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_vGD7KjKdhqbGG4SX5ois6nEzcLzPhuvzyIBrvNCOYZvTegyHXzkt5XD07L80yBvAYiKIEGoog9sdbdlS35_9owSWNoPQpdD13Lx40=s0-d "A=20^{\circ}")
.
Sendo![[;D\hat{B}C=50^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vnKjzlL12lkaYT-kEtNdoU1xGKXblZUOroi8Obc9rUf4_xrSycf3NgbTIC1BjAhj2ueWfl3-It9XyiuIwRxN2-MORPMFsQWHpVFt6yQT5K4_7WYed50iGU3Cw=s0-d "D\hat{B}C=50^{\circ}")
e ![[;E\hat{C}B=60^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s9ZHvW_8BLv3h-_dPm7AsOVslJh0MWxIspIw29cheUqafcNttx313Vhe-BVwh_xCAAnoL90uyOel861df85h4LoiZGGDO8Pg6qneu9-mhX6FI-NXlzJzwoV5w=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_s6AIjlfUDQpx7Bx5qJlx3HpLpn7D9hNCrnJxzgN4-5qgmPJ7nBHajOPkyxLZ4Uiu4r1cIdzqv3AYc3EzX_RWVbS2cdgF-gYNE2EaG03xGp-vZ4nqTKvMrDd9xoaNBgKbM=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_ttGsSxZX_nHfYfqLhqA5ix_V0d_W4wvbQ63VuYPDCgsRuJBiuMY6CcNWwZFRABFYUBfflvEI_azOtF3w=s0-d "CDE")
...
![[;\frac{CE}{CD}=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tILPhYZliqiHsmEfESeb9tQpnMODSoLlZb0pg1Dk5JQGVZrxx0uzctL0NPVMKPLHGcojPcKdJmWbcvXqHxyacfxq1uKYeRS_IUxkw31XbnMQ=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_v-V4KEdlHHJJJwC0JQZNe6R6Xoz0NnG3zJ94DLplJZunTveikt8kFMr0kw9Ep_j-VHX8MCLpKrslJL=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_v3F89abWA2qlcaHgr0hXNBPeEuCa7yGf2HdfvulVKIzSfbOhgzXu1nFK6GHiKdFT32G3S6u4e88WiWdsTZkpG8feA5aXQ6nr8=s0-d "C\hat{B}D=")
![[;C\hat{D}B;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uwMw2UOis0-kJLDjCxBpHXO53O2VjzsvIJu78Lh0Y-KKKAL0Ve6kKm04f5juhH8FWq8WpQwcDwhgO1K5D7CZiRq_fKNDolcA=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_umOt66UQmm5rBFyjE2NtfEtzF-aJd0ET-yOWp1d5Ux_ArJcV87p0m2MvZz9SwNLkAnSec_l694rdD0I-wt3uTIweuzyMmO1edK=s0-d "50^{\circ}")
então ![[;\triangle{BCD};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vsVfr2dBlE3mW7ajLe7-3u7411DZkP8tHZQ6o4G76zUr4bP5J6BduM7xLojKm-mBJKFrNM-7tPXuZiVUEeZbuSTmiDHPQ_ZcZjBjk=s0-d "\triangle{BCD}")
é isosceles. Assim, ![[;CD=BC;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sidpf8Xv3VIT7ODXz70FrfKn151N--k6sNObZhuS9aSkBF2oM120C0lzQx5bttCUAZL_r3BKBrEEIfkIY=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_vhQ0Ov4GVFondrQa9xnGuIj_2uaZHrs_5WTMM8-Vp_w-cjzpsH55PzV677XtVegqmY0A4NV0KlrdkzXvyK1lAFOO53s94njJrm_mNboQpjGXl2m-R8n1KPBN37SFKtBw2ifWOkDnGn1n5HHDLltqGCLPNDkn0yAVlVPtFj1QWHku5ru8izuDqY3BHhEgFHwTiy9Zx5kPJ3Bn8fPJPLClc=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_tUTUiOAvNyk_JUuYExDJd2lZi0UXMNVRTzb1UrrkSyckACVdlCugibFuIrjk_BSS3wffGBzCMEnyOpysTP1CVTNtfRoRbuGWp5i-JcGbivLzpw42Y_qdkOJhwEz7E=s0-d "sen(160^{\circ} - x)=")
