On Miquel quadrilateral theorem

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Let quadrilateral $ABCD$ is inscribed in circle $k(O)$. $AB$ and $CD$ intersects at the point $E$. $AD$ and $BC$ intersects at the point $F$. If $k_1(O_1)$, $k_2(O_2)$, $k_3(O_3)$, $k_4(O_4)$ are the circumcirles of the triangles $ABF, CDF, ADE, BCE$. Prove that:
a) $k_1, k_2, k_3, k_4$ intersects at the point $P$ lying on $EF$. (Miquel quadrilateral theorem).
b) $O$, $O_1$, $O_2$, $O_3$, $O_4$ and $P$ are concyclic.
c) $OP \perp EF$.

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I hope it is a beautiful statement and not very well known and would like to see a beautiful solution.

Note that the second intersection point of the circle through O, O1, O2, O3, O4, P with EF is the intersection point of the line through O, parallel to the line through the middles of the diagonals with EF. So we have seven points lying on the same circle.

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(Оффтоп)

If someone is interested - it is the beginning of the solution:
http://www.math10.com/f/viewtopic.php?f ... 460#p49460

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http://www.math10.com/f/viewtopic.php?f=49&t=10665 the full solution :) enjoy
if it is not known problem it is good to be named MK theorem ;)
(MK - stands for - Mirchev-Kazakov-Karakehaiov)

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I'm sorry for loosing your time. Almost all of these facts are well known.

[Ilya Nek]

http://olympiads.mccme.ru/lktg/2010/1/index.htm - very nice!!!