Olá gente! Hoje aprensentarei um problema bastante conhecido em geometria e algumas de suas soluções também muito conhecidas.
O triângulo![[;ABC;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tDkqLdj8fLLc4PRMmhvx_S4osGuCAh-7PkIDzO2uwtegJrc2OesjelFKl2PAJKZkO90dGkvuadAXFoYg=s0-d "ABC")
ao lado é isósceles de vértice ![[;A=20^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ufu0UJ01WqdLoHY0WwcZMuG8H1a3b-DZ6sQWmiHHrtSGup5i3c9uJuirGp1vQOjJX70-YN5MPBqV9XzfeFKJAOFHmdHnWZ0C_1nxY=s0-d "A=20^{\circ}")
.
Sendo![[;D\hat{B}C=50^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ueGwg9BWi6K3za6kAPB5k3YWrtb2BsKGbAi7b3uytnnAr_IeUbiFHL7ZlU9N3lB8gcjvrojKz9Tdf8C0qP7XDC6uapqOi4FT4lms4Hy_HIhy1rntsPJzh3Lis=s0-d "D\hat{B}C=50^{\circ}")
e ![[;E\hat{C}B=60^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tGN00iHs0ZSJliz0cWVkf3gwQhbps4BanbOlzNtFQ2viLbXN3eFa4atoTUoCueqLW6mOb4wLdf5WIgoiGmJOqdZOo7UEY0rww1kZXg9UqT1Apuw6C4t__H3b4=s0-d "E\hat{C}B=60^{\circ}")
calcule ![[;C\hat{E}D;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ts8d61ZnD2qm_jGkL4loeWWNer1L3Sqn4qaTxfoNTN29jLqlJIXkyG7TMmYdRJI_Y0fzj3qki8CUVvfbum0wVqifKZ4Kh1zg=s0-d "C\hat{E}D")
.
Resolução 1 :
Vamos usar lei dos senos nessa solução. Clique aqui para ler o post do João sobre geometria com contas.
Se fizermos![[;C\hat{E}D=x;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t5bnye1x2Ziwzx9S8hoeQ9Im7QQt5Lc53qRq6eVSxzC_RczHmXnnKJ5ggCVLZEjDSIl2iSfZ730g3oYGcWGZmb81Sq13kdpOmQ=s0-d "C\hat{E}D=x")
então ![[;C\hat{D}E=160^{\circ} -x;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v1Srph47O7vTnOnIFUlMHBM5aH6uXdm7cxwyo_VCR93ovLA9kkZanrhubDjKG7Lse-LmT1jQN_X6SywtrCG4csma4B-kDURqPU5wjXX5h-67DIH-0XLFsA0rfFsqGmh9k=s0-d "C\hat{D}E=160^{\circ} -x")
Aplicando a Lei dos Senos no triângulo![[;CDE;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vVG2xcroKsVYB2UpobBi-zTD7Sa3lKS8UAVhY5-kWOwKIN9s2LAqejnSohLoGvrCeum50-f8214L7qeQ=s0-d "CDE")
...
![[;\frac{CE}{CD}=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_smamBpebQLFnwgvXgErofcmXNx4YltZMyUv8LfIApRUUbJfD-BLYLCSO0MVh8G01rXnriVfon7dVCg7a3PEs5aEuQGNui8vqxJGcTDxm_Gsg=s0-d "\frac{CE}{CD}=")
![[;\frac{sen(160^{\circ}-x)}{sen (x)};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sURKFFnUVHzx8hqBJ7LgjJpbnt29N4OcmNLky8yGP0Om79H7_Yj5pY7TF23LNB1Cq9IYoZCRV9HwMPt_hXEcUA28Zzrwax_onYUM08hvCe30AiZiofeGmMLxm1ECZHnQPcV1Z9R0ItA03aq5KlALjCnFNS2wH5=s0-d "\frac{sen(160^{\circ}-x)}{sen (x)}")
(I)
e no triângulo![[;BCE;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sFmVjaivdpWXl8_4lyM2i_mz2hD1CxoGNNB-VsYmwhx11k22AM1RJpLp9KGIVcqZIUZhP9jqo_9zNU=s0-d "BCE")
obtemos:
![[;\frac{CE}{BC}= \frac{sen(80^{\circ})}{sen(40^{\circ})} = 2 \cdot cos (40^{\circ});]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vABnLZkrYhUsfVSPBqjSZ5dfIgcWb9x_OXsTdu3Vrp1vgek9R86Duvky6jnguGJkSLsFEHD5FVTi5kQ7d4L5Nj_A8xHQq4SwW3ms2QS5pItv8AD3KefP-aOFdwXKDdr8rTe8yX9R4D0GQpO437SCEs5nGaMpn5D5g4TTQgC4XJ5qozN0sB9MSJL981wJbW_SKFF_WhRdbha6Dx0U-vntZpshnIuQMts7Lwna6T2Ud4a32VJQq6u7b1izFgidtIvV7mj-8gXrkiIM8cTubeQs4=s0-d "\frac{CE}{BC}= \frac{sen(80^{\circ})}{sen(40^{\circ})} = 2 \cdot cos (40^{\circ})")
(II) (só usar a fórmula do seno do arco duplo!)
