cyclical points on a conic section $P_1, P_2, P_3, P_4$ are four arbitrary points on $xy = 1$. $P_1P_4$ intersects $P_2P_3$ at $D_1$, and similarly define $D_2, D_3$. Prove that $O, D_1, D_2, D_3$ are cyclical, where $O$ is the origin. 
I have never seen a geometry problem asking to prove cyclical points on a conic section. So frankly not sure where to start. I suspect bashing coordinates would work, but that's quite a bit of efforts (and I am not sure how to go from coordinates to proving points are cyclical). 
Is there some shortcut I could use by some geometric properties of $xy=1$?

 A: Here's a proof without coordinates. I will relabel $D_1=P_1P_4\cap P_2P_3$, $D_2=P_2P_4\cap P_1P_3$ and $D_3=P_3P_4\cap P_1P_2$.
Lemma 1. The centre of any rectangular hyperbola $\mathcal H$ through points $A$, $B$, $C$ lies on the nine-point circle of $\triangle ABC$.
Proof. Let $(ABC)$ meet $\mathcal H$ again at $D$, and let $H_A$, $H_B$, $H_C$, $H_D$ be the orthocentres of $\triangle BCD$, $\triangle CDA$, $\triangle DAB$, $\triangle ABC$, which all lie on $\mathcal H$. There is a $180^{\circ}$ rotation mapping $ABCD$ to $H_AH_BH_CH_D$, and the centre of this rotation is both the centre of $\mathcal H$ and the midpoint of $\overline{DH_D}$, so it lies on the nine-point circle of $\triangle ABC$. $\square$
Lemma 2. The incentre $I$ and excentres $I_1, I_2, I_3$ (opposite $D_1$, $D_2$, $D_3$) of $\triangle D_1D_2D_3$ lie on the hyperbola.
Proof. Take a projective transformation sending $P_1P_2P_3P_4$ to a square. Then $D_1$ and $D_3$ go to infinity, so $I_1I_2I_3I$ becomes a rectangle with sides parallel to $P_1P_2P_3P_4$. Then $D_2$ maps to the common centre of rectangle $I_1I_2I_3I$ and square $P_1P_2P_3P_4$, so these eight points lie on a common conic. $\square$
Now $(D_1D_2D_3)$ is the nine-point circle of $\triangle I_1I_2I_3$, so our two lemmas imply the result.
A: The following is definitely not a shortcut, but gives a self-contained reasoning (with the help of computer algebra system). Note that rectangular hyperbolas (the one with axes perpendicular to each other, or of eccentricity $\sqrt{2}$) are well studied due to its applications in other areas such as the study of centers of a given triangle. Examples include Feuerbach, Jerabek, and Kiepert hyperbolas, among others. The argument below is based on coordinates and all the computation is straightforward. A byproduct is that this also proves the Feuerbach's Theorem as a corollary. Note that since all rectangular hyperbolas are similar (being of the same eccentricity), we can use the one with equation $xy=1$ as our model without loss of generality.
Theorem. Let $P_i,i=1,\cdots,4$, be four points on a rectangular hyperbola. Let $A_i=\{P_1,P_{1+i}\},B_i=\{P_1,\cdots,P_4\}\setminus A_i,i=1,2,3$. Denote by $\ell(A_i)$ (resp. $\ell(B_i)$) the line joining the two points in $A_i$ (resp. $B_i$). Let $D_i:=\ell(A_i)\cap\ell(B_i)$. The the four points $D_1,D_2,D_3$ and the center $O$ of the rectangular hyperbola are concyclic.
Proof. As mentioned above, we use $xy=1$ as our model. Let $P_i=(t_i,1/t_i), i=1,\cdots 4.$ Then solving linear equations yields the following coordinates for $D_i=(x_i,y_i),i=1,2,3,$ where $$x_1=\frac{t_1t_2t_3+t_1t_2t_4-t_1t_3t_4-t_2t_3t_4}{t_1t_2-t_3t_4},\qquad y_1=\frac{t_1+t_2-t_3-t_4}{t_1t_2-t_3t_4}$$
$$x_2=\frac{t_1t_2t_3+t_1t_3t_4-t_1t_2t_4-t_2t_3t_4}{t_1t_3-t_2t_4},\qquad y_1=\frac{t_1+t_3-t_2-t_4}{t_1t_3-t_2t_4}$$ and
$$x_3=\frac{t_1t_2t_4+t_1t_3t_4-t_1t_2t_3-t_2t_3t_4}{t_1t_4-t_2t_3},\qquad y_1=\frac{t_1+t_4-t_2-t_3}{t_1t_4-t_2t_3}.$$ Now to prove that four points ($D_1,D_2,D_3,D_0$ with $D_0=(x_0,y_0)$ ) are concyclic, it suffices to check the vanishing of the following determinant (this can be seen by working with an equation for a circle), namely $$\left|\begin{array}{cccc}x_1^2+y_1^2&x_1&y_1&1\\ x_2^2+y_2^2&x_2&y_2&1\\ x_3^2+y_3^2&x_3&y_3&1\\ x_0^2+y_0^2&x_0&y_0&1\end{array}\right|=0,$$ where since $D_0=O=(0,0)$ (the center of the hyperbola), the result is reduced to showing $$\left|\begin{array}{cc}x_1^2+y_1^2&x_1&y_1\\ x_2^2+y_2^2&x_2&y_2\\ x_3^2+y_3^2&x_3&y_3\end{array}\right|=0,$$ which is true as can be checked directly by a computer algebra system (for example, in SAGE, use the factor command: it returns that a zero cannot be factored). QED
Lemma. For any triangle $P_1P_2P_3$ inscribed in a rectangular hyperbola $\mathcal{C}$, the orthocenter $H$ of $P_1P_2P_3$ lies on $\mathcal{C}$.
Proof. Without loss of generality, use the model $xy=1$ and the same parametrizations of $P_i, (i=1,2,3)$ as above. Then by direct computation, the orthocenter is given by $$H=(-1/(t_1t_2t_3),-t_1t_2t_3),$$ which indeed lies on $\mathcal{C}.$ QED
Corollary. (Feuerbach's Theorem) For a triangle $ABC$ inscribed in a rectangular hyperbola $\mathcal{C}$, its nine-point circle passes through the center of the hyperbola.
Proof. Given an inscribed triangle $ABC$ in $\mathcal{C}$, by the above lemma, the orthocenter $H$ of $ABC$ lies on $\mathcal{C}$. Now let $A,B,C,H$ be the four points $P_i$'s as in the Theorem. It is clear that the three points $D_i$'s correspond to the three feet of altitude from $C,B$ and $A$, therefore they lie on the nine-point circle of $ABC$. Since by the Theorem, this circle passes through $O$, the result is clear. QED
Remarks. The following gives some references without details (please look up undefined terms).


*

*In "Projective Geometry" by H.S.M. Coxeter, the author says "If $4$ points in a plane are joined in pairs by $6$ distinct lines, they are called the vertices of a complete quadrangle, and the lines are its $6$ sides." And a result on polarity induced by a conic is the following (See 8.21 of the same reference):
If a quadrangle is inscribed in a conic, its diagonal triangle is self-polar.

*In "Introduction to Plane Geometry" by H.F. Baker (1943), it was mentioned in Ex. 12 on page 158 that "..., if $PQR$ be a self-polar triangle in regard to the rectangular hyperbola, the circumcircle of $PQR$ contains the center $C$" (of the hyperbola). The proof was given (but not mentioned here because it depends on other results).
