CD AB = BA AB - Matematica pe Net
CD AB = BA AB - Matematica pe Net
CD AB = BA AB - Matematica pe Net
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• Aplicand proprietatea centrului de greutate punctului O, avem:<br />
3 3<br />
2<br />
⋅ OG = OA + OB + OC = 3⋅<br />
( OH + HG)<br />
= 3⋅<br />
( OH − OH ) = OH (relatia lui Sylvester);<br />
• Daca notam O1 mijlocul segmentului OH atunci OA + OB + OC = 2OO1<br />
(relatial lui Euler);<br />
Metode in rezolvarea problemelor de paralelism:<br />
1) Daca dreptele <strong>AB</strong> si <strong>CD</strong> sunt paralele atunci exista a ∈ R * astfel incat <strong>AB</strong> = a ⋅<strong>CD</strong><br />
; reciproc concluzia este sau <strong>AB</strong> || <strong>CD</strong>, sau<br />
A,B,C,D coliniare;<br />
2) Daca exista a, b ∈ R * astfel incat a ⋅ <strong>AB</strong> + b ⋅<strong>CD</strong><br />
= 0 , atunci <strong>AB</strong> ||<strong>CD</strong> sau A,B,C,D coliniare;<br />
3) Daca u = a ⋅i<br />
+ b ⋅ j;<br />
v = a'⋅i<br />
+ b'⋅<br />
j , atunci u, v au aceeasi directie daca si numai daca coordonatele sunt proportionale,<br />
b<br />
a'<br />
b'<br />
a = .<br />
4) u, v au aceasi directie daca si numai daca u + v = u + v .<br />
Metode in rezolvarea problemelor de colinaritate:<br />
1) A, B, C sunt coliniare daca si numai daca exista R *<br />
<strong>AB</strong><br />
dreapta); conditia se poate exprima si prin raportul = a ;<br />
AC<br />
a ∈ astfel incat <strong>AB</strong> a ⋅ AC<br />
2) A, B, C sunt coliniare daca si numai daca exista a, b ∈ R * astfel incat a ⋅ <strong>AB</strong> + b ⋅ AC = 0 ;<br />
= (relatia furnizand si pozitiia punctelor <strong>pe</strong><br />
3) Daca in sistemul ( O, i,<br />
j)<br />
avem A(<br />
x A,<br />
y A ), B(<br />
xB<br />
, yB<br />
), C(<br />
xC<br />
, yC<br />
) , atunci A, B, C coliniare daca<br />
4) Intr-un re<strong>pe</strong>r fixat , A, B, C coliniare daca si numai daca relatia a r + b ⋅ r + c ⋅ r = 0 implica a+b+c=0;<br />
⋅ A B C<br />
5) A, B, C coliniare daca si numai daca oricare ar fi puncul O, avem ca relatia a ⋅OA + b ⋅OB<br />
+ c ⋅OC<br />
= 0 implica a+b+c=0.<br />
Conditia ca trei vectori sa formeze un triunghi.<br />
u v,<br />
w<br />
, corespund vectorilor laturi ale unui triunghi daca u ± v ± w = 0 si suma oricaror doua module este mai mare decat al treilea.<br />
Conditia ca patru puncte sa formeze un paralelogram (degenerat).<br />
• Patrulaterul (degenerat) <strong>AB</strong><strong>CD</strong> este paralelogram (degenerat) daca si numai daca <strong>AB</strong> = DC ;<br />
• Segmentele <strong>AB</strong>, <strong>CD</strong> pot fi laturi opuse intr-un paralelogram (degenerat) daca si numai daca <strong>AB</strong> = ± <strong>CD</strong> , conditie ce se poate<br />
evidentia si astfel: <strong>AB</strong> + <strong>CD</strong> = 0 sau <strong>AB</strong> − <strong>CD</strong> = 0 ;<br />
Conditia ca patru punct sa formeze un tra<strong>pe</strong>z (degenerat).<br />
• Patrulaterul <strong>AB</strong><strong>CD</strong> este un tra<strong>pe</strong>z (degenerat) daca si numai daca exista a ∈ R * \{ 1)<br />
astfel incat <strong>AB</strong> = a ⋅ DC sau<br />
AD = a ⋅ BC .<br />
• Patrulaterul <strong>AB</strong><strong>CD</strong> este un tra<strong>pe</strong>z (degenerat) daca si numai daca exista a, b ∈ R+<br />
*, a ≠ b astfel incat a ⋅ <strong>AB</strong> + b ⋅<strong>CD</strong><br />
= 0 sau<br />
a ⋅ AD + b ⋅CB<br />
= 0 .<br />
+<br />
x<br />
x<br />
B<br />
C<br />
− x<br />
− x<br />
A<br />
A<br />
y<br />
=<br />
y<br />
B<br />
C<br />
− y<br />
− y<br />
A<br />
A<br />
;