CAPS Match, 2025

Denote $\angle ABC = \beta$ and $\angle ACB = \gamma$. The conditions translate as $\angle BAD = 45^\circ - \gamma$ and $\angle EAC = 45^\circ - \beta$. Denote by $G$ the intersection point of $BD$ and $CE$. Clearly $\angle BAG = 90^\circ - \gamma = 2\angle BAD$, and so $AD$ is the angle bisector of $BAG$. Similarly, $AE$ is the angle bisector of $GAC$.
Let $D', E'$ be the midpoints of $AD, AE$, respectively. It is enough to show that the circle through $A, D', E'$ also passes through the midpoint of arc $BAC$. Consider the circumcircle of $AD'E'$ and denote its second intersection points with $AB, AG, AC$ by $P, Q, R$, respectively.

First, we will show that $BP = CR$. Notice that due $\angle ABD$ being right, we have that $D'$ is the circumcenter of $ABD$, and so $D'B = D'A$. Then we get $\angle D'BA = \angle PAD' = \angle D'AQ$, and also $\angle AQD' = \angle BPD'$. Together with $D'B = D'A$, we have that triangles $D'AQ$ and $D'BP$ are congruent, and so $BP = AQ$. Similarly, we can show $CR = AQ$, and so $BP = CR$ as we wanted.

We will now show that the circle through $A, P, R$ passes through the midpoint of arc $BAC$. Denote by $M$ the second intersection of this circle with the circle $ABC$.

Clearly $\angle MBP = \angle MCR$ and $\angle MPA = \angle MRA$, and also $BP = RC$, so triangle $MBP$ and $MCR$ are congruent, giving $MB = MD$, which is enough.

Similarly to the previous solution, we consider homothety with center $A$ and coefficient $1/2$ to obtain points $D', E', M$ and prove that $AD'B, AE'C, BMC$ are isosceles triangles. Moreover, by angle chasing we can get $\angle BMC = \alpha, \angle AD'B = 90^\circ + \gamma$ and $\angle AE'C = 90^\circ + \beta$. Let us notice that the sum of these angles $\angle BMC + \angle AD'B + \angle AE'C = 360^\circ$. We may view these three isosceles triangles as three rotations (for example, triangle $AD'B$ corresponds to the rotation around $D'$ by angle $AD'B$ and sends point $B$ to point $A$). We will call them green, blue, and red.
Because the sum of the three angles is $360^\circ$, the composition of these three rotations is a translation. Moreover, if we follow the image of $C$ we notice that green rotation maps it to $B$, then blue maps it to $A$, and finally red maps it back to $C$. Hence, the translation is actually an identity. Let $M'$ be the image of $M$ under the blue rotation. Then $MD'M'$ is similar to $AD'B$. And because $M$ is the center of the green rotation, the composition of blue and red rotations has to map $M$ back to $M$. Hence, $M'E'M$ has to be similar to $AE'C$. And so $\angle D'ME' = \angle BAD' + \angle CAE' = \alpha/2 = \angle D'AE'$. Thus, $AME'D'$ is cyclic and we are done.

[sketch] Let $G, D', E'$ be as in the original solution. Denote $M$ the midpoint of arc $BAC$. Moreover, let the line $AM$ meet the lines $BD$ and $EC$ at $X$ and $Y$, respectively. Since $AM$ is the external angle bisector, we get $\angle XAB = \angle CAY = 90^\circ - \alpha/2$. Therefore, $\angle GXY = \angle GYX = \alpha/2$, so the triangle $GXY$ is isosceles. Since $AG$ is a diameter of the circumcircle of $ABC$, $GM \perp XY$, thus $M$ is the midpoint $XY$.
We can calculate $\angle XAD = 180^\circ - \alpha/2 - (45^\circ + \gamma/2) = 45^\circ + \beta/2$. Similarly $\angle CAY = 45^\circ + \beta/2$. This gives us that the triangles $DAX$ and $ACY$ are spirally similar. From this spiral similarity we get that also $D'E'M$ is similar to them. So, $\angle D'ME' = \alpha/2$.
We can calculate $\angle D'AE' = 180^\circ - (45^\circ + \beta/2) - (45^\circ + \gamma/2) = \alpha/2$, hence $A, D', E', M$ are concyclic. Homothety with center $A$ and coefficient 2 maps this circle to the desired circle.