Explicație pas cu pas:
punctele A,B,C,D pe cercul C(0,r)
notăm cu E,F,G,H, mijloacele arcelor AB,BC,CD,AD
m(arc AE) ≡ m(arc EB) = ½•m(arc AB)
m(arc BF) ≡ m(arc FC) = ½•m(arc BC)
m(arc CG) ≡ m(arc GD) = ½•m(arc CD)
m(arc DH) ≡ m(arc HA) = ½•m(arc DA)
m(arc AB) + m(arc BC) + m(arc CD) + m(arc DA) = 360°
notăm EG ∩ FH = {S}
[tex]m(\angle GSH) = \frac{m(arcGDH) + m(arcEF)}{2} \\ [/tex]
m(arc GDH) + m(arc EF) = m(arc GD) + m(arc DH) + m(arc EB) + m(arc BF) = ½•m(arc CD)+ ½•m(arc DA) + ½•m(arc AB) + ½•m(arc BC) = ½•[m(arc AB) + m(arc BC) + m(arc CD) + m(arc DA)] = ½•360° = 180°
[tex]\implies m(\angle GSH) = \frac{180\degree}{2} = 90\degree[/tex]
=> EG ⊥ FH
q.e.d.
