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Question Number 91836 by  M±th+et+s last updated on 03/May/20

Commented by  M±th+et+s last updated on 03/May/20

let A B C D be four points on a circle  let four more cirvces pass through   AB , BC , CD , DA Respectively,  meeting infurther points P,Q,R,S.  show that P,Q,R and S are consyclic.

letABCDbefourpointsonacircleletfourmorecirvcespassthroughAB,BC,CD,DARespectively,meetinginfurtherpointsP,Q,R,S.showthatP,Q,RandSareconsyclic.

Answered by mr W last updated on 04/May/20

Commented by mr W last updated on 04/May/20

Commented by mr W last updated on 07/May/20

α_1 +α_4 =α_2 +α_3 =π, since ABQP is cyclic.  β_1 +β_4 =β_2 +β_3   γ_1 +γ_4 =γ_2 +γ_3   δ_1 +δ_4 =δ_2 +δ_3   (α_1 +δ_2 )+(α_4 +δ_3 )=(α_2 +δ_1 )+(α_3 +δ_4 )  (β_2 +γ_1 )+(β_3 +γ_4 )=(β_1 +γ_2 )+(β_4 +γ_3 )  (α_1 +δ_2 +β_2 +γ_1 )+(α_4 +δ_3 +β_3 +γ_4 )=(γ_2 +δ_1 +α_2 +β_1 )+(α_3 +δ_4 +β_4 +γ_3 )  α_4 +β_3 +γ_4 +δ_3 =α_3 +δ_4 +β_4 +γ_3   2π−θ_B +2π−θ_D =2π−θ_A +2π−θ_C   ⇒θ_B +θ_D =θ_A +θ_C   since θ_A +θ_B +θ_C +θ_D =2π  ⇒θ_B +θ_D =θ_A +θ_C =π  i.e. PQRS is cyclic.

α1+α4=α2+α3=π,sinceABQPiscyclic.β1+β4=β2+β3γ1+γ4=γ2+γ3δ1+δ4=δ2+δ3(α1+δ2)+(α4+δ3)=(α2+δ1)+(α3+δ4)(β2+γ1)+(β3+γ4)=(β1+γ2)+(β4+γ3)(α1+δ2+β2+γ1)+(α4+δ3+β3+γ4)=(γ2+δ1+α2+β1)+(α3+δ4+β4+γ3)α4+β3+γ4+δ3=α3+δ4+β4+γ32πθB+2πθD=2πθA+2πθCθB+θD=θA+θCsinceθA+θB+θC+θD=2πθB+θD=θA+θC=πi.e.PQRSiscyclic.

Commented by  M±th+et+s last updated on 04/May/20

great work sir

greatworksir

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