This problem was published in the American Mathematical Monthly as problem E2553, Vol. 82 (1975).

Note that, Simson's line of $A $ wrt $ \Delta BCD $ is perpendicular to $ AA* $ where $ A* $ is the isogonal conjugate of $ A $ wrt $ BCD $. So $ AA* $ is parallel to the Euler line of $ BCD $. So $ A $ lies on the neuberg cubic of $ BCD $. So $ B $ lies on the Neuberg cubic of $ ACD $ and so. So done. You can see Bernard Gilbert's site for reference.

my Classical Solution we get $E$ and $F$ are the foots of $B$ to the line $AC$ and $DC$ Respectively.and $B^*$ is intersection the altitude from $B$ to $DC$ and circumcircle of the quadrilateral $ABCD$.and we get $A^*$ same way to $B^*$ .and $M$ is midpoint of $AB^*$ and $M^*$ is midpoint of $A^*B$ .and $H1$ is orthocenter of the trangle $ADC$ and $H2$ is orthocenter of the triangle $BDC$. lemma- the Simson line of $B$ to the triangle $ACD$ is parallel to $AB^*$ . proof-$<BB^*A=<BCA=<BFE$ and so $EF//AB^*$ . The same way we can prove the simson line of $A$ to the triangle $BCD$ is parallel to $A^*B$ . so easy too see $M,O,H1$ are on a line .so the triangle $AH1B^*$ is isocele.so $<H1AB^*=<HB^*A$ . we see $H1H2//AB$ this means the quadrilateral $H1H2B^*A^*$ is cyclic and we know that $<AA^*B=<AB*B$ so $<H1B^*A=<H2A^*B$ and so $<A^*AB^*=<A^*BB^*$ and $<AB^*H=<H2A^*B$ and so the triangle $BH2A^*$ is isocele and $O,H1,H2,M^*,M$ are on the same line and we are done.

Let $O$ be the circumcentre of $ABCD$ and $H$ be the orthocentre of $BCD$. Let the circumcircle of $ABCD$ be the unit circle and let the complex number of a point be its lower case affix. Let $P$ be the foot of the perpendicular from $A$ to $BD$ and let $Q$ be the midpoint of $AH$. It is well known that the Simson Line of $BCD$ from $A$ is $PQ$. It is well known that $p=\frac{1}{2} (a+b+d-\frac{bd}{a})$. $q=\frac{a+h}{2}=\frac{a+b+c+d}{2}$. So $\frac{p-q}{\overline{p} - \overline{q}} = \frac{\frac{bd}{a}+c}{\frac{a}{bd}+\frac{1}{c}} = \frac{b^2cd^2+abc^2d}{a^2c+abd} = \frac{bcd}{a}$. $\frac{o-h}{\overline{o}-\overline{h}} = \frac{b+c+d}{\frac{1}{b}+\frac{1}{c}+\frac{1}{d}} = \frac{bcd(b+c+d)}{bc+cd+db}$. Thus the Simson line of $BCD$ from $A$ is perpendicular to the Euler line of $BCD$ if and only if $\frac{bcd}{a} = -\frac{bcd(b+c+d)}{bc+cd+db} \Leftrightarrow ab+ac+ad+bc+bd+cd=0$. As this is symmetric wrt $a$, $b$, $c$ and $d$, the Simson line of $ACD$ from $B$ is perpendicular to the Eulder line of $ACD$ if and only if $ab+ac+ad+bc+bd+cd=0$. So we are done.

My solution: Let $ O $ be the center of $ (ABCD) $ . Let $ H_a, H_b $ be the orthocenter of $ \triangle BCD, \triangle ACD $, respectively . Let $ A'=AH_b \cap (O), B'=BH_a \cap (O) $ and $ M $ be the midpoint of $ CD $ . Since $ AH_b=2OM=BH_a $ , so $ AH_bH_aB $ is a parallelogram . Since $ A'B $ is parallel to the Simson line of $ A $ WRT $ \triangle BCD $ , so we get $ A'B \perp OH_a $ and $ \angle OH_aH_b=90-\angle ABA'=\angle H_bAO $ , hence from the well known property of parallelogram we get $ \angle H_aH_bO=\angle OBH_a $ . ... $ (*) $ Since $ \angle OBH_a=90- \angle B'AB $ , so from $ (*) $ we get $ B'A \perp OH_b $ . ie. The Simson line of $ B $ WRT $ \triangle ACD $ is perpendicular to the Euler line $ OH_b $ of $ \triangle ACD $ Q.E.D

Let the isogonal conjugate of $A$ wrt $BCD$ be $A^*$ and the isogonal conjugate of $B$ wrt $ACD$ be $B^*$. If the reflection of $O$ on $BC$ is $O'$ then $AH_aO'O$ and $DH_bOO'$ are parallelograms where $H_a$ is the orthocentre of $BCD$ and $H_b$ is the orthocentre of $ACD$. If $AB^* \cap BA^* = O''$, then we have $O'' \in OO'$ and so $AO''OH_a$ is a parallelogram $\implies O''$ is the reflection of $O'$ in $O \implies O''D \equiv BA^*\parallel OH_b$. Or just neuberg cubic.