Points P and Q lie on side BC of acute-angled triangle ABC so that ∠PAB=∠BCA and ∠CAQ=∠ABC. Points M and N lie on lines AP and AQ respectively, such that P is the midpoint of AM, and Q is the midpoint of AN. Prove that lines BM and CN intersect on the circumcircle of triangle ABC.
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I'll give a solution using complex numbers, hopefully no mistakes were made.
The triangles QCA and ACB are similar, so (C-Q)/(A-Q)=[(C-A)/(B-A)]£, where £ refers to complex conjugation.
Also, triangles ABC, PBA are similar, so (A-P)/(B-P)=[(C-A)/(B-A)]£ . But it is obvious that M-P=P-A and N-Q=Q-A, using these relations we get that (C-Q)/(N-Q)=(M-P)/(B-P), which means that the triangles CQN and MPB are similar.
Now we observe that angle(ABM) +angle(ACN)=180 degrees, which means that ABCD is cyclic, so obviously D is on the circumcircle of the triangle ABC.
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a_{i-1}
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I'll give a solution using complex numbers, hopefully no mistakes were made. The triangles QCA and ACB are similar, so (C-Q)/(A-Q)=[(C-A)/(B-A)]£, where £ refers to complex conjugation. Also, triangles ABC, PBA are similar, so (A-P)/(B-P)=[(C-A)/(B-A)]£ . But it is obvious that M-P=P-A and N-Q=Q-A, using these relations we get that (C-Q)/(N-Q)=(M-P)/(B-P), which means that the triangles CQN and MPB are similar. Now we observe that angle(ABM) +angle(ACN)=180 degrees, which means that ABCD is cyclic, so obviously D is on the circumcircle of the triangle ABC.
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Alternatively, you could note that △BPM∼△CQN and then perform a not too tedious trig bash.
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My friend said that he solved it with Barycentric coordinates in 5 minutes :O