Question

Let ABC be an acute-angled triangle and let D,E,F be the feet of the perpendiculars from A,B,C respectively to BC,CA,AB. Let the perpendiculars from F to CB, CA, AD, BE meet them in P,Q,M,N respectively .Prove that P,Q,M,N are collinear.

Answer 1

Explanation:

Observe that C, Q, F, P or concyclic. Hence

∠CQP = ∠CQP= 90° - ∠FCP = ∠B Similarly the concyclicity of F,M.Q,A gives 

∠AQN = 90° + ∠FQM = 90° + ∠FAM = 90° + 90° - ∠B =180° - ∠B. 

Thus we obtain ∠CQP + ∠AQN = 180° . It follows that Q, N, P lie on the same line.

We can similarly prove that ∠CPQ + ∠BPM = 180° . This implies that P, M, Q, are collinear. Thus M, N both lie on line joining P and Q.