Here's a sketch (I don't want to go into all the details, but I will if you really want me to ). Try to show that when $B,C$ are fixed, $A$ moves on the hyperbola having $D$ as its center and $B$ as one of its vertices, which has eccentricity $2$. To be precise, $A$ lies on the branch of this hyperbola which doesn't pass through $B$. $E$ also lies on this ellipse, being the "antipode" of $A$. Let $F$ be the symmetric of $E$ wrt $BC$. This means that $F$ is also on the hyperbola, on the other branch (the branch which doesn't contain $A$). From this condition we find that in the triangle $FCB$, if $G$ is the point where the perp bisector of $BC$ cuts $FC$, then $\frac{CG}{GF}=\frac{CB}{CF}$. Let $X$ be the point where the bisector of $\angle FCB$ cuts $FB$. We then get $GX\|BC$. At the same time, from $(*)$, we find $GX$ to be the bisector of $\angle BGF$. Now show that $GX=GB=GC$. What interests us is that $GX=GB$. A quick angle chase gives what we want. I'm sure there's a simpler solution, but I can't see anything right now, sorry.

grobber wrote: Here's a sketch (I don't want to go into all the details, but I will if you really want me to ). Yes, go into the details, please.

A fairly annoying proof: Let X be the point on BC such that XB = BD. Let Y be the midpoint of CD. Let Z be the point such that ZA || BC and ZB = AC. Note that since $\angle ACB = 2 \angle ABC$, WLOG CA = AZ = ZB = 1. Let $\theta = \angle ACB$. We calculate $XC = \dfrac{4}{3} BC = \dfrac{4}{3} (1 + 2 \cos \theta ) = \dfrac{4+8 \cos \theta }{3}$. Then applying the law of cosines, we calculate $AX^2 = CA^2 + CX^2 - 2 \cdot CA \cdot CX \cdot \cos \theta$ $= 1 + \Big( \dfrac{4+8 \cos \theta }{3} \Big)^2 - \dfrac{8 \cos \theta + 16 \cos^2 \theta }{3}$ $= \dfrac{9}{9} + \dfrac{16 + 64 \cos \theta + 64 \cos^2 \theta }{9} - \dfrac{24 \cos \theta - 48 \cos^2 \theta}{9}$ $= \dfrac{25 + 40 \cos \theta + 16 \cos^2 \theta }{9}$ $= \Big( \dfrac{5+4 \cos \theta }{3} \Big)^2$ $\implies AX = \dfrac{5 + 4 \cos \theta }{3} = 1 + \dfrac{2 + 4 \cos \theta }{3} = AZ + \dfrac{XC}{2} = AZ + XD$ Now let AX and ZD intersect at P. Since AX = AZ + XD, it follows that DXP and PAZ are isosceles, or $2 \angle ZDX = 180 - \angle AXD$. Note that Y and D are symmetric within trapezoid AZBC, so $2 \angle AYB = 360 - 2 \angle AYC = 360 - 2 \angle ZDX = 180 + \angle AXD$. Now we're essentially there. Just note that AYD ~ EBD, and AXD ~ ECD, and we get equivalently that $2 \angle EBD = 180 + \angle ECD$ which is another way of saying $2 \angle EBC = 180 + \angle ECB$.

