⇒ΔDAF≡ΔBDC
AD≡DB(ip)
FD≡DC(ip)
∧FDA≡∧BDC ( opuse la varf)
cazul de congruenta (L.U.L)
atunci ∧DFA≡∧BCD⇒FA||BE intersectat de secanta FC
F,A,G coliniare
ΔAEG≡ΔBEC (L.U.L)
AE≡EC
BE≡EC
∧AEG≡∧BEC (opuse la varf)
AG||BC
in ΔABC ∧A+∧B+∧C=180*
dar ∧B≡∧FAD
∧C=∧EAG
deci
∧FAD+∧EAG+∧A=180* deci punctele F,A,G coliniare⇔Bafta! Sper ca te-am ajutat!!!