Let \(O_1, M\) denote the circumcenter of \(\Delta ABC'\) and its reflection in side \(AB\) define \(O_2,N\) similarly. Clearly, \(M,N \in \perp\) bisector of \(BC.\)Now midpoints of the line joining (\O_1O_2 = P, MN = Q, AB= R, CD= S\) (say) Note that \(PQRS\) is a \(||\) gm so, \(P\) is the reflection of \(Q\) in mdpoint of \(RS\) so, the reflection of a point lying on the \(\perp\) bisector of \(BC\) w.r.t a point on \(\perp\) bisector of \(RS\) lies on \(\perp\) bisector of \(AD\) \(\quad \boxed {Q.E.D}\) ...

RC. wrote: Let \(O_1, M\) denote the circumcenter of \(\Delta ABC'\) and its reflection in side \(AB\) define \(O_2,N\) similarly. Clearly, \(M,N \in \perp\) bisector of \(BC.\). I'll proceed from this. Let $K$, $L$ be foots of perpendiculars from $O_1$, $O_2$ on $AD$. Let $Q$ be midpoint of $BC$, $T$ - midpoint of $AB$. It's clear that $T$ - center of symmetry that maps $O_1KA$ to $MQB$, so $KA=QB=BC/2$. Similarly $LD=BC/2=KA$ and result easily follows.

Lemma. Let $XYZW$ be any quadrilateral (not necessarily convex), then midpoint of $XY, YZ, ZW, WX$ form a parallelogram. Proof. Denote midpoint of point $A$ and $B$ as $M_{AB}$. Now, notice that $M_{XY} M_{XW} \parallel YW \parallel M_{ZY} M_{ZW}$. Similarly, $M_{XY}M_{YZ} \parallel XZ \parallel M_{WY} M_{YZ}$. Now, let $E'$ be the reflection of $E$ wrt $AB$, $F'$ be the reflection of $F$ wrt $CD$. Therefore, if we denote $K$ and $L$ as midpoint of $AB, CD$ respectively, we know that $(E,K,E'), (F,L,F')$ are collinear. Claim. $E'F'$ is the perpendicular bisector of $BC$. Proof. To prove this, notice that $E'$ lies on the perpendicular bisector of $BC$ as $E$ lies on the perpendicular bisector of $BC'$ and this is preserve under reflection. Similarly, $F'$ lies on the perpendicular bisector of $BC$ as $F$ lies on the perpendicular bisector of $CB'$. Let $G$ be the midpoint of $EF$ and $J$ be the midpoint of $E'F'$. Apply our lemma above to quadrilateral $EFF'E'$, then $KGLJ$ is a parallelogram. Therefore, we have $JG$ passes through the midpoint of $KL$ and $KL$ bisects $JG$. Now, let $H$ be the midpoint of $BC$ and $I$ be the midpoint of $AD$. Notice that applying our lemma above for quadrilateral $ABCD$, we get $KIHL$ being a parallelogram. This gives us $HI$ passes through the midpoint of $KL$, and $KL$ bisects $HI$. Combining both result, we get that $JGHI$ is a parallelogram, therefore, $JH \parallel GI$. However, notice that $JH \equiv E'F' \perp BC$, which means that $JH \perp AD$. Therefore, $GI \perp AD$ as well. Since we define $I$ to be the midpoint of $AD$, then $G$ lies on the perpendicular bisector of $AD$.

Let $O,O'$ be circumcenters of $ABC'$ and $CDB'$. Let $AD$ meet $ABC'$ and $CDB'$ at $X,Y$. Let $AC,BD$ meet $ABC'$ and $CDB'$ at $K,L$. Let $E,F$ be projections of $O,O'$ on $AD$. Let $M$ be midpoint of $OO'$. Since $AM^2 = \frac{2AO^2+2AO'^2-OO'^2}{4}$ and $DM^2 = \frac{2DO^2+2DO'^2-OO'^2}{4}$ we need to prove $AO^2 - DO^2 = DO'^2 - AO'^2$ or $AE = DF$ or in fact $AX = DY$. Note that $\angle AXB = \angle AC'B = \angle ACB$ so $AX = BC$ and similarly $DY = BC$ so $AX = DY$ as wanted.