Let $F$ be a midpoint of $BD$. Take $G$ so that $EG || EB$ and $CEGB$ is cyclic. Let $\angle ACB = \gamma$. Now our thesis is equivalent to proving that $EG=EB$ (just an angle chasing). Now by simple angle chase (and some similar triangles + using that $ACEF$ is parallelogram) we obtain following relations: 1) $BG=CE$ 2) $EB=CG$ 3) $FE=b$ 4) $CE=AF$ By Ptolemy we have $a \cdot EG+BG \cdot CE=EB \cdot CG$ or using above relations: 5) $a \cdot EG + CE^2 = EB^2$ I'll prove $a \cdot EB + CE^2 = EB^2$ which will imply thesis: Start with: $CE^2=\frac{4}{9}a^2+b^2-\frac{4}{3}ab \cos \gamma$ (cosine theorem in $CEF$ and above relations) $EB^2=b^2+\frac{1}{9}a^2+\frac{2}{3}ab \cos \gamma$ (cosine theorem in $FEB$ and above relations) so we can rewrite 5) as: $a\sqrt{b^2+\frac{1}{9}a^2+\frac{2}{3}ab \cos \gamma}=b^2+\frac{1}{9}a^2+\frac{2}{3}ab \cos \gamma - (\frac{4}{9}a^2+b^2-\frac{4}{3}ab \cos \gamma)$ or equivalently $\sqrt{b^2+\frac{1}{9}a^2+\frac{2}{3}ab \cos \gamma}=-\frac{1}{3}a+2b\cos \gamma$ Square it and clear out to get: $2a \cos \gamma = b(4 \cos ^2 \gamma -1)$ which can be rewritten as (using sine theorem in $ABC$ and $cos^2 \gamma =1 - \sin ^2 \gamma$): $\sin 3\gamma = \sin \gamma (3 - 4\sin ^2 \gamma)$ which is obviously true. Hence the result P.S. I consider it as quite nice without nasty calculations, but I'd like to see Grobber's solution in full

I like this problem. Thank Megus, probability1.01, grobber for your post. But I think your solution are long. I think we can prove this problem by other way. Every body will try.

Hi Leonhard Euler. You can see some way to done this problem at link: http://steiner.math.nthu.edu.tw/usr2/imo_98/

It's been a while since I solved an IMO problem . Let $ \angle ABC$=x, $ \angle ACB$=2x, BD=y, CD=2y. Extend AB to P and AC to Q so triangle APQ has double the sides of the original triangle, so E lies on PQ because AD=DE. Let the perp from B to PQ be F the perp from C to PQ be D, and the perp from A to BC be R. WLOG, let AR=BF=CD=1. Let EF=m. BCDF is a rectangle, so DF=3y. By similarity, PE=2y. From triangle ABC, we have $ (1/tan x) + (1/tan 2x) = 3y$, so simplification yields $ y = (3 - tan^2x)/(6tanx)$. From triangle PBF, $ tan x = 1/(2y + m)$. Subbing in gives $ y = (3 - (1/(2y + m))^2)/(6/(2y + m))$, and simplifying gives $ 3m^2 + 6my - 1 = 0\implies1/(3y + m) = 2m/(1 - m^2)\implies tanCED = tan(2EBF)\implies tan(BCE) = tan(2(EBC - 90))$.

Is there a nice pure geometric solution to this problem?

Full calculation for sketch by grobber: Let $ V$ be midpoint of $ CD,$ so that $ BD = DV = VC.$ Denote $ a = BC$ and $ \gamma = \angle ACB$ $ \Longrightarrow$ $ \angle ABC = \frac {_1}{^2}\gamma.$ Let circle $ (A)$ with center $ A$ and radius $ AC$ cut $ BC$ again at $ X$ $ \Longrightarrow$ $ \triangle AXC$ is isoceles with $ XA = CA$ $ \Longrightarrow$ external angle of $ \triangle ABX$ is $ \angle AXC = \gamma$ $ \Longrightarrow$ $ \triangle ABX$ is isosceles with $ XA = XB$ $ \Longrightarrow$ $ a = 2CA \cos \gamma + CA$ $ \Longrightarrow$ $ CA = \frac {a}{2 \cos \gamma + 1}.$ Considering $ A, \gamma$ values of variables $ P, \varphi = \angle PCB$ $ \Longrightarrow$ $ CP = \frac {a}{2 \cos \varphi + 1}.$ This is polar equation of a hyperbola branch $ \mathcal H_1$ with focus $ C,$ excentricity $ \varepsilon = 2$ and major axis line $ BC.$ Equation of the other branch $ \mathcal H_2$ is $ CP = \frac {a}{2 \cos \varphi - 1}.$ For $ \varphi = 0,$ $ CV, CB$ satisfy equations of $ \mathcal H_1, \mathcal H_2,$ respectively $ \Longrightarrow$ $ V, B$ are hyperbola vertices $ \Longrightarrow$ midpoint $ D$ of $ BV$ is the hyperbola center. Since $ A \in \mathcal H_1,$ its reflection in the hyperbola center $ D$ is on the other branch, $ E \in \mathcal H_2$ $ \Longrightarrow$ $ CE = \frac {a}{2 \cos \theta - 1},$ where $ \theta = \angle ECB.$ Let circle $ (E)$ with center $ E$ and radius $ EC$ cut $ BC$ again at $ Y$ $ \Longrightarrow$ $ \triangle ECY$ is isosceles with $ CE = YE.$ Using equation of $ \mathcal H_2,$ $ YB = YC - BC = 2 CE \cos \theta - BC = CE = YE$ $ \Longrightarrow$ $ \triangle YBE$ is also isosceles and $ \angle ECB = \angle ECY = \angle EYC = \angle EYB = 180^\circ - 2 \angle EBY = 2 \angle EBC - 180^\circ.$

