Δ ABC isoccel=> ∡ABC=∡ACB (il voi nota cu x)
AB=AC
daca M∈AB si N∈AC astfel incat AM=AN din ultimele doua =>MB=NC
daca AM=AN =>ΔAMN isocel=> ∡∡ANM
Notez ∡ABN cu l₁ si ∡MCN cu l₂
∡NBC t₁si ∡MCB cu t₂
ACUM:
∡X₁=∡X₂
∡X₁=∡l₁+∡t₁=> ∡l₁=∡X₁-∡t₁
∡X₂=∡l₂+∡t₂=> ∡l₂=∡X₂-∡t₂
daca ∡X₁=∡X₂=>∡l₁=∡t₁ si ∡l₂=∡t₂
si ∡l₁+∡t₁=∡l₂+∡t₂
∡l₁+∡l₁=∡l₂+∡l₂
∡2l₁=∡2l₂=> ∡l₁=∡l₂ adica ∡ABN=∡ACM
b) ∡X₁=∡X₂
∡X₁=∡l₁+∡t₁=> ∡l₁=∡X₁-∡t₁
∡X₂=∡l₂+∡t₂=> ∡l₂=∡X₂-∡t₂
daca ∡X₁=∡X₂=>∡l₁=∡t₁ si ∡l₂=∡t₂
∡ t₁+∡t₁=∡t₂+∡t₂
∡2t₁=∡2t₂=> ∡t₁=∡t₂ adica ∡NBC=∡MCB
sau mai poti demonstra ca MNBC este trapez isoscel