![[; sen(180^{\circ} - (160^{\circ} - x)) =;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sWPjea6JtxmwcDztsnVwevbQ4Qtv-3dqgmgrNpDV1GJpp4-K_p8osSQ2ixEB2ML3Qqz1iNmhZtQO1fxTg2jMt9vp8amWRrkOvEcG1XXBoXuoe6gyawc7BjaBq7oveEJr_58GOy4UXcKyJvrQW67GWi66y4-PU_EZ-I_iUCD_TQH-Rh1g=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_vhSFwgf5TBYlzO0gaMKj-xb42AhwnDiqlW0bl-XZvLKObBwU6octLQHbuqX-RkVg73arzorgw66JfpocLDCn6Vx7W9I61pv5uWj_wrvPqtdLVvO5KhlKFTuGETtbb6AaMdSjLgw68y9k6rlwyTbZebOh-oVIdjDY7CUrgt_aR1lj2FWgJT4FkFHJTc8iRHaA1DP9fpgrn417jmeK3HFqDr6ZnSWXPvU7vfA8KrQXKixggXhA9-EPn_PrKrxzlPcdTqffA0bkQnd4jMgvMUHipWIY0LPc0g-deIGH5CsnicFf7DT2chffOCRuZmlXaf5_lIwCgr=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_vTux5tkqWBHxEbo35DPHj2uCuZmO1e0TVS1Z7NAGKszHJMnNiQHODqY5HXhZwTtc7uREzo9Jrci-6leqr8Db2aPu5dSm4sDGVatEphYPa317VHDQ=s0-d "cos(20^{\circ})")
![[;\cdot;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sPlAqB9Kgx6hsLVF0DptDvX6uMqOKLoO3SbkttYsWuKqAFIEHDqsfdNOEGF3qXek2ZnGTmVaHMLlRXpC_UNvc=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_us1_xKIEM-INx8ihwN3hI03-si1fzz-pdTBsFTjotlmh-Ux_Xu4zYx-R7HPZzd_5AXUs6larOdF6eKcfsw6Oi47IhTg7cr5dEaxP_6xQY=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_vmRPSX_SuPAyT3QHcOL8cEBWt9Wn82U63JgIBfQVC3cqJ3OhCi8woFzoTttbWUMNdN9OQYh_Hpf2nYePXPU2zvWk6DkPD4kiwlyvH8Z_f7etYgYeA3FGYgCWEQl8x4CZO6-i-tDYwzVwrTvohxkdzN09MrfhdsaG7oNNcUOv6HQWdcPUXhJHPqI_0dPyF2GlV9YXklqm6uOBCZIDalPNRL2hNTW2YPTarvHLbuc9AspCEe=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_sNX2GRYUAEbUBbQ4ZDq1gNqviBqc3RtMMPzcDFZ-0jO9GV0EMhaKsJsyHyeob8ySqdO7ccYCuxBDp4rCaKkjHJJYYYjaysYqJP3xB4ekPAurc=s0-d "ctg(x)=\sqrt{3}")
daonde![[; x=30^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uyMvSvMyFPWlNnSofssLeS9gu_JkmffRMSyetuEPn-QalI06CpybHiTqR5VCf6XLwhzHk8-3IGSs3VB68Gpjdjfp_zEiuHNppmMkiZ2KfS=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_u1K3IP51laoNzKW68dTrvoAcCg-fDngzuNKlbJpiyq-WmQWwemoj839tvAqepTXufn7W-RQJ0hWyA=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_vnNIKE8P60-e8tvPOoNVe8s-PFcLTg1WDnigaqeFRAcw-M-9uJEXPlKPkkodzUOeyx6cNMNJWdWVRG=s0-d "CE")
. Assim, ![[; \triangle{BCP};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sCHjAmZOVnjPnOZEL6J7D0Fik6RTpUi45xLWhZYd4R36pk-qtctwQeFxMrTeN7L8IKknF8WNi6a6ewtQNQor9G8hG7NdGM90CcXiPxyrgB=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_v69QMLZaUF_YzdcP0QMTnSJyBxONyOTD1MmYWyNP8a_8aEVrG82cbdJNeTA5pU393g1srJ90P0s_K7ThG7=s0-d "CP=BC")
