MathLabs

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 4 of 6: Apply the cosine rule to PBRPBR, QARQAR, and PCQPCQ
Previous stepNext step
PR2=PB2+RB2−2PB RBcos⁡(B+60∘),QR2=QA2+RA2−2QA RAcos⁡(A+60∘),PQ2=PC2+QC2−2PC QCcos⁡(C+60∘)PR^2=PB^2+RB^2-2PB\,RB\cos(B+60^\circ),\quad QR^2=QA^2+RA^2-2QA\,RA\cos(A+60^\circ),\quad PQ^2=PC^2+QC^2-2PC\,QC\cos(C+60^\circ)
Detailed analysis

The cosine rule in each of the three triangles, using Step 3 for the included angle and Step 2 for the adjacent sides, gives exactly the three displayed identities.