The circumcircle to XYZ intersects again the same sides in U,V,W

Lines AU, BV, CW are concurrent or parallel |

This property is still true if M is "at infinity", that is MX, MY, MZ parallel.

This is a direct consequence of the Ceva theorem :

AX, BY, CZ parallel or concurrent ⇔ (XB / XC)(YC / YA)(ZA / ZB) = -1 [1] |

As XYZUVW are concyclic, we get BX.BU = BZ.BW,
that is XB / ZB = WB / UB,

And the similar relations with A and C :
ZA / YA = VA / WA and
YC / XC = UC / VC

Substituting these relations into [1], and rearranging the elements, we get :

(UC / UB)(WB / WA)(VA / VC) = -1

Which proves, from Ceva theorem, that AU, BV, CW are also concurrent or parallel.

Construct (with compass and straightedge) these lines so that AU, BV, CW are also parallel lines.