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_vK7VG4i7tOhTt2vCAs6TtJ5qCaj9rHRJhQAoXL2G6jAk3gbBqTSxEOgFa8iXoU0GSq8VyGv1jqRaLjFw=s0-d "ABC")
ao lado é isósceles de vértice ![[;A=20^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_seGDVHholy0tRUv_OQxoyIpYLtKD8QFNoJTHipPsNJ5wa5tNrfDqFTv2YQSRg0rICSEpuRQOS6zVe96tnRDQvHW4gh8Sm4F6BT0oU=s0-d "A=20^{\circ}")
.
Sendo![[;D\hat{B}C=50^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vxdOstkT5Tth0M9zm4p5SbFdmPRdugHZT2pzmW0J1ew4ts75UoSniE5icixutC3VjyHATE_fORyN1pSZhea4LltxPm8X64q6w1d9BoKfVhnpPBBZDxxclU88A=s0-d "D\hat{B}C=50^{\circ}")
e ![[;E\hat{C}B=60^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ueLAiyKv_NrQ4hzXE98_DBPlyYpWQqRUnFWS638sBMsC82Z6X-OrSoPH4c41YA_i3GbvddxKtIG8eXSxgnRRukfHaSUBEgTzS2NnE0TF-1ltysLWFR3nnMedg=s0-d "E\hat{C}B=60^{\circ}")
calcule ![[;C\hat{E}D;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tEl-sWgHFSY70IYE1J9gWx39ZxtiBlNOz61ua9LQaF6IgIMxzr7neNtjgXE1LSyQmZdPOA6_uIwTVsTeqRGu1NZAbbg4ppuQ=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_t7peYvuQ_lR3cFFUCTYardTUlzj4JaKcvbEvqNPABaCriz9O_IGAtjnZPwysSc2H2Rm6NcNyznCDzvFXC7Kga7yh9hboPElwoX=s0-d "C\hat{E}D=x")
então ![[;C\hat{D}E=160^{\circ} -x;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tnN0BmffPD001C8Q_003l38vJJRVOSQwfd8NY_U4AIndgCDn-CH_tL7SpHrlQwIBQAKgq2LXi2KcoxynWvXGJkiHJ3mscvPcE6CTUnfVimMjZ6jBl0Ya1ZSQHdTt1NB6E=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_vrEES1wQxMM0XlhPbCV2Hym5iLqE113TL7vjRGbP2TxmU9rJkfBB36y-DXQttxVW6_gshdTjbjCZ_7CA=s0-d "CDE")
...
![[;\frac{CE}{CD}=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tQuv31KkuW7UWFDumIzwovY30K11VyhRqoiijkV4lg1MYg-u_nrTwJw9kS71GBfMc5lDbTk4BSgcgeF6-MCC41x7PLZFpeMUOwmHHWDy89Sw=s0-d "\frac{CE}{CD}=")
![[;\frac{sen(160^{\circ}-x)}{sen (x)};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vMQsB4fD2Eza-p18JriMXETdaNA6gj530iRDK8tRByNAwa_73hLLdG4V9HJD6BKsolDxPwUtc_DZcsgWxPpmiswQLmgYyr6LVmImEwnSAG-y57nCKGr_tfCzld6HYyftj_ScgGAt0Epm66MBfNNWaL5xMu1Ej9=s0-d "\frac{sen(160^{\circ}-x)}{sen (x)}")
(I)
e no triângulo![[;BCE;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sEieWhM7eCAe66v_YNzWkUgNR4tCBdN6ot9vHRoMjKfkBXPdA2zRGgE_3d1NTjADK6uQYbnXmSl748=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_up7Y0PJ3aAJ74NI3nl_oRKk2L8YPQeQbT_R-Jd402fTwiXGLBixR7A_Q2PyZcY5EUvPU6meIUcBBA4gXsO-s6Idf0zCnCGxVXR0MmIEgUMJxNcgN0xPuJP1Aiy2tu-bsxdLWz2nPt3lAxXaGI_c2A3W8NTIbhYAO5IVrmPsEci9p4kZiI6dYlD3jnJ3dBI5zldiMi9sQss6IAhVvI3QBAHUD3GdUgid9g2S0HAek--I0QkIg78KFewCsryA6IcBVI0J1n_ukoTW9y0vhs1Hyo=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_t0W9HPViYYwjUF6ewIYq5JX2zq9Z76MbkMJsXjEHeoAlpe50JLkbRSC3azk6acBiRDuCMgxbbjpuvSgTrTbQ9_D0JAXdvlQtA=s0-d "C\hat{B}D=")