nttu wrote: Let $ ABC$ be a triangle such that $ \angle ACB = 2\angle ABC$. Let $ D$ be the point on the side $ BC$ such that $ CD = 2BD$. The segment $ AD$ is extended to $ E$ so that $ AD = DE$. Prove that \[ \angle ECB + 180^{\circ } = 2\angle EBC. \] Let M be the midpoint of DC. Another solution is applying Apolonius's theorem for triangles $ ACD, ABM, CDE$ and note that $ c^2=b(a+b)$

For simplicity's sake, assume $AC=b$ and $BC=3$. Reflect $B,C$ in $D$ to construct new points $G,F$. Then $FB=BD=DG=GC=1$. We need to show that $\angle AGF= 90+\angle AFG$. If $H$ is the foot of the altitude from $A$ to $BC$ and $G'$ the reflection of $G$ in $H$, then it will suffice to show that $AF$=$FG'$. Let $T$ the point on $BC$ such that $\angle ABC =\angle TAB$. The $BT=AT=AC=b$, so $FG'=FB+BT+TG'=FB+BT+CG=b+2$. Now $AF^2=FH^2+AC^2-HC^2=b^2+(FH-HC)(FH+HC)=b^2+FC*FT=b^2+4(b+1)=(b+2)^2.$

Proof outlines The geometry seems nice but I don't find a solution using pure geometry.Any ways here are my outlines: First evaluate $AD^2=ED^2$ by cosine rule in $\triangle{ACD}$.Now calculate $CE^2,BE^2$ by appollonius theorem in $\triangle{CDE},\triangle{BDE}$.The rest is annoying but computable.

First, note that quadrilateral $ABEC$ is convex, since its diagonals intersect. Moreover, from $CD > BD$ and symmetry we get $\angle ABE > \angle ACE$; combined with $\angle ACB > \angle ABC$ we conclude $\angle EBC < \angle BCE$. Thusly, \[ 0^{\circ} < 2 \angle EBC - \angle ECB < 360^{\circ}. \] We now use complex numbers to show the above quantity is a multiple of $180^{\circ}$. We let $C = 1$, $A = z$, $B = z^3$ for some $z$ on the unit circle, so $D = \frac{2z^3+1}{3}$ and hence $E = 2D -A = \frac{4z^3-3z+2}{3}$. Thus, \begin{align*} \frac{\left( \frac{C-B}{E-B} \right)^2} {\frac{E-C}{B-C}} &= \frac{(B-C)^3}{(E-B)^2(E-C)} = \frac{(z^3-1)^3}{\left(\frac{z^3-3z+2}{3}\right)^2 \cdot \frac{4z^3-3z-1}{3}} \\ &= \frac{27(z-1)^3(z^2+z+1)^3} {\left[ (z-1)^2(z+2) \right]^2 \left[ (2z+1)^2(z-1) \right]} = \frac{27(z^2+z+1)^3}{(z-1)^2(z+2)^2(2z+1)^2} \\ &= 27 \left( z + \frac 1z + 1 \right)^3 \left( z + \frac 1z - 2 \right)^{-2} \left( 4 + \frac 4z + 5 \right)^{-2} \end{align*}which is real since $z+1/z$ is real. This implies the result.