Como![[;C\hat{B}D=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_stYrqSrxP2_ZyvZzMhuttFbP5jESmXwc35-et_DYtTQUgFk-RJ2A5bi-QIJqSne4rexk0bB1DW96Ve2zYJwFgpqhyvAffrpPE=s0-d "C\hat{B}D=")
![[;C\hat{D}B;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vGr1fxmLVWp_ZsEJvlne4OnVqkzdL-PYdpyYo7whSktBOeTq7acSUo7-u5kTc8Ip5HenoHfeO5_Mk2Kg87Uw0RiHI1nmRzkg=s0-d "C\hat{D}B")
![[;=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sN4nM66deQK4lF7CjLMHyK__et9dHjWPWRJ2RToBMx68BP-432Ka7wUXEoG7wG45WZDiofspXJDY8=s0-d "=")
![[;50^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uvU0MqP2OaIfcC8iys_1U9bal1gKRU_20bhS6qy7IovBd1df0t7JWZoZ5qE8iL0mynUN8MTW68gJnrhKCslbaL1pu_fmdq_rzJ=s0-d "50^{\circ}")
então ![[;\triangle{BCD};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uLda7dnRmzFRdcvQjjbDBUSkqmFFRB1u555oChOBuONjHgQwZUpkSbh6GJfj8AP6REQvMhH7sWN6tQCv3bGQWMKiafuTOA0roHNM8=s0-d "\triangle{BCD}")
é isosceles. Assim, ![[;CD=BC;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_te1UlwTGtjIPy-HF39kOi_eWmDFG-Q-1lOjEFnKc2EyYyjPP0bwWj6gWUTuG7HmHizGmXTQMNdMx9YgII=s0-d "CD=BC")
Isso nos permite escrever (I)=(II)
Logo,![[;\frac{sen(160^{\circ} -x)}{sen(x)}=2 \cdot cos(40^{\circ});]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vwJfbwM4EYjAKnBAceOcTRBg7KBzsIDYo9BF0fxlOP6Dtv3wxkpB4iey2pAW2JooNA8Y9rCy2yMKxJHTlKnTWw80RJIfJm40rTpiD_fyd4CeQoJfSJ9beB8w0wAX7Fsh0vXoSoe_Y5kb17XLUdNu53qJTxFgBL-ARLdkp2J2q0A0I3htSPbu1veYxP6myeYcvTV3HcpQzTl6xWRd8Ke9Y=s0-d "\frac{sen(160^{\circ} -x)}{sen(x)}=2 \cdot cos(40^{\circ})")
Sendo![[;sen(180^{\circ} - x) = sen(x);]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tqXIIjjQb83bwxOS8D82M2alSPQsaUTcbRqJrb-09-iZ8AqRak3sCzFSgs7ZvvehYW72gM6o_DtykX58q1VGAWHwWAvUq32ckBBPTpEZujfJGs1-g7ceJU3lu3iR48TDQSTKX9kkGkP8PFBTY=s0-d "sen(180^{\circ} - x) = sen(x)")
Temos
![[;sen(160^{\circ} - x)=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s1O-ogxwwzLsbpWEzmPLLCvWJ0PTLHMLhlsjxL2TPn-7UjIIekZV1PNNvk9aCd7MNwz9QQ_DgYcYydHelgHLFJGHbbvbIsJxi3AfH8xda5I8GbZaZzSH46t4rcNmc=s0-d "sen(160^{\circ} - x)=")
![[; sen(180^{\circ} - (160^{\circ} - x)) =;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_stI5DfBwZE6hW1XcLgw3dG8rxe4iMs8UOxUQjeHD86ZiqgdwKanvyfZhUC7c6dZTYAaPEK1FO5bkJ7ATLzv7CcJTJGL_ZpMGNjw1P4E5NqeoJZkMmlopGVIrNaApCjiNUOd7k9P_L3HY9EJpHyrB4U_-IYGOE1MeDLN7PQkTnP_ZuL4Q=s0-d "sen(180^{\circ} - (160^{\circ} - x)) =")
