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Post Posted: Jun 07, 2008, 9:47 am • # 1 


Let ABC be a triangle with \measuredangle{BAC} < \measuredangle{ACB}. Let D, E be points on the sides AC and AB, such that the angles ACB and BED are congruent. If F lies in the interior of the quadrilateral BCDE such that the circumcircle of triangle BCF is tangent to the circumcircle of DEF and the circumcircle of BEF is tangent to the circumcircle of CDF, prove that the points A, C, E, F are concyclic.

Author: Cosmin Pohoata

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Cosmin Pohoata, Bucharest, Romania
 
 
Post Posted: Jun 07, 2008, 2:55 pm • # 2 


pohoatza wrote:
Let ABC be a triangle with \measuredangle{BAC} < \measuredangle{ACB}. Let D, E be points on the sides AC and AB, such that the angles ACB and BED are congruent. If F lies in the interior of the quadrilateral BCDE such that the circumcircle of triangle BCF is tangent to the circumcircle of DEF and the circumcircle of BEF is tangent to the circumcircle of CDF, prove that the points A, C, E, F are concyclic.

Solution
 
 
Post Posted: Jun 07, 2008, 11:09 pm • # 3 


Could anyone solve it without using "inversion"?
 
 
Post Posted: Jun 08, 2008, 7:26 am • # 4 


From fact that circumcircle of triangle BCF is tangent to the circumcircle of triangle DEF, we get \angle DFE=\angle FCD+\angle FBE and similarly \angle EFB=\angle FDE+\angle FCB. By adding,
\angle DFB=(\angle FCD+\angle FCB)+\angle FBE+\angle FDE=\angle DCB+\angle FBE+\angle FDE=\angle DEB+\angle FBE+\angle FDE. Hence D,F,B are collinear. So \angle DFE=\angle FEB+\angle FBE. Comparing with \angle DFE=\angle FCD+\angle FBE, we get \angle FEB=\angle FCD which imply A,C,E,F are concyclic.
 
 
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