Here's an incomplete start to a potential synthetic route -- maybe someone better than me can have a go at finishing it... Let $F$ be the point such that $ACBF$ is an isosceles trapezium, and let $EB$ intersect $AF$ at $G$. Let $GD$ intersect $EC$ at $H$. Note that because $\angle ACB = 2\angle ABC$, $CF$ bisects $\angle BCA$ and so $CA$ = $AF$ = $FB$. Taking a homothety at $E$, scale factor $2$, $D$ is transformed to $A$ and so $B$ is transformed to $G$, so $B$ is the midpoint of $EG$. Then $D$ is the centroid of triangle $CEG$ because $DC$ = $2DB$. So $H$ is the midpoint of $CE$, and $H$ therefore gets transformed to $C$ under this homothety. Hence $AC$ is parallel to $GD$ and since $AG$ and $DC$ are parallel, $ACDG$ is a parallelogram. So $\angle BDG = \angle BCA = \angle BFG$ and $BDFG$ is cyclic, with $DG = AC = BF$ so $BDFG$ is an isosceles trapezium. Then let $I$ be the intersection point of $FD$ and $CE$. After projecting line $AFGP_\infty$ ($P_\infty$ being the point at infinity along $AFG$) through $D$ to line $EIHC$, and cancelling lengths, $\frac{FA}{GA} = \frac{EI}{CI}$. [It turns out that actually $EI$ = $FA$ and $GA$ = $CI$ is equivalent to what we want to prove: proving that $CID$ is isosceles is, after a bit of angle chasing, equivalent to the problem statement.]

We must first bound the value of $x=2\angle EBC - \angle ECB$. We clearly have $ABEC$ is convex, so \[2\angle EBC - \angle ECB \leq 2\angle EBC< 360\]Let $A'$be the foot of A to BC A' is always "closer" to C than $D$ since $A'$ is to the left of the midpoint of $BC$ which is to the left of $D$. Thus, $E$ will always be to the right of $M$ and which gives $\angle EBC > \angle ECB$, so: \[2\angle EBC - \angle ECB > \angle EBC > 0\]Thus, we have that if $\angle x$ is determined by a complex number that's real, then x is a multiple of 180 and due to our bounds we will know that it is exactly 180. Set the unit circle to be $(ABC)$ and set our two free variables as $a$ and $c$. Note that the angle condition can be reinterpreted as \[\left(\frac{c}{a}\right)^2 = \frac{a}{b} \Longrightarrow b=\frac{a^3}{c^2}\]Now, due to length condition we have \[d=\frac{c+2b}{3}=\frac{c^3+2a^3}{3c^2}\]Next, \[e=2d-a=\frac{2c^3+4a^3}{3c^2}-\frac{3ac^2}{3c^2} = \frac{2c^3-3ac^2+4a^3}{3c^2}\]Thus, \[\angle EBC := \frac{c-b}{e-b}=\frac{c-\frac{a^3}{c^2}}{\frac{2c^3-3ac^2+4a^3}{3c^2}-\frac{a^3}{c^2}} = \frac{3(c^3-a^3)}{2c^3-3ac^2+a^3} = \frac{3(c^3-a^3)}{(a-c)^2(a+2c))}\]And, \[\angle ECB = \frac{e-c}{b-c}=\frac{\frac{2c^3-3ac^2+4a^3}{3c^2}-c}{\frac{a^3}{c^2}-c} = \frac{-c^3-3ac^2+4a^3}{3(a^3-c^3)} = \frac{(a-c)(2a+c)^2}{3(a^3-c^3)}\] We now consider the complex number associated with $2\angle EBC - \angle ECB=\arg(k)$. This is \[k= \left(\frac{3(c^3-a^3)}{(a-c)^2(a+2c)}\right)^2 \div \frac{(a-c)(2a+c)^2}{3(a^3-c^3)}=\frac{-27\cdot (c^3-a^3)^3}{(a-c)^5 \cdot (a+2c)^2\cdot (2a+c)^2}\]Now, we have \[\overline{k} = \frac{-27\cdot (\left(\frac{1}{c}\right)^3-\left(\frac{1}{a}\right)^3)^3}{(\frac1a-\frac1c)^5 \cdot (\frac1a+2\frac1c)^2\cdot (2\frac1a+\frac1c)^2} \cdot (ac)^9= \frac{-27\cdot (a^3-c^3)^3}{(c-a)^5 \cdot (c+2a)^2\cdot (2c+a)^2}=k\]Thus, $k$ is real and we are done.$\blacksquare$