![[;sen(20^{\circ}+ x) = 2\cdot cos(40^{\circ}) \cdot sen(x)=2\cdot cos(60^{\circ} - 20^{\circ}) \cdot sen(x);]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ura4Rc6mrZky3cWyS9u6IkQjhO82bhHNoOSkEIrF_NMiT0Yyz5NLLKhVhx7dSoZdL4QwOBOeYMCvwC2x-HzWakODJydfbvHVKtdCBzTZIFipuObA6_civmOYZDaJccQpQhh1it-t11XGFak_XXWYBg4g01PlXq8GKYxLSQELBszoaHAtpKiny4cut3FMVsoyRloVb91_0ixmubSajDbblWIqUp_Fnv0cYs94A4tpdgTH4s9fNVdQg-y6qd-r_XF2tdkxkKHVLlprOK-8Y3mDh5djBgHXk6h-xH-uRXBk-Hnme38iohXYYdb7lWmSypZ9HB9vwE=s0-d "sen(20^{\circ}+ x) = 2\cdot cos(40^{\circ}) \cdot sen(x)=2\cdot cos(60^{\circ} - 20^{\circ}) \cdot sen(x)")
Usando a fórmula da soma de arcos para o seno e da subtração de arcos para o cosseno obtemos:
![[;sen(20^{\circ}) \cdot cos (x) +;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tyKu2pDmdUaSJ09CAEzec5z-WdJCWFD2UWgYfLI_8SaQF4Njna6q4MM6SHqdaxbzZWAlOQObtUDaFLIpxxFE-Xcb2arNRC_GERmCiY_OKe7Bd4oK5TZg8eM8tf0jOdC_aWhqngZmTHXf43Z5_y5DXWCA=s0-d "sen(20^{\circ}) \cdot cos (x) +")
![[;cos(20^{\circ});]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vk8tSUY5HSAViFTwt7FM1ns3pZBFNd2j9DOYy3QVzrlJ4N3dZVVj4eMDHVu1cgIp_pf-SWbQUU3Jzr5mqGrRz6BgUCMiJnv9YE1E7Rljvyn7stPg=s0-d "cos(20^{\circ})")
![[;\cdot;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s6oXzXyZS6xToWxn7x5e_zsfnxYjNnOWpEiCYUTt776-rwlmx51EFXxk1GYOSTGBl5UQrfDSmQI-9F2yZ28Ps=s0-d "\cdot")
![[; sen(x)=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vt-CJrqKsxoeFkKBpy6icaXq3g8jcX424TE5jpny7pFwqt1Qj-sPcqsFnySJbYRIP-91_n0oHWH-NVTTWWWNR_3_9HKwfT=s0-d "sen(x)=")
![[; \cdot sen(x);]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tj8n391CxYAGKV1EGkmubU5UnJ00WsJyLXRkdtYJbARjWB8zUrrdwszd320xxAtT29sFLF4iAlBDHFu9Icy51_8YV9tznGDuVu5iePLdQ=s0-d "\cdot sen(x)")
![[; + \sqrt{3};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tYIYKbIBS-lELMh17vRk1got2Qx0rFXG69Bxi0XQ3OwKg3RgpLw1w07hPuaZbvH2hB0XMKDmUwaGr2bU6T3POtv8LOMik2Oiiy4JZW=s0-d "+ \sqrt{3}")
![[;sen(20^{\circ});]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ubXZcBYHR3HF_Xdgg_rUIVXWRmZd--knTD8dtmksaeqU3ftXXm2oEOl8xY3e_MFyW1JvEbPB9_5l47K5QOyip98rUlCAie72ZUmbx-eKxl1UQPAw=s0-d "sen(20^{\circ})")
