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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 6 of 6: Finish: PQRPQR is isosceles right
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PR=QR,PR2+QR2=PQ2  ⟹  ∠PRQ=90∘PR=QR,\quad PR^2+QR^2=PQ^2\implies \angle PRQ=90^\circ
Detailed analysis

The first identity in Step 5 gives PR=QRPR=QR because lengths are nonnegative. The second is the Pythagorean relation in triangle PQRPQR, with hypotenuse PQPQ; equivalently, the cosine rule gives cos⁡∠PRQ=(PR2+QR2−PQ2)/(2PR QR)=0\cos\angle PRQ=(PR^2+QR^2-PQ^2)/(2PR\,QR)=0. Thus ∠PRQ=90∘\angle PRQ=90^\circ and the required result follows.