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Question Number 207362 by mr W last updated on 12/May/24

Answered by A5T last updated on 12/May/24

Ptolemy′s theorem: AC×BP=AP×BC+AB×PC  AB=AC=BC⇒PB=PA+PC

Ptolemystheorem:AC×BP=AP×BC+AB×PCAB=AC=BCPB=PA+PC

Answered by sniper237 last updated on 13/May/24

Let Q ∈ [BP] such as PQ=PC  As mesBPC=mesBAC=60 then  QPC is equilateral as ABC  So  r(A)=B and r(P)=Q with r=R(C;(π/3))   Thus r is an isometry, PA=BQ  Then PB=PQ+QB=PC+PA

LetQ[BP]suchasPQ=PCAsmesBPC=mesBAC=60thenQPCisequilateralasABCSor(A)=Bandr(P)=Qwithr=R(C;π3)Thusrisanisometry,PA=BQThenPB=PQ+QB=PC+PA

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