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Problem 3

On the sides of an arbitrary triangle ABCABC, triangles ABRABR, BCPBCP, CAQCAQ are constructed externally with ∠CBP=∠CAQ=45∘\angle CBP = \angle CAQ = 45^\circ, ∠BCP=∠ACQ=30∘\angle BCP = \angle ACQ = 30^\circ, ∠ABR=∠BAR=15∘\angle ABR = \angle BAR = 15^\circ. Prove that ∠QRP=90∘\angle QRP = 90^\circ and QR=RPQR = RP.
Step 3 of 6: Track the three angles for the cosine rule
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∠PBR=B+60∘,∠QAR=A+60∘,∠PCQ=C+60∘\angle PBR=B+60^\circ,\quad \angle QAR=A+60^\circ,\quad \angle PCQ=C+60^\circ
Detailed analysis

Because the three auxiliary triangles lie externally on BCBC, CACA, and ABAB, the angle between BPBP and BRBR is B+45∘+15∘=B+60∘B+45^\circ+15^\circ=B+60^\circ. Cyclically, the corresponding angles are A+60∘A+60^\circ and C+60∘C+60^\circ.