![[;C\hat{D}B;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vIzUqILLeit5y-MVxG63XZP_RtHjrW5cUHLlUCHzj7FBMeVXtEsB1vHpRhj9jujeixAEBog231W_9DKk4VIoHIuqv81dxS2Q=s0-d "C\hat{D}B")
![[;=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uq-If1zDWaqgd0ZxDOSNUmvhaNDQ-tRVUGdgoL3WcFwe-wbfZpCpfy0ryDV1H_4O03El7R7lWEOms=s0-d "=")
![[;50^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ujt2L0IUNVEBNhO-6ZkFvCHVYqDr7x3Yrvoumw4AK5jPey24Ebbi80uOwA6Yyile1YhWiTCukOFoMS8sv2x3JhZUKQj4HlsgMV=s0-d "50^{\circ}")
então ![[;\triangle{BCD};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tt8jY_pQqwBIi0FwbgK6UhG6Qy8rR0EfQIs6Ea6Cc1kBRCXmykgXdV4lM3b06vk4hRtQ2HLBvRUDB_xClFldKllEhsGMmifCBsO5E=s0-d "\triangle{BCD}")
é isosceles. Assim, ![[;CD=BC;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vti_J1JYwHQTvTxadEvTTfWvz1zLVyRao1D8vSZS_CAmia0efFnt0TWSQtR8tWZ432s571HKgac93RVmY=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_tdDeHNZcXqHh6QM3AonF0BCey9kMaTlzHOYwwEkRSL5NbxBQ520JL9wCr9LYhOXUapaS9R-lQvlqtv6k4v8GhNbzF_dDTjT5GEEVQ2i4tVMs41tT_2zO9yk2GSzWehDsoVv7DnoZ7DszWaRzyCf12IlQAI91jkX8Ksexbfd30b9V3L5Ysc89DJM4CfSRnzp5wBlOUWLPcmGsAuhDEc9U0=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_scWTb853iMcY4JBk2TmkduhHM0nhRzSlU0sXRp_iO5yaJFfKgxx9UQJQNbo5U7dUJKh9aR0VFn-BmPoxwjgXPKxAPkrbidpyskjOMuWMVfUYtbg5XjBn8fIcSSlrMoozEQ4X0p1HTZSwXXNZc=s0-d "sen(180^{\circ} - x) = sen(x)")
Temos
![[;sen(160^{\circ} - x)=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u7WRqrTZ9ol7qCt4gGvufD_Ny6iiJKewt_pmdaVZHr4dA0YKvFgVfOsqcuaf1XgsEY-epKm1SAHnIMJsHVhu24YRALi4bIwQyVRaHhO_FRA2U8pKGRjr66Yf14igw=s0-d "sen(160^{\circ} - x)=")
![[; sen(180^{\circ} - (160^{\circ} - x)) =;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sj2BGgjqKjaUL38tZka1Ekfvzx9IinK6Y8wDGMvf3iklTxp6TEZ4GEbmDjyRHx7Vi2WX5STsaruUL0p8Gh7LzLxEQ0kOJOcn4lePg1XXfOHEv35Gc2RQUAT5wWQbg1OzUZ7nzLBY9avUSGFx9fHt807wQvf40aXabAs8wzY99cFkk3wg=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_v5PMpLoCEayAOFCr0lTD5UmeZ1aSW4593VWpDw7aFWgEIP8ivAtbuNiihbQQhRFoepV4mIokuKczvhaWsgyWH8mfqxvRR-9VRSpmRIO4ajtX3a8ChN9E4f_Dx-zJqAp0v8JfwDx7DCEHrHCPWpZdmCCV8QeQHnbyWUdYbVgKfuQzMstvJmA8h3hTr3K4wZUO4LCgZ-Z7RHn6mGSiKmkc2tLajIwKoVENGexuAOEsk2RkoHgxPB6DgrC2kn5Dk9C7LowZj1To2oP3jqqEO8xbJzJsYhJQgGR3qMh8grIqAFViFG3ANFbFEPhgAYWtvCZoF6v-MP=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_uJMiGc8w75Bl2AnYubj9-iWyhomborJoWKApvE1dXuMkXj_bPMXr1s0eY7adOr8v06WIz_2SjtFH8et1s4sc3s6cE3_vrzoBH8b6rI6MgCeylay8w_FXHTGtN2FQkLSuOy0tG-Ouh-F8uTUoAxxHvF2g=s0-d "sen(20^{\circ}) \cdot cos (x) +")