![[;sen(20^{\circ}) \cdot cos(x) = \sqrt{3}sen(20^{\circ}) \cdot sen(x);]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sY_YpOWRQ416YhJMMw2Lz6IJMzyWiXIupheafuP5dcnmjIuol3X7Yv1si-ThvIovGFsyUhj1qDVe4-aPUCG2liDM09v5BWWNqFgnpYzZTwxcSk0XPLTcruNSoMHlaE6n3g5nYeAd-t-RIGmVdEFH0nXLZaJMo_cJ4i29fCfNfR0LaolyL5JeO5dQ7KRc1ELIQIP3YfXyNJSF2qL7lA0QMDW_ZZyY_K2ZXHs0UuGgVxrwJ4=s0-d "sen(20^{\circ}) \cdot cos(x) = \sqrt{3}sen(20^{\circ}) \cdot sen(x)")
![[;ctg(x)=\sqrt{3};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tr7nm3s55fa9zG6orQxPfy6RfbD_0pBlWxiz8LyLL3IaiOQGy99qhE6wvYBytClhmhFiGdKfs8o-D62rJrtJItuhcrWBAznMqHx5kuMgFEZDQ=s0-d "ctg(x)=\sqrt{3}")
daonde![[; x=30^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_umzkbdVkAgoC-MhkoNMFQ-2AN-eX4Pj2oxmXPJsCH_1gp5Vv6BX7-FYagApR82RUdLNhGWIWHeSvurlkXdt04dmi5-8CH3rJKjebQxYoca=s0-d "x=30^{\circ}")
.
Uma outra solução é obtida de forma mais bonita.
Resolução 2:
Observe a construção abaixo
Na figura abaixo![[;ED'\parallel BC;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t_TOFqqV3ORhm-M4pUYGx_byfe5h_VNk1TWlN_v-VZPYegA3cgCrlbsjAkQ6aaDmd-1vnTBHIZzBeSyxVLEWhUw5HX-I-wIRLQl8X7=s0-d "ED'\parallel BC")
. Seja ![[;P;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tIXYNn3rn5cWiBifD9ReaX6duiMX0NmZEqaYwEF2fSfYMgb8_f2ruSWuxADzkOjw1ll7iVhQD2u60=s0-d "P")
o ponto de encontro de![[;BD';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tfiqvVrLiiQRlhQ7MMdDbpsCQhwE3mecT60oBre2SBa7_0G7_vI4n0_X09SkkqrrCWrJwWHWpZ7ocMMO6R=s0-d "BD'")
com![[;CE;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tHiblV56QvKS7tyWsZWAYRRiF446AzNr0TRcr8_iXR0IHUUfWqQiM8P7MNNv3uDvRkvyxH3ryT_QS3=s0-d "CE")
. Assim, ![[; \triangle{BCP};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t9cPHrwclCxvrcdN6dpa7ViHp-QP1dAa0tsOBMpup-5a4spj0xXqYZYJKLt0upuYOBmfDcyVsfWQ01W8HCZBXO2s9Q2COXIoIyM6CXMCve=s0-d "\triangle{BCP}")
e ![[;\triangle{ED'P};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sbAjaYuqixDPZSEpLy_N49A43husFCiktxFo6Vw-hruGL4-EwP_iomYFo88AW6rt3jtghk-ZFfM9SvdENudZqAuZjWIzumqVBJMm246WeC=s0-d "\triangle{ED'P}")
são triângulos equiláteros. Ou seja, ![[;CP=BC;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s9kPfU2pQ5r76QXyXyFsAsZ0qJ_duCjkM0x-lqXnY9sx2-C8gqkzU5qyS2Wlo2gi9qcsaz8eX1pO7AKdJ7=s0-d "CP=BC")