Let $F$ be the midpoint of $CD$ and $G$ the reflection of $C$ across $D$. Note that $ABEF$ and $AGEC$ are parallelograms. Therefore, it suffices to show that $\angle AGF = 2\angle AFG - 180^\circ$. Let $H$ be the foot of the perpendicular from $A$ to $CD$. Let $I$ and $J$ be reflections of $F$ and $C$ across $H$, respectively. We know that since $\angle ACJ=\angle AJC=2\angle ABJ$, $AC=AJ=BJ$. We have \[AB^2-AC^2=BH^2-CH^2=(BH-CH)(BH+CH)=BJ\cdot BC=AC\cdot BC\]We claim that $AG^2-AF^2=AG\cdot GF$. Once we show this, we have \[AG\cdot GF=AG^2-AF^2=GH^2-FH^2=GF\cdot GI\]which shows $AG=GI$ which implies the angle condition we want. Now, time to actually prove it. Let $CF=FD=DB=BG=x$, $AB=b$, $AC=c$, $AD=d$, $AF=f$, and $AG=g$. We have $b^2-c^2=3cx$. By Appolonius's Theorem, we have that $b^2=\tfrac{d^2+g^2}{2}-x^2$. Also by Appolonius's Theorem we know $d^2=\tfrac{c^2+g^2}{2}-4x^2$. Together we obtain that $b^2=\tfrac{c^2+3g^2}{4}-3x^2$. We have \[\frac{3g^2-3c^2}{4}-3x^2=3cx\]We want to show that $g^2-f^2=3gx$. Similarly, we have $f^2=\tfrac{g^2+3c^2}{4}-3x^2$. Thus, \begin{align*} g^2-f^2 &= \frac{3g^2-3c^2}{4}+3x^2 \\ &= 3cx+3x^2+3x^2 \\ &= 3x(c+2x) \end{align*}So it only remains to show that $g=c+2x$. We know this because $\tfrac{3g^2-3c^2}{4}-3x^2=3cx$. We're done.

We will employ complex numbers. Note that on the arc $BC$ containing $A$, $A$ is 2/3 of the way from $B$ to $C$. This means that $$a^3=bc^2$$$$b=\frac{c^2}{a^3}.$$Now, we have $D=\frac{2}{3}b+\frac{1}{3}c$ and $$e=2d-a=-a+\frac{4}{3}b+\frac{2}{3}c.$$The condition we are trying to show is equivalent to $$\frac{c-b}{c-e}\div (\frac{b-e}{b-c})^2\in R$$$$\frac{(c-b)^3}{(3a-4b+c)(3a-b-2c)^2}\in R.$$At this point we substitute our expression for $b$ to get $$\frac{(c-\frac{a^3}{c^2})^3}{(3a-4\frac{a^3}{c^2}+c)(3a-\frac{a^3}{c^2}-2c)^2}$$$$\frac{(c^3-a^3)^3}{(3ac-4a^3+c^3)(3ac^2-a^3-2c^3)}.$$Note that $$3ac-4a^3+c^3=(a-c)(-4a^2-c^2+4ac)$$$$=-(a-c)(2a-c)^2$$and $$(3ac^2-a^3-2c^3)=(a-c)(-a^2+2c^2+ac)=(a-c)(2c-a)(c+a).$$Of course, ignore all negatives since we are only checking that it is a real number. Cancelling $(a-c)^3$ on the top and bottom, this just becomes $$\frac{(a^2+ac+c^2)^3}{(2a-c)^2(2c-a)^2(c+a)^2}.$$Now, if we take the conjugate of this expression and then multiply by $a^6c^6$, the top doesn't change, and we distribute $a^2c^2$ to each factor in the denominator, then in each factor the $a$ and $c$ will swap. Since it is already symmetric, it will be the same after we take the conjugate, hence done.