![[;cos(20^{\circ});]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sTco32HFNkgrYyjk54L2xEZkb3vZIcr0yJ3pH9iY8pT4a4Z0n4oTfNoBREkl0bbKaVSpcG0717ZTG3fTsCVDW13aG_zVlvWztCSTzbjrFLtlscDw=s0-d "cos(20^{\circ})")
![[;\cdot;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v4uhokWA6Da9i4uxNuKLOscCYA3yIQefYKlLUb4A7hliimbAOMxYOvcrYAVb2kDNUoNVEM-Kg3Ss9Yw8Nzbw4=s0-d "\cdot")
![[; sen(x)=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vRFRr4xyRKpMvr46Hgy0ADN1CCgfqErco_Fld4n_1HjxHWLq0kGvJDhyiI7KAxs0jLB6xKHHBrJp8zviFvcmgpeGpSsL2E=s0-d "sen(x)=")
![[; \cdot sen(x);]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_slRc128FD8qYGTRXCDft5uZ3wHR_zoRkmA3UCCUgjSbwK4mmlaID1YU-5Rfy-fiB3pVG3hlKBIKj15eJD01zxjWZyVjhXNEXxfk5OQ2RE=s0-d "\cdot sen(x)")
![[; + \sqrt{3};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vbMzwNIXEPCzV_n8POTeWeQcG04ll_mjF1bvHH9JF3w2omusRm3uA7Tu3-gbXh-hAhGpiFUXrIY-8G9afMAI8nH_mIjNQ55w8pLWoq=s0-d "+ \sqrt{3}")
![[;sen(20^{\circ});]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sYIHZN69si8vZ5lSB7zMTuMS5J83SFjrqdU8adeQtilXcZ0ayUlDMczGZdebMfiFnGtL9aXjTne210LNX5y9oNtHxbikxsuQ2_aFKe5gJv0V11Xg=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_vVF9UTm1ARqRbJGZppzE8TO0IlkcvMLq_QaQCPnN17U0ZS8NbA4P0EZo8b8IoiKF3g92mZr-gYXL5PPwjTuE3mRlaBym4Jcvk25HztmL04x0xoGHIuya-Gs6vOUbeAf-Rk0arIRQwx2itr78_NHB29NsNJWLkMiRzY14G3PqRhpqsXKMMj6Hvab6DLRY6jisIQxXnvG2i_uE7ZkKf2EGjlrUdDXZHItJdI0QfcJxdTbMNA=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_twDEjY4ZTqSdJ9zHUSKqZvn3EtLiXOuRM7E8TliDLWPyKv4-BS6iqf97AT1rCQnM4X3Vt_J4QW6qXxLPJycX-WNnUspjoEfgGfDzjPWEIaIRE=s0-d "ctg(x)=\sqrt{3}")
daonde![[; x=30^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sSFizUrkUHZpbxfs4dYo6vWyEZ_7-QXrMU_Os20xQmqXMvP8v6fx90cRxyTPLtaaJZRArSjtMCRTXxAKDhDkrAE8_e5nkjS0hZAIsPt5rA=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_v7wkQ2ETf5UZKnP4HqMOam1cBFVK8WlWftwTfcEC7Sf96JNWHui8BWrkk_trY70nmLs5OtnLWviuVwhd-NSAkhmKguWzpbthDFW1QM=s0-d "ED'\parallel BC")
. Seja ![[;P;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tU7FZu5q1Y3v2-L3k_W6CXS5Q6Zr0D21RexzFzM3TpBUQunaH3sbbZ8eUYN7ZL_Be6C6qCLFZsW7E=s0-d "P")
o ponto de encontro de![[;BD';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sXUUwOLTLExnKaJHQgup6H1FEHIdYNUGRSoSCSPSDtVnukvNyudFuERZgzE-s95Jakmfcb9BAsIHx2iaQG=s0-d "BD'")
com![[;CE;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tLuDT9TYPWkFzrqkwEpe_sk5YdoIMovGnOFU10YTdv-JyqkXgb2K401JlCdhN2Mb5vBdTxppY0i964=s0-d "CE")
. Assim, ![[; \triangle{BCP};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uhygXnOqkQ-57JACxLL-loQ5QXe0W8pzW9wDxsIT9eJHWfB6scWEqn17ekkJqszf-38EM-D62qyhMkuL4OTQi-EXDvzCwJnWP8m7Edn83g=s0-d "\triangle{BCP}")
