Question

Ajude preciso desse exercicio resolvido...é correção de prova

Answer 1

Observe a figura em anexo ao final desta resposta.

Observamos que os triângulos $ABC$ e $BDE$ são semelhantes. Logo, devemos ter

$\dfrac{AC}{AB}=\dfrac{DE}{EB}\\\\\\ \dfrac{2}{2\sqrt{3}}=\dfrac{DE}{EB}\\\\\\ \dfrac{1}{\sqrt{3}}=\dfrac{DE}{EB}~~~~~~\mathbf{(i)}$


Mas $DE=\dfrac{AD}{2}.$ Logo,

$\dfrac{1}{\sqrt{3}}=\dfrac{AD}{2\cdot EB}\\\\\\ \dfrac{1}{\sqrt{3}}=\dfrac{AB-BD}{2\cdot EB}\\\\\\ \dfrac{1}{\sqrt{3}}=\dfrac{2\sqrt{3}-BD}{2\cdot EB}$


O produto dos meios é igual ao produto dos extremos (multiplica em cruz):

$\sqrt{3}\cdot (2\sqrt{3}-BD)=2\cdot EB\\\\ 6-\sqrt{3}\cdot BD=2\cdot EB\\\\\\ EB=\dfrac{6-\sqrt{3}\cdot BD}{2}~~~~~~\mathbf{(ii)}$

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Aplicando o Teorema de Pitágoras ao triângulo $BDE\,,$ devemos ter

$(EB)^2+(DE)^2=(BD)^2\\\\\\ (EB)^2+(DE)^2=(BD)^2\\\\\\ (EB)^2+\left(\dfrac{AD}{2} \right )^{\!\!2}=(BD)^2\\\\\\ (EB)^2+\dfrac{(AD)^2}{4}=(BD)^2\\\\\\ (EB)^2+\dfrac{(AB-BD)^2}{4}=(BD)^2\\\\\\ (EB)^2+\dfrac{(2\sqrt{3}-BD)^2}{4}=(BD)^2\\\\\\ \left(\dfrac{6-\sqrt{3}\cdot BD}{2} \right )^2+\dfrac{(2\sqrt{3}-BD)^2}{4}=(BD)^2$

$\left(\dfrac{6-\sqrt{3}\cdot BD}{2} \right )^2+\dfrac{(2\sqrt{3}-BD)^2}{4}=(BD)^2\\\\\\ \dfrac{(6-\sqrt{3}\cdot BD)^2}{4}+\dfrac{(2\sqrt{3}-BD)^2}{4}=(BD)^2\\\\\\ (6-\sqrt{3}\cdot BD)^2+(2\sqrt{3}-BD)^2=4\cdot (BD)^2$


Fazendo $BD=x\,,$ temos

$(6-\sqrt{3}\,x)^2+(2\sqrt{3}-x)^2=4x^2\\\\\ (36-12\sqrt{3}\,x+3x^2)+(12-4\sqrt{3}\,x+x^2)=4x^2\\\\ 48-16\sqrt{3}\,x+\diagup\!\!\!\!\!\! 4x^2=\diagup\!\!\!\!\!\! 4x^2\\\\\\ 48-16\sqrt{3}\,x=0\\\\ 16\sqrt{3}\,x=48\\\\ x=\dfrac{48}{16\sqrt{3}}\\\\\\ x=\dfrac{3}{\sqrt{3}}\\\\\\ x=\sqrt{3}\\\\\\ \therefore~~\boxed{\begin{array}{c}BD=\sqrt{3} \end{array}}$

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Logo,

$AD=AB-BD\\\\ AD=2\sqrt{3}-\sqrt{3}\\\\ \boxed{\begin{array}{c}AD=\sqrt{3} \end{array}}$


Finalmente, aplicando o Teorema de Pitágoras ao triângulo $ACD\,,$ temos

$(CD)^2=(AC)^2+(AD)^2\\\\ (CD)^2=2^2+(\sqrt{3})^2\\\\ (CD)^2=4+\sqrt{3}\\\\ \boxed{\begin{array}{c}CD=\sqrt{4+\sqrt{3}} \end{array}}$


Bons estudos! :-)