Another bashy-bash. As usually, set $(ABC)$ be the unit circle. In addition WLOG, rotate your triangle so that $c=1$. Now, note that the given condition implies, \begin{align*} 2\angle ABC &= \angle ACB\\ \arg\left(\left(\frac{(a-b)}{(c-b)} \right)^2\right) &= \arg\left(\frac{b-c}{a-c}\right)\\ \arg\left(\frac{(b-c)^3}{(a-c)(a-b)^2}\right) &= 0 \end{align*}This implies, $\frac{(b-c)^3}{(a-c)(a-b)^2} \in \mathbb{R}$. So, we must have \begin{align*} \frac{(b-c)^3}{(a-c)(a-b)^2} &= \overline{\left(\frac{(b-c)^3}{(a-c)(a-b)^2} \right)}\\ &= \frac{(c-b)^3a^3b^2c}{b^3c^3(c-a)(b-a)^2}\\ 1 &= \frac{a^3}{bc^2}\\ a^3=b \end{align*}Now, let $D'$ be the midpoint of $DC$. Then, it is clear that $D$ is the midpoint of $BD'$. Thus, we have \begin{align*} \frac{b+d'}{2} &= d\\ \frac{c+d}{2} &= d' \end{align*}Solving this system of equations gives \[d=\frac{2b+c}{3}\]Thus, \[e=2d-a=\frac{4b+2c-3a}{3}\]Now, the required condition can be rewritten as, \begin{align*} 2\angle EBC - \angle ECB &= 180^\circ\\ \arg\left(\left(\frac{c-b}{e-b}\right)^2 \right) - \arg\left(\frac{e-c}{b-c} \right) &= 180^\circ\\ \arg\left( \frac{(b-c)^3}{(e-b)^2(e-c)} \right) &= 180^\circ \end{align*}Thus, we require $ \frac{(b-c)^3}{(e-b)^2(e-c)} \in \mathbb{R}$. We have, \begin{align*} \frac{(b-c)^3}{(e-b)^2(e-c)} &= \frac{(b-1)^3}{\left( \frac{4b+2c-3a-3b}{3} \right)^2 \left( \frac{4b+2c-3a-3}{3} \right)}\\ &= \frac{27(b-1)^3}{(b-3a+2)^2(4b-3a-1)}\\ &= \frac{27(a^3-1)^3}{(a^3-3a+2)^2(4a^3-3a-1)}\\ &= \frac{27(a^3-1)^3}{(a-1)^5(a+2)^2(2a+1)^2} \end{align*}and \begin{align*} \overline{\left(\frac{(b-c)^3}{(e-b)^2(e-c)} \right)} &= \frac{27\left(\frac{1}{a^3}-1 \right)^3}{\left(\frac{1}{a^3} -\frac{3}{a} +2 \right)\left( \frac{4}{a^3} - \frac{3}{a}-1\right)}\\ &= \frac{27(a^3-1)^3}{(2a^3-3a^2+1)^2(a^3+3a^2-4)}\\ &= \frac{27(a^3-1)^3}{(a-1)^5(a+2)^2(2a+1)^2} \end{align*}Thus, $\frac{(b-c)^3}{(e-b)^2(e-c)} = \overline{\left(\frac{(b-c)^3}{(e-b)^2(e-c)} \right)}$ which implies that $\frac{(b-c)^3}{(e-b)^2(e-c)} \in \mathbb{R}$, which means that we have \[\angle ECB +180^\circ = 2\angle EBC\]which was the desired conclusion.

