.183165181898985,7.68444004589862)--(-13.48423867322032,6.261443399799254)--cycle, linewidth(1.2) + ttttff); /* draw figures */ draw((-15.695823578133023,6.058332269947024)--(-14.183165181898985,7.68444004589862), linewidth(1.2) + ttttff); draw((-14.183165181898985,7.68444004589862)--(-13.48423867322032,6.261443399799254), linewidth(1.2) + ttttff); draw((-13.48423867322032,6.261443399799254)--(-15.695823578133023,6.058332269947024), linewidth(1.2) + ttttff); draw(circle((-15.23847727342894,7.149509779702296), 1.1831458068311769), linewidth(1.2)); draw((xmin, -0.41913098521124825*xmin-0.520273730057814)--(xmax, -0.41913098521124825*xmax-0.520273730057814), linewidth(1.2)); /* line */ draw(circle((-14.00751475720401,6.004428369260507), 1.6891691168547733), linewidth(1.2)); draw((-14.00751475720401,6.004428369260507)--(-15.695823578133023,6.058332269947024), linewidth(1.2)); /* dots and labels */ dot((-15.695823578133023,6.058332269947024),linewidth(4.pt)); label("$A$", (-15.669448040135638,6.108996645052543), NE * labelscalefactor); dot((-14.183165181898985,7.68444004589862),linewidth(4.pt)); label("$B$", (-14.135751546350104,7.7844634029695206), NE * labelscalefactor); dot((-13.48423867322032,6.261443399799254),linewidth(4.pt)); label("$C$", (-13.45912074026825,6.315207938334633), NE * labelscalefactor); dot((-14.59003112567667,6.159887834873139),linewidth(4.pt)); label("$D$", (-14.561062338744412,6.212102291693588), NE * labelscalefactor); dot((-12.785312164541656,4.838446753699888),linewidth(4.pt)); label("$E$", (-12.756713522526137,4.891061194105203), NE * labelscalefactor); dot((-14.00751475720401,6.004428369260507),linewidth(4.pt)); label("$O$", (-13.981093076388538,6.057443821732021), NE * labelscalefactor); dot((-14.851669167668517,6.031380319603766),linewidth(4.pt)); label("$M$", (-14.825270558262087,6.083220233392282), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy]

Let $X,M,N$ be the midpoint of $AO,AB,AE$, respectively. We can easily get that $MDCB$ is an isoscele trapezoid. $$\angle BMC=\angle BDC=\angle BAE$$Thus, $MC\parallel AE$ implying that $C$ is the midpoint of $BE$. Consider a homothety center at $A$ sending $\triangle AMN$ to $\triangle ABE$. Since $O$ is the circumcenter of $ABE$, we have that $X,D$ are the circumcenter of $AMN$ and the midpoint of $MN$, respectively, hence $\angle XDM=90^{\circ}$. Moreover, $$\angle MDB=\angle MCB=\angle AEB.$$Lastly, we have $\angle OAB=90^{\circ}-\angle AEB$. We finish this by noticing that $$\angle BAX+\angle BDX=\angle BAX+\angle MDX+\angle MDB=(90^{\circ}-\angle AEB)+90^{\circ}+\angle AEB=180^{\circ}$$giving the desired result.

After angle chasing,use median sine to get C is the midpoint of BE.This finishes the proof.

First note that $\angle BDC = \angle BAC$ since both are equal to half of arc $BD$, so $\triangle BDC \sim \triangle EAB$. This means that $\frac{BC}{DC} = \frac{2BC}{AB} = \frac{EB}{AB}$, so $2BC = EB$ which means that $C$ is the midpoint of $EB$. Now denote the intersection of $AO$ with the circumcircle of $ABD$ by $P$. We must show that $\triangle APD \sim \triangle AOC$. We also know that $\angle ADP = \angle OAE$ since they both are equal to half of arc $AP$, but $\angle OAE = 90 - \angle ABC = \angle ACO$. Thus, $\angle ADP = \angle ACO$, which means that $\triangle APD \sim \triangle AOC \implies \frac{AD}{AC} = \frac{AP}{AO} = \frac{1}{2}$ as desired.

