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Circle through equilateral triangle

Draw circle through 3 vertices of equilateral triangle.

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  Construct angle bisectors for A & B that cut BC & AC at F & E resp.

  AF & BE intercept at D

  ∠EAD = 30∘ = ∠FBD      (construct)
  ∠EDA = ∠FDB             (opp. ang.s of intersecting str. lines)
  
  ∠DAB = 30∘ = ∠DBA      (construct)
  ∴ ∆DAB is isos.         (2 angles equal)
  ∴ AD = BD
  
  ∴ ∆EAD ≣ ∆FDB          (2 ang.s & inc. side =)
  ∴ ED = DF & EA = FB
  
  But AC = BC             (equilateral triangle) 
  ∴   CE = CF             ( equals - equals are equal)  
  
  Join CD
  ∆CED ≣ ∆CFD              (all 3 sides =)
  ∴ ∠ECD = ∠FCD
         = 30∘             (∠ECF = 60∘ as ∆ABC equilateral)
         = ∠EAD
  ∴ ∆ACD isos. 
  ∴ CD = AD
    CD = DE               (proven) 
	  
    So D is centre of circle of radius AD and passes through A, B & C.            	
	
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