Fie m∡(CBE)=α
si m∡(BEC)=β
cum m∡(BCD)=90°⇒αsi β, complementare
DC≡CB ( ABCD patrat)
mas∡ FDC=mas∡ BCD=90 °( ABCDpatrat)
EC≡DF ( ipoteza=DC/2)
⇒(caz CC)ΔBCE≡ΔCDF⇒m∡(CBE)=m∡(DCF)=α
⇒m∡GEC+mas∡(GCE) =α+β=90°⇒mas ∡(EGC)=90°⇔BE⊥FC, cerinta
b)mas ∡(ABG)=90°-α=β (1)
ΔABF≡ΔBCE ( caz CC, analog cu punctul a))⇒mas ∡(AFB)=β(2)
dar FA⊥AB (ipoteza) si BG⊥FG (demonstratie, punctul a))⇒ABGFpatrulater inscriptibil (in cercul de diametru FB)⇒mas ∡(AFB) = mas∡( AGB) (subintind aceeasi coarda, AB
cum mas ( AFB)=β(realatia (2)) ⇒mas ∡(AGB)=β (3)
din mas ∡(ABG) =β (rel (1))
si mas ∡(AGB)=β ( rel (3) ⇒ΔABG isoscel de baza BG
