Question

Aratati ca daca intr-un triunghi de laturi a,b,c are loc relatia a+c/b = toata fractia sub radical a+c/a-c atunci triunghiul este dreptunghic

Answer 1

$\dfrac{a+c}{b} = \sqrt{\dfrac{a+c}{a-c}}\,\Bigg|\verb ^2 \\ \\\\ \Rightarrow \Big(\dfrac{a+c}{b}\Big)^2 = \dfrac{a+c}{a-c}\\ \\\\ \Rightarrow \dfrac{(a+c)^2}{b^2} = \dfrac{a+c}{a-c}\,\Bigg|:(a+c),\quad a+c\neq 0\\ \\ \\\Rightarrow\dfrac{a+c}{b^2} = \dfrac{1}{a-c}\,\Bigg|\cdot b^2(a-c),\quad b^2(a-c)\neq 0\\ \\ \\ \Rightarrow (a+c)(a-c) = b^2\\ \\ \Rightarrow a^2-c^2 = b^2$

$\Rightarrow\boxed{a^2 = b^2+c^2,\quad m(\widehat{b,c}) = 90^\circ}\quad \text{(Teorema lui Pitagora)}$

Answer 2

Răspuns:

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