A Cosine Law solution by Mouroukos here (I shall translate it and copy paste it when I use my pc) Any geometric solution is welcome for (b)

Cute problem! Solution for $b)$. Let $\omega$ denote the circumcircle of $\odot (ABC)$. Let $X$ be the intersection point of $\omega$ with the perpendicular bisector of segment $BC$ and with $\triangle XBC$ acute. Let $l$ be the tangent line to $\omega$ that contains point $C$. Let $Y$ be the projection of $X$ to $l$. I will prove $A = X$. So, let's say $A \neq X$. By $a)$, we get that $CY = \frac{BC}{2}$. But, $CD = \frac{BC}{2}$, so point $D$ is point $Y$ or the simmetric of point $Y$ wrt $C$. Case $I$. We have $D = Y$ From this, we have that $X$, $A$ and $Y$ are on the same line. So, point $A$ is the intersection point of $\omega$ with $XY$. Now, since $Y$ is on the opposite side of $XC$ as $B$ and $Y$ is in the exterior of $\omega$ (if it is in the interior of the circle, then $l$ cannot be tangent to $\omega$), so $A$ is on the segment $XY$. But this means $A$ is on the arc $XC$, so $\angle ABC < \angle ACB$, thus $AB < AC$, contradiction. So, we must have $A = X$, from where $AB = AC$, what we wanted to prove. Case $II$. Point $D$ is the simmetric of $Y$ wrt $C$. Let $Z$ be the simetric of $Y$ wrt $C$ and let $P$ be on $\omega$ such that $XP$ is a diameter. Let $O$ be the center of $\omega$. We have $O$ is the midpoint of $PX$. Since $OC || XY$ (are perpendicular to $l$) and $C$ is the midpoint of $YZ$ and $O$ the midpoint of $XP$, we get $PZ || OC$. We need to prove that $D \neq Z$. So, let's say $D = Z$. Thus $A$ is $P$ or the second intersection of $PZ$ with $\omega$. Since $\angle BPC = 180 - \angle BXC > 90$, $A$ cannot be $P$. We have $2$ cases. $1.$ Point $A$ is on arc $BC$. We have $\angle BAC = \angle BPC > 90$, contradiction $2.$ Point $A$ is not on arc $BC$. Let $C'$ be the simmetric of $C$ wrt $O$. Since $l$ is tangent to $\omega$ and $Y, C, Z$ lie on $l$ in this order, then $X, C', A, P$ must lie on $\omega$ on this order. Thus, $A$ is on arc $C'P$. Since $A$ is not on arc $BC$, then $A$ is on arc $C'B$. So, $\angle ABC = \angle ABC' + \angle C'BC = \angle ABC' + 90 > 90$, contradiction. The diagrams are for case $I$ and for case $II$ (subcase $2$).

@above, I do not see why Y and D cannot be symmetric wrt C since CD=CY gives 2 cases for point Y

parmenides51 wrote: @above, I do not see why Y and D cannot be symmetric wrt C since CD=CY gives 2 cases for point Y Sorry for not including a proof (I was in a little bit of a hurry). I will edit the above post to include this. Here is the lost part: I will use the same notations as I did above. Let $Z$ be the simetric of $Y$ wrt $C$ and let $P$ be on $\omega$ such that $XP$ is a diameter. Let $O$ be the center of $\omega$. We have $O$ is the midpoint of $PX$. Since $OC || XY$ (are perpendicular to $l$) and $C$ is the midpoint of $YZ$ and $O$ the midpoint of $XP$, we get $PZ || OC$. We need to prove that $D \neq Z$. So, let's say $D = Z$. Thus $A$ is $P$ or the second intersection of $PZ$ with $\omega$. Since $\angle BPC = 180 - \angle BXC > 90$, $A$ cannot be $P$. We have $2$ cases. $1.$ Point $A$ is on arc $BC$. We have $\angle BAC = \angle BPC > 90$, contradiction $2.$ Point $A$ is not on arc $BC$. Let $C'$ be the simmetric of $C$ wrt $O$. Since $l$ is tangent to $\omega$ and $Y, C, Z$ lie on $l$ in this order, then $X, C', A, P$ must lie on $\omega$ on this order. Thus, $A$ is on arc $C'P$. Since $A$ is not on arc $BC$, then $A$ is on arc $C'B$. So, $\angle ABC = \angle ABC' + \angle C'BC = \angle ABC' + 90 > 90$, contradiction. Here is the diagram for case $2$.