man what the heck is the old geo shortlist :sob: We present a purely synthetic solution. Let's define a lot of points: Let $M$ be the midpoint of $CD$, hence $CM=MD=DB$. Let $E'$ be the reflection of $E$ over line $BC$. Let $T$ be the reflection of $A$ across the perpendicular bisector of $BC$. Let $P = AB \cap CT$. Since $AMEB$ is a parallelogram it follows from reflecting that $TD \parallel BE'$, and moreover what we want to show is equivalent to proving that the acute angle between $AM$ and $TD$ is the same as the acute angle between $E'C$ and $BC$. It would thus be sufficient to prove that $\measuredangle (AM, CE') = \measuredangle (TD, BC)$, since then we get a cyclic quadrilateral to finish From symmetry now it suffices to prove that the triangle formed by $CE'$, $AM$, and $BC$ is isosceles. Take a reflection across the perpendicular bisector of $BC$, and using even more parallelograms it suffices to show that if $N$ is the midpoint of $AD$, the triangle formed by $BN$, $TD$, $BD$ is isosceles. We take a homothety at $D$ with scale factor 2. Let $F$ be the reflection of $D$ over $B$ and let $X = AF \cap TD$. It suffices now to show that $FXD$ is isosceles. We apply phantom points. Let the circle with diameter $PT$ intersect the circle centered at $A$ with radius $AT$ at $X'$. A homothety at $P$ reveals that $PX'$ intersects $BC$ at a point $C'$ such that $BC' = BC$. Now since \[ \measuredangle CX'P = \measuredangle CX'T - 90^\circ = \measuredangle CTA = \measuredangle CBP, \]$X'$ lies on $(CBP)$. Morover since it is easy to see that $AC$ is tangent to $(CBP)$, it follows that $AX'$ is also tangent to $(CBP)$. Let $AX' \cap BC = F'$. Perspectivity at $X'$ reveals that \[ (C, F'; B, C') = -1 \]which immediately implies that $3BF' = BC$. It is also not hard to angle chase the fact that $BX'$ bisects $AT$ since the midpoint of $AT$ lies on $(PTX')$. Therefore, it follows that $TX'$ intersects $BC$ at a point $D'$ such that $2BD' = CD'$, and hence $X' = X$. Since $AX = AT$, by homothety at $X$ we are done.

Lemma: The locus of points $A$ such that $\angle ABC$ is half of $\angle ACB$ is a branch of the hyperbola of eccentricity $2$ going through $B$ with foci at $C$ and the point $F$ such that $FB$ is one fourth of $FC$. Proof: This follows by complex numbers. Anyways, applying this lemma we get the equivalent problem. Equivalent wrote: Let $F_1$ and $F_2$ be points with midpoint $O$. Let $B_1$ and $B_2$ be the midpoints of $F_1O$ and $F_2O$ respectively. Let $\mathcal{H}$ be the hyperbola with foci $F_i$ through $B_i$. Then if $A_1, A_2$ are taken on $\mathcal{H}$ collinear with $F_1$, then $\angle A_1MA_2$ is a right angle. Note that $A_1M$ and $A_2M$ both are degree $2$, so it remains to find five cases by MMP and rotation lemma to show that they are the same point. We can take $A$ as $B_i$ for two cases. For the asymptote case, we get a $30-60-90$ triangle by asymptotes which finishes. Then by symmetry this gives another $2$ so we are done mhm

The complex solution is so clean? Rename $(A, C, D, B) \to (P, O, X, Y)$ and select $W$ and $Z$ so that $\overline{OWXYZ}$ are collinear and equally spaced in this order. Then if we reflect triangle $EBC$ through $D$ we get triangle $PWZ$. The problem rewrites as the following: Problem: Let points $O$, $W$, $X$, $Y$, and $Z$ be collinear and equally spaced in this order. Show that for any point $P$ not on this line, \[ \angle POY = 2 \angle PYO \implies \angle PZW + 180^{\circ} = 2 \angle PWZ. \][asy][asy] unitsize(1cm); real r = 1.4; real t = 1; real y = sqrt(3)*sqrt(t^2-4r*t+3r^2); pair O = (0r, 0); pair W = (1r, 0); pair X = (2r, 0); pair Y = (3r, 0); pair Z = (4r, 0); pair P = (t, y); draw(P -- O -- Z -- cycle ^^ W -- P -- Y); dot("$P$", P, N); dot("$O$", O, S); dot("$W$", W, S); dot("$X$", X, S); dot("$Y$", Y, S); dot("$Z$", Z, S); [/asy][/asy] Proof. We use complex numbers with $o = 0$, $w = 1$, and so on. The angle conditions, mod $180^{\circ}$, rewrite as \begin{align*} \measuredangle POY + 2 \measuredangle PYO = 0 &\iff (p-0) (p-3)^2 = p^3 - 6p^2 + 9p \in {\mathbb R}, \\ \measuredangle PZW + 2 \measuredangle PWZ = 0 &\iff (p-4) (p-1)^2 = p^3 - 6p^2 + 9p - 4 \in {\mathbb R}. \end{align*}Clearly these are equivalent, and furthermore $\angle PWZ > \angle PZW$, so $2 \angle PWZ - \angle PZW = 180^{\circ}$. $\blacksquare$