More truths unearthed.

This seems simple enough:

**1. The diagonals of the quadrilateral ABCD meet at X. The circumcentres of triangles ABX, BCX, CDX and DAX are P, Q, R, and S respectively. Then PQRS is a parallelogram.
**However, very interesting things happen when ABCD is cyclic.

**2. ABCD is inscribed in a circle with centre O. The diagonals of PQRS meet at M. Then M is the midpoint of OX.**

(There is an interesting corollary at the end of this post.)

Diagram of 1, with ABCD convex:

(There is an interesting corollary at the end of this post.)

Diagram of 1, with ABCD not convex:

Diagram of 2

**Proof of 1. (the easy part)**

This simple proof works for both convex and non-convex quadrilaterals. All you have to do in the latter case is extend a few lines.

We will use the fact that the circumcentre of a triangle is located at the intersection of its perpendicular bisectors. (*)

Let ∠QPS = t. Let PQ cut BX at J; let PS cut AX at K; let RS cut DX at L.

By (*), ∠PJX, ∠PKX, and ∠SLX = 90

So ∠KXJ = 180 – t (angles in a quadrilateral)

∠KXL = t (angles on a straight line)

∠KSL = 180 – t (angles in a quadrilateral)

So ∠QPS + ∠PSR = 180. Hence PQ and RS are parallel by co-interior/allied angles.

We can repeat to show PS and RQ are parallel. This proves that PQRS is a parallelogram.

**Proof of 2. (the hard part)
**I found this difficult. But when I finally managed to prove it, I found that I had uncovered some really nice geometry en route (in my opinion anyway).

We claim that QOSX is a parallelogram.

Let’s use a stripped down diagram to prove this.

Let OQ cut BX at U, and let OS cut AX at V.

Now of course, we will need to find some handy cyclic quadrilaterals!

Let ∠SVD = t. As OS is the perpendicular bisector of AD, we can use it as a line of symmetry to conclude that ∠AVS = t.

VS is perpendicular to AD, so ∠XAD = 90 – t (angles in a triangle)

∠XSD = 180 – 2t (angle at centre)

But SX = SD (equal radii) so SXD is isosceles.

So ∠SXD = t = ∠SVD

Therefore SVXD is a cyclic quadrilateral by the converse of the same segment theorem.

Similarly, UQCX is cyclic.

Also, ∠COD = 2∠CAD = 180 – 2t

But, using same segment in our new cyclic quadrilateral SVXD, ∠XSD = ∠XVD = 180 – 2t

So ∠COD = ∠CVD – so COVD is cyclic by the converse of same segment.

But we can use an identical argument to show that CUOD is cyclic.

Therefore, as U and V lie on the circumcircle of COD, CUOVD is a cyclic pentagon.

(

*Corollary**: Let t*

*he circles BXC and DAX intersect at X and at Y. You will notice that the circle CUOVD passes through Y. This makes CUYOVD a cyclic hexagon! Furthermore, XY and YO are the internal and external angle bisectors of angle CYD respectively, meaning that ∠OYX is always 90 degrees. Proof of these claims will be included at the end.)*

We will now exploit these cyclic quadrilaterals by using different angles.

Let ∠UQX = w.

∠UQX = ∠UCX = w (same segment in cyclic quad UQCX)

∠UCX = ∠UDV = w (same segment in cyclic quad CUVD)

∠UDV = ∠XSV = w (same segment in cyclic quad SVXD)

So ∠UQX = ∠XSV = w.

Furthermore, ∠QOS = 180 – ∠UCX = 180 – w (opposite angles in cyclic quad CUOV)

So ∠OQX + ∠QOS = 180, so QX and OS are parallel. Also ∠QOS + ∠OSX = 180, so OQ and SX are parallel.

So QOSX is a parallelogram.

Now how does this help us?

We finally use the fact that the diagonals of a parallelogram bisect each other.

M is the midpoint of QS as it is the intersection of the diagonals of parallelogram PQRS.

But QS is a diagonal of QOSX.

Therefore M must bisect the other diagonal of QOSX: OX.

So M is the midpoint of OX!

**Corollary**

Extend CY to meet the circle ABCD again at E. Join DE.

**
1. Y lies on circle CUOVD.
**∠CYD = ∠CYX + ∠XYD

∠CYX = ∠CBX (same segment in cyclic quad CBYX)

∠XYD = ∠XAD (same segment in cyclic quad AYXD)

∠CBX = ∠CED (same segment in cyclic quad CBED)

∠XAD = ∠CED (same segment in cyclic quad CEAD)

So ∠CYD = 2∠CED.

But 2∠CED = ∠COD (angle at centre)

So ∠CYD = ∠COD

Hence CYOD is cyclic by the converse of the same segment theorem.

So Y lies on circle CUOVD.

**2. ∠OYX = 90.**

To do this we will prove that OY and XY are internal and external angle bisectors of angle CYD, respectively.

∠CBX = ∠XAD (same segment in cyclic quad ABCD)

But ∠CBX = ∠CYX and ∠XAD = ∠XYD (same segment)

So ∠CYX = ∠XYD. So XY internally bisects ∠CYD.

Now let’s show OY is the external angle bisector.

Let ∠OYD = m.

∠OVD = 180 – m (opposite angle in cyclic quad OYDV)

∠SVD = m (straight line)

∠AVS = m (by symmetry, as SV is the perpendicular bisector of AD)

Radius SV and chord AD are perpendicular, so by angles in a triangle, ∠CAD = 90 – m.

∠CED = 90 – m (same segment)

Radius OY and chord ED are perpendicular, so by angles in a triangle, ∠OYE = m.

Hence ∠OYD = ∠OYE.

So OY is the external angle bisector of ∠CYD.

Internal and external angle bisectors are perpendicular, so ∠OYX = 90 as desired.

There it is, without all the clutter.

This could be counted as a result in itself:

**The quadrilateral ABCD is inscribed in a circle with centre O. The diagonals of the quadrilateral meet at X. Circles ABX and CDX intersect at X and at Y. Then ∠OYX = 90.**