Question

help mee !!!
16 puntos

Answer 1

[tex]1) ~~ DF \perp~AB \\ ~~AC\perp~AB \\ ------------------------------ \\ Rezulta~ca~AC||DF \\ Avem:~D-mijlocul~lui~BC \\ ~~~~~~~~~~~~AC||DF \\ Din~astea~doua~rezulta~ca~F-mijlocul~lui~AB. \\ Deci~DF-linie~mijlocie~cu~proprietatea~ca~:DF= \frac{AC}{2} \\ Analog~se~arata~ca~DE||AB,D-mijl.=\ \textgreater \ DE~linie~mijlocie \\ cu~proprietatea~ca~DE= \frac{AB}{2} \\ Deoarece~DE||AB~si~AC||DF,iar~AB\perp~AC \\ rezulta~ca~DE\perp~DF \\ Iar~cu~Teorema~Pitagora~avem~ca~: \\ DE^2+DF^2=EF^2. \\ DE= \frac{AB}{2} \\ DF= \frac{AC}{2} \\ =\ \textgreater \ EF^2= \frac{AB^2+AC^2}{4} \\ EF^2= \frac{BC^2}{4} \\ 4EF^2=BC^2|^{ \sqrt{}} \\ \\ \\ \circled{2EF=BC} [/tex] [/tex]

Answer 2

AD mediana coresp.ipotenuzei in tr.dr ABC
AD=BC/2
BC=2AD (0)
DF⊥AB(ipoteza)⇒DF||AC (1)
DE⊥AC(ipoteza)⇒DE||AB (2)
din (1) si (2)⇒AEDF paralelogram
  Dar cum m∡A=90°,   AEDF paralelogram cu un unghi drept, AEDF  dreptunghi⇒
[EF]≡[AD] (diagonale in dreptunghi)
 cum BC=2AD (0)⇒BC=2EF, C.C.T.D.
as simple as that!!