MathLabs

Problem 4

Points PP and QQ lie on side BCBC of an acute-angled triangle ABCABC so that ∠PAB=∠BCA\angle PAB=\angle BCA and ∠CAQ=∠ABC\angle CAQ=\angle ABC. Points MM and NN lie on lines APAP and AQAQ, respectively, such that PP is the midpoint of AMAM, and QQ is the midpoint of ANAN. Prove that the intersection of lines BMBM and CNCN lies on the circumcircle of triangle ABCABC.
Step 5 of 5: Conclusion: S sees BC at the supplement of angle A
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∠CSB=∠BPM=β+γ=180∘−∠BAC\angle CSB = \angle BPM = \beta+\gamma = 180^\circ - \angle BAC
Detailed analysis

From △BPM∼△BSC\triangle BPM\sim\triangle BSC, corresponding angles give ∠CSB=∠BPM\angle CSB=\angle BPM; combined with ∠BPM=β+γ=180∘−∠BAC\angle BPM=\beta+\gamma=180^\circ-\angle BAC from an earlier step (since β+γ\beta+\gamma is the supplement of ∠BAC\angle BAC via the angle sum in △ABC\triangle ABC), we get ∠CSB=180∘−∠BAC\angle CSB=180^\circ-\angle BAC. This is exactly the condition for SS to lie on the arc BCBC of the circumcircle of ABCABC not containing AA, completing the proof.