Triunghiul ABC are latura BC inclusa intr-un plan a si AB =24cm , AC=32 cm si BC=40 cm .Fie E € (AB) si F €(AC) astfel incat AE=8cm si AF=6cm .
a) Daca dreapta EF intersecteaza planul a in P ,aratati ca punctele P, B si C sunt coliniare.
b) Calculati perimetrul patrulaterului BEFC .
BC²=AB²+AC²(se verifica) atunci EF²=AE²+AF² EF=10 cm 2.ΔABC≈ΔAEF AE/AC=AF/AB=EF/BC 8/32=6/24=10/40=1/4 atunci ∧AEF≡∧ACB ∧AFE≡∧ABC Trebuie demonstrat ca PBA+ABC=180° ∧ABC=90°-∧C PBA=180°-P-∧PEB ∧PEB=∧AEF=∧C ∧PBA=180°-(∧P+∧C) dar ∧ABC=∧AEF⇒ ∧AFE+∧PFC=180°⇒ ∧PBA+∧ABC=180° 2.perimetrul EA=24-8=16 AF=32-6=26 EF=10 P=10+40+16+26=92 cm
