Triangulo Stewart

  • May 2020
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Construamos um triangulo ∆ACJ, tal como na figura abaixo:

Repare que propositalmente o lado BJ foi construido com tamanho x. Nisso repare que o ponto D ´e o baricentro desse novo triangulo ∆ACJ, nisso pela rela¸c˜ ao de Stewart no triˆ angulo ∆CDJ. w2 k2 x2 (2x)·x + 2x·x − x2 = 1, que fica: w2 k2 + (2x)·x 2x·x − 1 = 1 w2 k2 (2x)·x + 2x·x = 2 w2 + k 2 = 4x2 2 2 2

w + k = (2x)

Que ´e justamente a rela¸c˜ ao de Pit´agoras! Logo o triˆangulo ∆CDJ tem de ser retˆ angulo, e o ˆ angulo β deve ser reto! Como α ´e o suplementar de β, ele ´e reto tamb´em(que corresponde ao ˆangulo b C DE). Provado! This paper was written with LATEX 2ε

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