Let ABC be a triangle.
The A-excircle of triangle ABC has center Oa and touches the side BC at the point Aa.
The B-excircle of triangle ABC touches its sidelines AB and BC at the points Cb and Ab.
The C-excircle of triangle ABC touches its sidelines BC and CA at the points Ac and Bc.
The lines CbAb and AcBc intersect each other at some point X.
Prove that the quadrilateral AOaAaX is a parallelogram.
Remark. The A-excircle of a triangle ABC is defined as the circle which touches the segment BC and the extensions of the segments CA and AB beyound the points C and B, respectively. The center of this circle is the point of intersection of the interior angle bisector of the angle CAB and the exterior angle bisectors of the angles ABC and BCA.
Similarly, the B-excircle and the C-excircle of triangle ABC are defined.
[hide="Source of the problem"]Source of the problem: Theorem (88) in: John Sturgeon Mackay, The Triangle and its Six Scribed Circles, Proceedings of the Edinburgh Mathematical Society 1 (1883), pages 4-128 and drawings at the end of the volume. geometryparallelogramratioangle bisectorexterior anglegeometry proposed