Quadrilateral and perpendculars 2

[ins-]

It is given an inscribed in a circle quadrilateral $ABCD$. E is the intersection point of $AD$ и $BC$. $K$ and $L$ are the feets of the perpendiculars from $E$ to $AC$ and $BD$. $M$ and $N$ are the middles of $AB$ and $CD$. $P$ is the intersection point of $KL$ and $MN$. Prove that $KL \bot MN$ and $KP=LP$.

[Dimoniada]

Change in quadrilateral $ABCD$ points $B \leftrightarrow C$ and look answer on you previous "Quadrilateral and perpendculars" .

[ins-]

Can you be more detailed?

[Dimoniada]

Решение, данное Liouville, универсально. Ведь прямые Гаусса и Обера никуда не денутся, подобие треугольников останится. Вообще, возможна любая перестановка вершин $(ABCD)$.

[ins-]

Thank you for the valuable remark - it is a solution using well known facts:
http://www.artofproblemsolving.com/Foru ... 8&t=434239