Question

Puteti sa ma ajutati la ex 17?

Answer 1

Răspuns:

Vezi atasament

Answer 2

$M - Mijlocul~[BC] \implies [BM] \equiv [MC] \\ \\<(AMC) \equiv <(BMD) \text{~~Opuse la varf}\\ \\M - Mijlocul~[AD] \implies [AM] \equiv [MD]  \\ \\ \bigtriangleup AMC~~|~~[BM] \equiv [MC]~~~~~~~~~~~~~~|~L.U.L \\\bigtriangleup BMD~~|~~[AM] \equiv [MD]~~~~~~~~~~~~~~| \implies \bigtriangleup AMC \equiv \bigtriangleup BMD\\~~~~~~~~~~~~~~~|~~<(AMC) \equiv <(BMD)~|<(AMC) \equiv <(BMD \\  \\b) \text{ABCD este paralelogram are laturile paralele si are diagonalele congruente }~\implies~\text{ABCD este patrat}$

$\text{Notam AB=AC=DC=BD=x} \\\text{Aplicand teorema lui Pitagora in triunghiul ADC scoti } AD=x\sqrt{2} \\ \\AM = \dfrac{AD}{2}=\dfrac{x\sqrt{2}}{2} \\ \\\\2AM < AB + AC\\ \\2*\dfrac{x\sqrt{2}}{2} < x+x \\ \\x\sqrt{2} < 2x \iff (x\sqrt{2})^2 < (2x)^2 \iff 2x^2 < 4x^2 \iff x^2<2x^2 \implies 2AM<AB+AC$