Question

Considere a circunferência de centro O em que são traçadas duas cordas AB e CD que se cruzam no ponto E. Sabendo que as medidas de AE = 8, EB = 24, CE = 6, calcule o valor da medida ED.

Answer 1

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$\large\green{\boxed{\rm~~~\gray{ED}~\pink{=}~\blue{ 32 }~~~}}$

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$\bf\large\green{\underline{\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad}}$

$\green{\rm\underline{EXPLICAC_{\!\!\!,}\tilde{A}O\ PASSO{-}A{-}PASSO\ \ \ }}$✍

❄☃ $\sf(\gray{+}~\red{cores}~\blue{com}~\pink{o}~\orange{App}~\green{Brainly})$ ☘☀

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☺lá, Mellytzza, como tens passado nestes tempos de quarentena⁉ E os estudos à distância, como vão⁉ Espero que bem❗ Acompanhe a resolução abaixo, feita através de algumas manipulações algébricas, e após o resultado você encontrará um resumo sobre Teorema das Cordas que talvez te ajude com exercícios semelhantes no futuro. ✌

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➡ $\sf\blue{ AE \cdot EB = CE \cdot ED }$

➡ $\sf\blue{ 8 \cdot 24 = 6 \cdot ED }$

➡ $\sf\blue{ 192 = 6 \cdot ED }$

➡ $\sf\blue{ ED = \dfrac{192}{6} }$

➡ $\sf\blue{ ED = 32 }$

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$\large\green{\boxed{\rm~~~\gray{ED}~\pink{=}~\blue{ 32 }~~~}}$ ✅

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$\sf\large\red{TEOREMA~DAS~CORDAS}$

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☔ Sejam as duas cordas $\overline{AB}~e~\overline{CD}$ da circunferência c, formadas pelas retas secantas r e s, concorrentes de forma que se cruzem no ponto P.

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$\setlength{\unitlength}{0.8cm}\begin{picture}(6,5)\thicklines\qbezier(4,3)(6.9,2.8)(7,0)\qbezier(1,0)(1.1,2.8)(4,3)\qbezier(4,-3)(6.9,-2.9)(7,0)\qbezier(1,0)(1.1,-2.7)(4,-3)\put(0,-1){\line(3,1){8}}\put(0,2){\line(4,-3){7}}\put(2.8,-0.1){\circle*{0.18}}\put(4,0){\circle*{0.18}}\put(4.2,-0.5){O}\put(1.08,-0.65){\circle*{0.18}}\put(1.2,1.1){\circle*{0.18}}\put(6.8,1.25){\circle*{0.18}}\put(5.94,-2.44){\circle*{0.18}}\put(0.6,0.7){A}\put(0.6,-1.4){C}\put(7.2,0.7){D}\put(6.4,-2.7){B}\put(2.7,-0.8){P}\put(7.2,-3.2){r}\put(7.9,1.2){s}\end{picture}$

$\sf (Esta~imagem~n\tilde{a}o~\acute{e}~visualiz\acute{a}vel~pelo~App~Brainly$ ☹ )

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$\setlength{\unitlength}{0.8cm}\begin{picture}(6,5)\thicklines\qbezier(4,3)(6.9,2.8)(7,0)\qbezier(1,0)(1.1,2.8)(4,3)\qbezier(4,-3)(6.9,-2.9)(7,0)\qbezier(1,0)(1.1,-2.7)(4,-3)\put(0,-1){\line(3,1){8}}\put(0,2){\line(4,-3){7}}\put(2.8,-0.1){\circle*{0.18}}\put(4,0){\circle*{0.18}}\put(0.6,0.7){A}\put(0.6,-1.4){C}\put(7.2,0.7){D}\put(6.4,-2.7){B}\put(4.2,-0.5){O}\put(1.08,-0.65){\circle*{0.18}}\put(1.2,1.1){\circle*{0.18}}\put(6.8,1.25){\circle*{0.18}}\put(5.94,-2.44){\circle*{0.18}}\put(5.89,-2.4){\line(1,4){0.9}}\put(1.12,-0.6){\line(0,1){1.6}}\qbezier(5.2,-1.9)(5.6,-1.5)(6,-1.8)\qbezier(5.7,0.9)(5.8,0)(6.55,0.3)\put(6.2,0.6){ \alpha }\put(5.6,-2.1){ \beta }\qbezier(1.17,0.3)(1.7,0.2)(1.85,0.6)\qbezier(1.15,0.1)(1.8,0)(1.8,-0.4)\put(1.2,0.6){ \alpha }\put(1.2,-0.4){ \beta }\put(7.2,-3.2){r}\put(7.9,1.2){s}\put(2.7,-0.8){P}\end{picture}$

$\sf (Esta~imagem~n\tilde{a}o~\acute{e}~visualiz\acute{a}vel~pelo~App~Brainly$ ☹ )

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☔ Sabemos que ambos os triângulos APC e o triângulo BPD são congruentes semelhantes já que ambos possuem os 3 ângulos congruentes (sabemos disto pois temos um par de ângulos opostos pelo vértice e dois pares que vêem, cada um deles, um mesmo arco). Por esta semelhança sabemos que, pela lei dos senos

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➡ $\sf\orange{ \dfrac{\overline{CO}}{sen(\alpha)} = \dfrac{\overline{AO}}{sen(\beta)} }$

➡ $\sf\orange{ \dfrac{sen(\beta)}{sen(\alpha)} = \dfrac{\overline{AO}}{\overline{CO}} }$

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e

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➡ $\sf\orange{ \dfrac{\overline{OB}}{sen(\alpha)} = \dfrac{\overline{OD}}{sen(\beta)} }$

➡ $\sf\orange{ \dfrac{sen(\beta)}{sen(\alpha)} = \dfrac{\overline{OD}}{\overline{OB}} }$

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donde obtemos que

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➡ $\sf\orange{ \dfrac{sen(\beta)}{sen(\alpha)} = \dfrac{sen(\beta)}{sen(\alpha)} \iff \dfrac{\overline{AO}}{\overline{CO}} = \dfrac{\overline{OD}}{\overline{OB}} }$

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e que finalmente nos dá a relação

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$\large\red{\boxed{\pink{\boxed{\begin{array}{rcl}&&\\&\orange{\rm \overline{AO} \cdot \overline{OB} = \overline{CO} \cdot \overline{OB}}&\\&&\\\end{array}}}}}$

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$\bf\large\red{\underline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}}$☁

☕ $\bf\large\blue{Bons\ estudos.}$

($\orange{D\acute{u}vidas\ nos\ coment\acute{a}rios}$) ☄

$\bf\large\red{\underline{\qquad \qquad \qquad \qquad \qquad \qquad \quad }}\LaTeX$✍

❄☃ $\sf(\gray{+}~\red{cores}~\blue{com}~\pink{o}~\orange{App}~\green{Brainly})$ ☘☀

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$\gray{"Absque~sudore~et~labore~nullum~opus~perfectum~est."}$