Mathematical and Physical Journal
for High Schools
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# 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