The bisectors of the angles ∢CAB and ∢BCA intersect the circumcircle of ABC in P and Q respectively. These bisectors intersect each other in point I. Prove that PQ⊥BI.
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Tags: geometry, circumcircle
22.07.2016 23:39
Extend BI to hit the circumcircle at a point M. Note that \overarcBP+\overarcAQ+\overarcAM2=∠A+∠C+∠B2=1802=90.So PQ⊥BI, as desired. QED
23.07.2016 04:28
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23.07.2016 11:23
Dear Mathlinkers, a consequence of the A, C-Mention's circle... Sincerely Jean-Louis
15.02.2021 23:58
Attention. This stunt is performed by professional stoopid (i.e. me). Let R be the midpoint of arc AC. We claim that QR2+PB2=PR2+QB2. Using Law of Sines for each length, we need to show that sin2(α)+sin2(β+γ)=sin2(β)+sin2(α+γ),where α=∠ACI, β=∠BAI and γ=∠ABI, thus α+β+γ=90∘. Note that for any x, we have sin(90∘−x)=cosx, but since we have, sin2α+cos2α=1=sin2β+cos2β,we are done.
28.06.2023 10:40
AQPC is cyclic angle APQ= angle ACQ=C/2 ACQB is cyclic angle ACB= angle BAQ=C/2 ACBP is cyclic, angle PAB= angle BCP=A/2 AQPC is cyclic, so angle QPC=180°-angle QAC = 180°-(A+C/2)=B+C/2 angle APC= angle QPC- angle APQ =B+C/2-C/2=B IN ∆ IPC, angle PIC= 180°-( angle ICP+ angle IPC) =90°-B/2 so, angle BIP= angle BIC- angle IPC= 180°-{(B+C)/2}-(90°-B/2)=90°-C/2 Let QP intersect BI at O angle IOP =90°-C/2+C/2=90° So, PQ is perpendicular to BI