Isosceles trapezium

[ins-]

It is given trapezium ABCD with base AB and intersection point of the diagonals E. M is the middle of AE, N is the middle of CD. P is the foot of the perpendicular from E to AD. If MP is perpendicular to NP find the angles of the trapezium.

Is it correct problem as is?
What if?
1) trapezium is isocsceles.
2) BC=CD=DA

[Dimoniada]

ins- в сообщении #485854 писал(а):

Is it correct problem ...?

No.

[ins-]

I had a reason to ask this question.

I tried to "discover" a beautiful problem about midline in triangle and I observed the following:

It is given a regular pentagon ABCDE. F is the intersection point of the diagonals AD and BE. M and N are the middles of the segments AE and BF. Prove that if P is the foot of the perpendicular from F to AB then MP and NP are perpendicular.

It is very easy problem and can be solved with inscribed angles or using midline. Based on this information I tried to "discover" a problem about finding angles in trapezium that have not standard values.

[Dimoniada]

Take a square insted of trapezium. Satisfy your condition, you can bild any more trapezium either.
(Извиняюсь за мой корявый английский )