Ceva and beyond...

Given a triangle ABC and a variable point M. Lines MA,MB,MC intersect the lines of opposite sides in X,Y,Z.
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.

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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.


Considr now the case with AX, BY, CZ being parallel.
Construct (with compass and straightedge) these lines so that AU, BV, CW are also parallel lines.

Hint    Solution



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