1) Am atasat desenul.
AB = 4 cm
BC = [tex] \frac{3}{2} [/tex]*AB = [tex] \frac{3}{2} [/tex]*4 = 3*2 = 6 cm
EF = 6 cm
DE = [tex] \frac{2}{3} [/tex]*EF = [tex] \frac{2}{3} [/tex]*6 = 2*2 = 4 cm
Deci in ΔABC si ΔDEF avem:
AB=DE=4 cm
BC=EF=6 cm
m(<ABC)=m(<DEF)=50 grade
deci ΔABC ≡ ΔDEF (L.U.L.)
prin urmare AC≡DF
<BAC≡<EDF
<ACB≡<DFE
2) Cum ΔABC≡ΔDEF rezulta ca avem congruentele:
BC≡EF (1)
<ABC≡<DEF, deci
m(<ABC)=m(<DEF) si deci
[tex] \frac{m(<ABC)}{2} = \frac{m(<DEF)}{2} [/tex]
adica:
m(<MBC)=m(<NEF) (2)
respectiv:
<ACB≡<DFE, deci
m(<ACB)=m(<DFE) si deci
[tex] \frac{m(<ACB)}{2} = \frac{m(<DFE)}{2} [/tex]
adica:
m(<MCB)=m(<NFE) (3)
Din (1), (2) si (3) rezulta:
ΔMBC≡ΔNEF (U.L.U.)
