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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the problem 730 details.
http://img196.imageshack.us/img196/7341/problem730.pngFor clarity see sketch on the left of above pictureWe will use Ceva’s theorem in this case.Quadrilateral AA3A1A2 is cyclic with diameter AA1We have ∠ (AA3A2)= ∠ (AA1A2)= ∠ (ABA1)= ∠B ( angles face the same arc AA2)Similarly ∠ (AA2A3)= ∠AA1A3)= ∠ (ACA1)= ∠CSo sin(A4AA3)/sin(A4AA2)=cos(AA3A2)/cos(AA2A3)=cos(B)/cos(C) (angles complement) ----(1)Similarly sin(B4BB3)/in(B4BB2)=cos(C )/cos(A) ----- (2)And sin(C4CC3)/sin(C4CC2)=cos(A)/cos(B) -----(3)Multiply expressions (1) x(2) x(3) we get RHS= cos(B)/cos(C) x cos(C )/cos(A) x cos(A)/cos(B) =1So Cevians AA4, BB4 and CC4 will concurrent per Ceva’s theorem
Join C₁A₁,A₁B₁,B₁C₁.∆BC₁A₁~∆BCA with ∠BA₁C₁=A=∠B₃B₂B,∠BC₁A₁=C=∠B₂B₃BSo C₁A₁∥B₃B₂. Given B₂B₃⊥BB4. Thus C₁A₁⊥BB4.Let C₁A₁^BB4 = Y, A₁B₁^CC4 = Z, B₁C₁^AA4 = X(^ denotes intersection)BB4 divides C₁A₁ (at Y) in the ratio C₁X:XA₁ =BXcotC:BXcotA = cotC:cotA.SimilarlyCC4 divides A₁B₁(at Z)in the ratio cotA:cotB,AA4 divides B₁C₁(at X)in the ratio cotB:cotC.Product of the above three ratios being +1, it follows by Ceva,AA4X, BB4Y, CC4Z are concurrent.
To Pravin:Refer to your solution . "Product of the above three ratios being +1" is not applicable to Ceva's theorem in this case.please clarify.
Thanks Peter. I have recast my proof as follows.Let BB4 be extended to meet AC at Y’ (similarly CC4 to Z’ on AB, AA4 to X’ on BC)BB4 ⊥ B₃B₂ So ∠ABY’=∠B₃BB4=90°-∠B₂B₃B=90°-C and similarly ∠CBY’=90°-AAY’:Y’C=(ABY’)/(BY’C)where (...)denotes area. = [AB.BY’.sin∠ABY’]: [BC.BY’.sin∠CBY’] = (AB:BC)(cosC:cosA) = (sinC:sinA)(cosC:cosA)= sin2C:sin2AIn the same way, CX’:X’B = sin2B:sin2C and BZ’: Z’A = sin2A:sin2BSo(AY’:Y’C).(BZ’:Z’A).(CX’:X’B) = 1Hence by Ceva’s theorem applied to ∆ABC the cevians AA4X’, BB4Y’, CC4Z’ are concurrent.
Hint:It's cut at circumcenter