Trivial by Power of point

Let $P$ be the intersection of the tangents to $\odot(O)$ at $A$ and $B$. Also let $X$ and $Y$ denote the circumcenters of $\triangle{AOC}$ and $\triangle{BOD}$ respectively. Define $X'$ as the reflection of $O$ over $AC$ and $P'$ as the projection of $O$ on $AC$. Observe that \begin{align*} XB\cdot XC &=OX^2-R^2 \\ &=OX^2-OP\cdot OP' \\ &=OX^2-OX\cdot OX' \\ &=XX'\cdot XO. \end{align*}Thus we have $X'\in\odot{(BOC)}\implies Y\in AC$ since $AC$ is the perpendicular bisector of $OX'$.

Use coordinate geometry

Let $X$ be circumcentre of $AOC$ which lies on $BD$. Let $Y$ be circumcentre of $BOD$. Let $\mathbb{H}$ be the homothethy with centre $O$ and ratio $2$. The image of $Y$ is the pole of $BD$ which lies on polar of $X$. Hence it's enough to show that image of $AC$ under $\mathbb{H}$ is the polar of $X$ which is well known. Edit: just noted it's straightforward by complex numbers...Take the circumcircle le as unit circle.The condition is equivalent to $(ac+bd)=(a+c)(b+d)$ hence the symmetry

A brute-force approach, because why not? Let the circle be the unit circle centred at $(0,0)$. Let $A = (x_a,y_a), \cdots, D=(x_d,y_d)$. The circumcentre of $\Delta AOC$ is the intersection of the perpendicular bisectors of lines $AO$ and $CO$. Now the line $AO$ has equation $y_ax = x_ay$, so the perpendicular bisector has the form $x_ax+y_ay = k$, and it passes through $\left ( \frac{x_a}{2}, \frac{y_a}{2} \right )$, so $k = \frac{x_a^2 + y_a^2}{2} = \frac{1}{2}$. Thus the perpendicular bisector of $AO$ has equation $$2x_ax + 2y_ay - 1 = 0$$Similarly, the perpendicular bisector of $CO$ has equation $$2x_cx+2y_cy-1 = 0$$The equation of line $BD$ is given by $$(y_b - y_d)(x-x_b) = (x_b - x_d)(y-y_b)$$which rearranges to $$(y_d - y_b)x + (x_b - x_d)y + (x_dy_b - x_by_d) = 0 $$Call $ (x_dy_b - x_by_d)$ by $f_{db}$. The three lines intersect at a common point iff $$\left | \begin{array}{ccc} 2x_a & 2y_a & -1 \\ 2x_c & 2y_c & -1 \\ y_d - y_b & x_b - x_d & f_{db} \\ \end{array} \right | = 0 $$ or $$\left | \begin{array}{ccc} x_a-x_c & y_a - y_c & 0 \\ 2x_c & 2y_c & -1 \\ y_d - y_b + 2x_cf_{db} & x_b - x_d + 2y_cf_{db} &0 \\ \end{array} \right | = 0 $$ or $$(x_b - x_d + 2y_cf_{db})(x_a - x_c) - (y_d - y_b + 2x_cf_{db})(y_a-y_c) = 0$$ or $$(x_b - x_d)(x_a - x_c) + (y_b - y_d)(y_a - y_c) - 2f_{ca}f_{db} = 0$$ which has the symmetry $a \leftrightarrow b$, $c \leftrightarrow d$. Remember that the iff in bold allows us to finish the problem here.

Let $O'$ be the reflection of $O$ in $BD$. Since, the center of $\odot(AOC)$ lies on $BD$, $O'$ lies on $\odot(AOC)$. Invert around $O$, $O'$ gets sent to the centre of $\odot(BOD)$ and $\odot(AOC)$ gets sent to $AC$. So, the centre of $\odot(BOC)$ lies on $AC$.

Set $(ABCD)$ as the unit circle. Then if $P$ is the circumcenter of $AOC$, we get that $p = \frac{ac(\overline{a}-\overline{c})}{\overline{a}c - \overline{c}a}$, which simplifies to $p = \frac{ac}{a+c}$ The condition for $D,B,P$ to be collinear is that $\frac{d-b}{d-p}$ is real, which simplifies to $bd-bc-ab=ad+cd-ac$ which is just $ac+bd=(a+c)(b+d)$ Similarly, if $Q$ is the circumcenter of $BOD$, we get that the condition for $A,C,Q$ to be collinear is also $ac+bd=(a+c)(b+d)$ Therefore, if $B,D,P$ are collinear, so are $A,C,Q$, as required

Straightforward application of $Complex$ $ Bashing$... Its really easy !!

L567 wrote: $\frac{d-b}{d-p}$ is real, which simplifies to $bd-bc-ab=ad+cd-ac$ can someone explain how this works?