# Cool orthocentre thing

Let ABCD be a cyclic quadrilateral that is not a trapezium and is inscribed in a circle with centre O. Suppose its diagonals meet at T. Also let AB and CD produced intersect at P, and let AD and BC produced intersect at Q. Then T is the orthocentre of triangle OPQ.

Proof
This proof actually builds on some of the stuff from some previous posts but I have found some simplifications to the proofs which I will show here.

Let the circumcircles of triangles BCT and DAT meet at X.

∠TXC = ∠TBC (same segment)
∠DBC = ∠DAC (same segment)
∠DAT = ∠DXT (same segment)
So TX is the internal angle bisector of ∠DXC. (The red angles above show what is going on)
Now produce CX to intersect the circle ABCD again at Y.

We have ∠DXC = 2∠TXC
But ∠TXC = ∠TBC (same segment) as before
And ∠DBC = ∠DYX (same segment)
So ∠DXC = 2∠DYX
Since ∠DXC is an exterior angle of triangle DYX, ∠DXC = ∠DYX + ∠XDY = 2∠DYX so ∠XDY = ∠DYX
Hence triangle DYX is isosceles, with YX = DX. So the perpendicular bisector of DY passes through X. But the perpendicular bisector of DY also passes through O as DY is a chord of the circle ABCD. So OX is the perpendicular bisector of DY and since triangle DYX is isosceles, it follows that OX bisects ∠YXD. Hence OX is the external angle bisector of ∠DXC, so OX and XT are perpendicular.
Also note that ∠DOC = 2∠DYX (angle at centre is twice angle at circumference) so ∠DOC = ∠DXC.
Hence DOXC is a cyclic quadrilateral by the converse of the same segment theorem.

Now let us extend TX to a point S beyond X, just to aid with explanation.

Firstly, ∠BXS = ∠BCA (exterior angle of cyclic quad)
Similarly ∠SXA = ∠BDA (exterior angle of cyclic quad)
Since ∠BCA = ∠BDA (same seg), we can write ∠BXA = ∠BXS + ∠SXA = 2∠BCA
But ∠BOA = 2∠BCA (angle at centre is twice angle at circumference)
Thus ∠BOA = ∠BXA so BXOA is cyclic by the converse of the same segment theorem.

Now PC × PD = PB × PA by power of a point. Thus P has equal power with respect to the two pink circles, so it must lie on their radical axis. Hence P lies on OX.

Also, QC × QB = QD × QA (by power of a point), so it has equal power with respect to the two new pink circles. Thus Q lies on their radical axis, XT.
Thus we have shown that OXP and XTQ are straight lines, and that ∠OXT = 90. Thus QT is perpendicular to the line OP.
But we can repeat exactly the same steps, but considering the circumcircles of triangles BTA and DTC, to show that BT is perpendicular to the line OQ.
Thus, since T lies at the intersection of two altitudes, it is the orthocentre of triangle OPQ.