como é isosceles, então ![[;\triangle{PCD};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u8hBrWW9zuDYvc_7M_gjrM3bg_0Pq89mQaysbd3eBnxw5kDXOZQ0ebkCknVEXlNuhWnogJNO-49Yo9oueCDrqyr3x4fxk2_9yfkXs=s0-d "\triangle{PCD}")
é isosceles com ângulo do vértice ![[;\hat{C}=20^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tEaIodFzK_WFI1oIa2Loa4Xgqxv3W5z5ncJkLilv7ExT-CjqwsAz5NsWpe1qi9b81EgAzIzh8SXhgu3km0lGmbmjEjkCsKUXZaCANYx-xlMGgt-Q_ydD68=s0-d "\hat{C}=20^{\circ}")
, assim![[;C\hat{P}D=80^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sQaKfnq_GXh7EaxnU9edR0yqcJ_UoQpB0S0jN9Sy9XtADZYlfZYSBjZxdHvyeHXa5DiyfI0K8rxncgZuCNuAD4HC-e91aDTwIqii2lpjDcKRBKp7bRCpSpouE=s0-d "C\hat{P}D=80^{\circ}")
como ![[;E\hat{P}D'=60^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v7L7pcXg0BC6-hb_wkkHXY9nxo83pZCLp65I6Of0Mp1cLVdAuSYgriVi74jPU0MtR5rCS9doSsKaN3-DRXjMSNmkd6sE-wHfn1rO9EBqRdC3wIFevTZH1X--u47X0=s0-d "E\hat{P}D'=60^{\circ}")
e ![[;E,P;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v5bTq86irjvJasNuQA2n6d3bQzRk5wq8CbHNXBK6OGsA34zvoHjj7O9tPWQJV3tQ5P5NVLpE8L-ig_HA=s0-d "E,P")
e ![[;C;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s35Di_mzT38BSNiV-bqvkpYMPCakBrNRfz6fGEHIhoxivK9O1_ZFNWcuJIZQbxE-IxaFg1IQ3zBA=s0-d "C")
são colineares, ![[;D'\hat{P}C=40^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vbeKN_8rh_qLy_lJwwUhYbZsWr5z9vCn84eLBd2UEgg1mcHfOfXTqIE4Cq06-G5M1ayB_nqrH7y5VFHPMcIwclqBRAoH5gZxYm0jcAVdAhvIUvo_D5hh8-pVKX3Jk=s0-d "D'\hat{P}C=40^{\circ}")
.
Como e ![[;B\hat{D'}C=40^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sZbkvVKJOaVo1qitl8Fmp5dqdR6knGwgP2xKG3Od0rqPK2OAJn4J2Jpr4wofZr9KobG9Xd5hVlW3I-HdyirMxyCzYPV4sU-Pu8GLq96JAz6Ks9h7LolBgRCLTK3cU=s0-d "B\hat{D'}C=40^{\circ}")
![[;\triangle{PDD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u6CttUteWWa-bsCIXuGHCSN-kR1KrNVz2W6AqOshA09nofaJwlXH9AZlfhFpEFuDX44Trq7MVf-5s8jQi34lfRpuHxSz-3qLJMVQXGwrwb=s0-d "\triangle{PDD'}")
é isosceles com angulo do vétice ![[; \hat{D}=100^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vE0QPFREURMSUAPfSMyDqf-CKuOKK6oMDD7LixRjA482dUExm0kMppTiRqtLAS9R3FUXovbtPCvtSG-JgdtKxO7mooTV8F6opjda7A2XH9wHJqyuOxgrwXfMUB6w=s0-d "\hat{D}=100^{\circ}")
.
