Problem B. 4673. (December 2014)

B. 4673. \(\displaystyle E\) is the intersection of the diagonals of a cyclic quadrilateral \(\displaystyle ABCD\), and \(\displaystyle K\) is the centre of the circumscribed circle. The intersection of the lines of sides \(\displaystyle AB\) and \(\displaystyle CD\) is \(\displaystyle F\), and the intersection of the lines of sides \(\displaystyle BC\) and \(\displaystyle DA\) is \(\displaystyle G\). The second intersection of the circumscribed circles of triangles \(\displaystyle BFC\) and \(\displaystyle CGD\) is \(\displaystyle H\). Prove that the points \(\displaystyle K\), \(\displaystyle E\) and \(\displaystyle H\) are collinear.

Suggested by Sz. Miklós, Herceghalom

(4 pont)

Deadline expired on January 12, 2015.

Statistics:

21 students sent a solution.
4 points:	Gál Boglárka, Geng Máté, Imolay András, Keresztfalvi Bálint, Khayouti Sára, Kocsis Júlia, Nagy-György Pál, Németh 123 Balázs, Szebellédi Márton, Szőke Tamás, Vághy Mihály, Williams Kada.
3 points:	Gyulai-Nagy Szuzina, Heinc Emília.
2 points:	2 students.
0 point:	5 students.

Problems in Mathematics of KöMaL, December 2014