The problem is proposed by me, Gazeta Matematica, 10/1998, E:11637.

To Dinu Teodorescu: Congratulation! It is such nice classic problem.

I am very sorry! The initial statement of problem was wrong. The first part of statement is "Let ABC be a triangle with angle BAC

Thank you so much for your kindly edited the problem.

To Dinu Teodorescu: Humans make mistakes - there is nothing unusual about it, and researchers are humans.

Indication : Let T be a point on AB so that MT is orthogonal to AB. Clearly MT=(1/2)CN.

Dear Dinu,

it seems that the assumptions concerning the angles are redundant.

Solution to the problem:

Consider B´such that B is a midpoint of AB´. We obtain two pairs of similar triangles:

Putting 1. and 2. together we obtain BP=2PN.

Dear Viera,

You are right! Good remark, beautiful solution! Congratulations! In fact, you solved the following extension of initial problem:

"Let ABC be a triangle and M on AC so that MA=MC. Let N be a point on segment AB so that BM=CN. Let P be the intersection point of BM and CN. Prove that BP=2PN."