como é isosceles, então ![[;\triangle{PCD};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v5il63kRBfR5cuNdALNUfz-9nDW0bqFbYUurG6qp2g-2x7mwqvgMxEtOOUOa8P0QbyMr074DKB7vQ9YILfAnC4YWGUcbi3qrLKPLA=s0-d "\triangle{PCD}")
é isosceles com ângulo do vértice ![[;\hat{C}=20^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vRnH0BwF1udTbWUlnvPAFitKfE0kOjlfKvhOnJS_ZBq0SO0jNR65Hn5_IoUsWK6tEyVnfRu4gPNEEupEOXUKmj2XfudZJgemVacAv4sppSsX-3EQIpSyr_=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_u8p5WjnuEk3fPOYa0Suiva9joGp-VmZRALT6N2sqOSIPGW7_y8veaek6juH2T4BAN3RagLRmgMcFERd5nsXA-3sIHvKZBuUllBep_RuYTHDkVtnbR4VXSjc3Z3x1c=s0-d "E\hat{P}D'=60^{\circ}")
e ![[;E,P;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vqFyXHcZmfN1ouuk8vyvmrBzebwqS9ttDQ6-RWkeKKqxbd8uAymAWIgunxhU_pwKG7Pfq0JYHWpwSBIw=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_vVIBWWuzd7YKe4Lixfj8Ub1PgTpad8EhbWLw_neeYJgq0R1tAst4eG4t91fn3Pcb691Z53xx3k_sEFC_cOy8GM93XuCjpThMIs1yWNRGliEZby8VgOhF-wzctWs1w=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_trLhHN51ss3CKa0XVpQGrEmlo4V9JlQwpaqNEHSE2NOYE1aoHJJO8jfW-HXpqQRhQ39anLpO5qNSPfYgaqoBNi=s0-d "ED'=EP")
e ![[;DP=DD';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vPUBuAgXiVrH0oWEIjWDhcb35QA8NOP1KcMSAAFJO6_zXhOFBYG2stfOdzxU6KINMlyUKtDbV7lHNP42CFiXh2=s0-d "DP=DD'")
assim, pelo caso LLL, ![[;\triangle{EPD}=\triangle{EDD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t_ydKNdaGdNGYkMrL6VDCbWXWLPmlQn6t6gk_GUMMqtOvGFW-tRf4v1u86YX-JdUdJefOL1KPjDkTkNrfs9sBwxKlX69LOCjbY3Mi_mT8LxaKZmvSWVpvQdincVtjqXiLealmA=s0-d "\triangle{EPD}=\triangle{EDD'}")
. Por isso, ![[;ED;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vctvz6GFZY5TGpINcAtR-c0omhahFkcjvWn6_6mEruUyJx0eKdGyDHK0Ksp29Y02U6x2oeNzta7kpq=s0-d "ED")
é bissetriz de ![[;P\hat{E}D';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vLI0xGxkeLib45N5Ceeuyqd6J8sZX2U_4cXbRuSQwrq8tuN_61Guhsq9zSfpPrtaUkZaY6hnx3wQjM0zDZdQtHMW8kWJiSrl8Xfg=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?
Lembre-se: Para melhorar a qualidade de nossas postagens avalie-as logo abaixo. Se você gostou do blog recomende aos seus amigos e se inscreva por e-mail para receber nossas atualizações.
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?
Lembre-se: Para melhorar a qualidade de nossas postagens avalie-as logo abaixo. Se você gostou do blog recomende aos seus amigos e se inscreva por e-mail para receber nossas atualizações.
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