Lembrando que![[;\triangle{EPD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_srysLqsYbpAfKrFB-jd18CqKIOVNTwdbY-MSFyk5ePPR8xSv_4iKSxSXpCwqSEKQs-MMCU64Hbe2ktA5Gdmf-pZpSZENTpZbfdm0kd_egU=s0-d "\triangle{EPD'}")
é equilatero. Temos ![[;ED'=EP;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uX0f7pyWnIWBZUPTTvuJ-V9rW4FM4V7Vx0R9q9ck8x7pxCM6661ijbUDHXGf7weoOTUsYLmGlWRDH86FX1Xm2e=s0-d "ED'=EP")
e ![[;DP=DD';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uwQ9HWHYBPc4AS7o92FXlAeoDcg6nnhttSPHYrZbHri3ubp_VeKbTuauhmCUj6vO08Thcapi7tnKLQvdSrsAiG=s0-d "DP=DD'")
assim, pelo caso LLL, ![[;\triangle{EPD}=\triangle{EDD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tYd0OC52RY5HRRTiIo0x-dToWdwmo5l-mHh1QBZQhcPKMEH42UeNb4QUxkeSEgkXS_wuZdRs2nVDJX9zpK-GftnmxFZUHwe_wzkwJt8W9JL_zVKUlDP9AWsZJn8aAhAT4GJL-X=s0-d "\triangle{EPD}=\triangle{EDD'}")
. Por isso, ![[;ED;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vMrtjbfk3LiXzPbDvEFUcEgj56zi1xsec8AKFW6kByBCDHEyEu_LjQ5HTXNz-OwoBGP_FZLm9VC2L2=s0-d "ED")
é bissetriz de ![[;P\hat{E}D';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s6gaUxHN9xOGwhEPnyaP3oslk6o6E_8mAit0jAm6IQWabd0W0GWD7ySb78CiUFlBteDrdpCIJpPG5hPD5gU9jhtAZBDnYZOAVQSg=s0-d "P\hat{E}D'")
, logo, ![[;C\hat{E}D=30^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tbqQqmwPoG2t_rMjGzE2CG8aBVwxaVMJMjr6Gg9OxEK8DAScyObR5Kwvr3N6V4rTVHyXPgph-uBJwTzy50BA9wySa0k8G0MCLKdywT3atA20cj_17yFLglHA=s0-d "C\hat{E}D=30^{\circ}")
.
Há muitas outras soluções para esse problema. Por que você não tenta a sua?
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Até mais!
O triângulo
ao lado é isósceles de vértice .Sendo
e calcule .Resolução 1 :
Vamos usar lei dos senos nessa solução. Clique aqui para ler o post do João sobre geometria com contas.
Se fizermos
então Aplicando a Lei dos Senos no triângulo
...(I)e no triângulo
obtemos: (II) (só usar a fórmula do seno do arco duplo!)Como
então é isosceles. Assim, Isso nos permite escrever (I)=(II)
Logo,
Sendo
Temos
Usando a fórmula da soma de arcos para o seno e da subtração de arcos para o cosseno obtemos:
daonde
.Uma outra solução é obtida de forma mais bonita.
Resolução 2:
Observe a construção abaixo
Na figura abaixo
. Seja o ponto de encontro de com. Assim, e são triângulos equiláteros. Ou seja, como é isosceles, então é isosceles com ângulo do vértice , assimcomo e e são colineares, .Como
e é isosceles com angulo do vétice .Lembrando que
é equilatero. Temos e assim, pelo caso LLL, . Por isso, é bissetriz de , logo, .Há muitas outras soluções para esse problema. Por que você não tenta a sua?
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Até mais!
Belas Demonstrações para este problema clássico e, na minha opinião, belíssimo, pois seu enunciado é bastante simples mas sua resolução necessita de um pouco de engenhosidade, gostei bastante!
ResponderExcluirAté mais !
Olá diego! Obrigado pelo seu comentário! De fato, embora eu não seja muito bom em geometria, acho incrível como algumas soluções podem ser tão elegantes. Continuaremos postando problemas interessantes.
ResponderExcluirAbraço e até mais !
Legal ! O blog está muito bom ! Parabéns :D
ResponderExcluirObrigado, Lucas! Fazemos o melhor para passar o conhecimento!
ExcluirPorq n tm um campo onde os problemas apresentados fornecem as soluções?
ResponderExcluirPuxa vida, o blog é bem interessante! Meu professor o indicou em sala, parabéns!
ResponderExcluirCumprimentos,
Lu