W problem. The key claim is the following. Claim: $\triangle BMC \sim \triangle EAB$ Proof. This follows since $\angle MCB=\angle ACB=\angle ABC=\angle ABE$ from the isosceles triangle and $\angle EAB=\angle BMC$ from the tangency. $\blacksquare$ From this, as $\frac{MC}{AB}=\frac 12$, we find $C$ is the midpoint of $EB$. Now, let $A'$ be the reflection of $A$ over $B$; by homothety it suffices to show $A'AOC$ is cyclic. Then, $\angle BCA'=\angle MBC=\angle AEB$, so $\angle AOB=2\angle BCA'$ from $O$ being the circumcenter. As $OAB$ is isosceles, we find $\angle BOA=90^{\circ} -\angle BCA'$. Thus it suffices to show $OC \perp BE$ which is true since $C$ is the midpoint of $EB$.

Let $F$ be the midpoint of $\overline{AB}.$ Then, we have $\angle FCA = \angle FBD = \angle ABD = \angle EAD = \angle EAC,$ so $\overline{FC} \parallel \overline{AE},$ thus by converse of midpoint theorem, we have that $C$ is the midpoint of $\overline{BE}.$ By taking a homothety of scale factor $2$ centered at $A,$ it suffices to show that $X, C, O, A$ are concyclic, where $X$ denotes the reflection of $A$ about $B$. But, we have $\angle AXC = \angle ABD,$ and $\angle AOC = \angle AOB + \angle BOC = 2\angle AEB + \angle BAE = 2\angle AEB + \angle BAC + \angle CAE = 2 \angle FCB + \angle FCA + \angle BAC = \angle DBC + \angle FCB + \angle FCA + \angle BAC = (\angle FCB + \angle FCA) + (\angle ABC - \angle ABD) + \angle BAC = \angle ACB + \angle ABC + \angle BAC - \angle ABD = 180 - \angle ABD = 180 - \angle AXC,$ so $X, C, O, A$ are cyclic, and we're done.

just sine bash and little angle chasing kill the problem we have median sine for $\triangle ABC$ and for $\triangle ABO$ and if you did angle chasing you know $C$ is midpoint of $BE$ you will obtain result

Wat. :

Here's a trig bash solution. For convenience's sake we now designate $\angle DBA = \alpha$, and $\angle ABC = \theta$. After fiddling around you get the feeling that $C$ might be the midpoint of $BE$. To prove this we use the fact that $\angle CAE = \angle DBA$ by the tangency condition, notice that $\angle AEC = 180 - \angle CAE - \angle ACE = \theta - \alpha$ and thus by the law of sines on $\triangle ACE$ we have \[\dfrac{CE}{\sin(\angle DBA)} = \dfrac{AC}{\sin(\angle AEC)} \Longrightarrow CE = \dfrac{AC\sin(\angle DBA)}{\sin(\angle AEC)} = \dfrac{AC\sin(\alpha)}{\sin(\theta - \alpha)}\]We use trig on $ABC$ and obtain \[BC = 2AC\cos(\theta)\]Thus it suffices to show that \[2\cos(\theta)\sin(\theta - \alpha) = \sin(\alpha)\]Notice that by the product to sum laws it suffices to show \[\sin(2\theta - \alpha) = 2\sin(\alpha)\]However by law of sines on $\triangle ABD$, we obtain $\dfrac{AD}{\sin(\alpha)} = \dfrac{AB}{\sin(2\theta - \alpha)}$ The fact that $D$ is the midpoint of $AC$ immediately proves that $C$ is the midpoint of $BE$. From here one just needs to notice that for homothety reasons it suffices to show $\measuredangle AOC = \measuredangle ABD$, however now that $\measuredangle BCO = 90$ this is a trivial angle chase.