Dinu Teodorescu,

How about applying Ceva's theorem twice (in trigonometric form)... to get a proof

https://www.cut-the-knot.org/triangle/TrigCeva.shtml

As it is given: "... perpendicular from B to NM ..." = perhaps,

this should be: "... perpendicular from B to PM ..." - correct ?!

Thus, a proof comes from:

once, applying Cheva's theorem in trigonometric form for the cevians AO, BO, CO.

And, a second time, applying Ceva's theorem for the potential cevians, i.e. for the perpendiculars from A, B, C to PN, PM, and MN, respectively.

Where both sets of cevians make equal angles with the sides of triangle ABC because each quadrilateral APON, BPOM, CMON is inscribed, and corresponding pairs of angles are spanned by mutually perpendicular rays.

Steftcho P. Dokov Yes Steftcho, you are right! Many thanks and sorry to all readers for my mistake ! I will modify in the statement!

Thanks, you are welcome, Dinu. A picture is attached for visualizing the whole proof.

Steftcho P. Dokov Your approach is correct, my solution is the same.

The essential is the fact that the two cevians from A , AO and the perpendicular from A to PN, make equal angles with the sides AB and AC of triangle ABC( we say that the two cevians are isogonal cevians) due to the fact that quadrilateral APON is inscriptible. Analogous, BO and the perpendicular from B to PM, CO and the perpendicular from C to MN are pairs of isogonal cevians.

Now all we have to make is to apply Ceva's theorem, Steiner's ratio theorem and the reciprocal Ceva's theorem.

Independent of current problem, the following theorem can be obtained using Ceva's theorem, Steiner's ratio theorem and the reciprocal Ceva's theorem:

THEOREM 1: In a triangle the isogonals of three cevians having a common point are three cevians which have a common point.

Steiner's theorem is a generalization( an extension) of bisector theorem in a triangle:

THEOREM (Steiner) Let AM, AN be two isogonal cevians in triangle ABC, with M,N on BC. Then (AB^2)/(AC^2)=(BM*BN)/(CM*CN).

Clearly, when M and N move towards each other( conserving the property of Isogonality), lines AM and AN will be identical, representing the bisector of angle BAC.

Steiner's ratio theorem can be proved constructing two pairs of similar triangles, or simpler, using areas of triangles( the most famous proofs).

Now I think is clear how THEOREM 1 can be proved.

Regarding the problem from above question, it's possible to obtain a solution intersecting two perpendiculars, from A and B, let us say, in a point S, then proving CS is orthogonal to MN. I think is not easy!

Steftcho P. Dokov Thanks for the picture! Congrats for solution! You prove a solid knowledge in geometry!

Steftcho P. Dokov It's not absolutely necessary to use the trigonometric form of Ceva's theorem, the classic one works too!