Respuesta :

[tex]\bigtriangleup ABD \cong \bigtriangleup BCD\Rightarrow \frac{CD}{BD} = \frac{BD}{AD} \Leftrightarrow CD = x = \frac{ {BD}^{2} }{AD} = \frac{ {8}^{2} }{15} = \frac{64}{15} \\ BC =y = \sqrt{ {BD}^{2} + {CD}^{2} } = \sqrt{ { {8}^{2} +\frac{64}{15} }^{2} } = \frac{136}{15} \\ \Rightarrow x = \frac{64}{15} , \: y = \frac{136}{15} [/tex]

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