e ![[;\triangle{ED'P};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tfjX6qMH8iap5Ez4qJqPKF2cm582upScAvctTwxH64xl8XXqwvV442OdDR7y6ITga_t0FZQ3IO-zZs1IWjCHGtVYG34SfQh7D7ajyML5t2=s0-d "\triangle{ED'P}")
são triângulos equiláteros. Ou seja, ![[;CP=BC;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v7sGzMVDI2Lb1HODwpzUNhyobWQKTZ9uvC4yR4NJFaSqr22EDcaz9EXR9cq8fHBTdDeH3k0fKnT1gj0Aoc=s0-d "CP=BC")
como é isosceles, então ![[;\triangle{PCD};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t103txFIxDt8B4pRI9HZ2MrHiLDX96X2XsyjjDGNsDbx4tDkgQ5TnEUGS1uO23mcadXkKl6HhdjqrU-AH9-WUl-eE33JvNoYpw10k=s0-d "\triangle{PCD}")
é isosceles com ângulo do vértice ![[;\hat{C}=20^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tbBQqiU9g-1tCW_RSNkiMu9E-T1CL2pXK9JQyEs-qXslKn7gCP98vYscg89KyWVUjzv897424wl2K3hvcknO-7DcvCAA7qermzol-tLSQ19Cq4l8R0gmXV=s0-d "\hat{C}=20^{\circ}")
, assim![[;C\hat{P}D=80^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v5th6cu6aXWVHWFMTIVkiUdIv9xFaYtwHaILen2hbzzYHdna4GYWIs9gaXJoodmc8ySjlOLsaeAQHTfOZQqJ9VWEcB3jUJE63TEVK8hfofRP7EUli1WGu0UAk=s0-d "C\hat{P}D=80^{\circ}")
como ![[;E\hat{P}D'=60^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tAtVFgQLkmzRPutQUSpbwyJ56ws1TRpRrq_no88PIEu5BEoSyZIt0WmSO8C-S7-Qkp0q34ni7-fFkRdpZni510rreS0nyhdbSAyYB0BMzXgKWExJMRleCxmI0735Y=s0-d "E\hat{P}D'=60^{\circ}")
e ![[;E,P;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tDyO9X1efogV8tScHJj2zghKK-74pLCT7LpefgziKgX-TvQFmsTIvXvUANGMZvLaOU3T3yMdcE6GyhUA=s0-d "E,P")
e ![[;C;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_varWQas5JOqlUYvd2U5cZSDIwjp_nYvslb6wp969HoXnIp-kUqumRnPfTsm8SRvkS1XyeOnZYYyw=s0-d "C")
são colineares, ![[;D'\hat{P}C=40^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t5KVuLwhYFOR6WfYrUnm_v-sLI2RtHAY85Edu3hScsUxVeQrMG_WxylwVq_hdq9qCbQh0d2Kjnt7vE5gz5DVeoD5J8Zkl75KK2NPBew8jRRYAP7rbL_4-m3f7y0BI=s0-d "D'\hat{P}C=40^{\circ}")
.
Como e ![[;B\hat{D'}C=40^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u_u0qecXDc6fG1UZetceokbeuhWMpBGzCZKaAt2xOOr-hKXoc9HBxTUVNBS30-bkAes-oV0kB_0qzvPB6EUKFrytRZxG_LQb7aSwY2YQXzNu8AoHq7NIS048oOdHk=s0-d "B\hat{D'}C=40^{\circ}")
![[;\triangle{PDD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sJEWdI5qqI35pEvD90cU9FmdqjCbJkIYqqmFUSteWhETsjXGeMKpVtcsioBYHrMne380Gi7D4D56ZSJDSYYqFiCcAYKekBXBLg15QPsz4-=s0-d "\triangle{PDD'}")
é isosceles com angulo do vétice ![[; \hat{D}=100^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uObNkIY2kGudDykSmsP0XjHhMipHr4Psq-7AFpAOhVdagXkKxwCe6q7OkAtrL7lW1TTszaC4LYCnpZD070H333GTqySKmFN8j4R7XnJT16vsPE1a7Pz8QETA86hw=s0-d "\hat{D}=100^{\circ}")
.
Lembrando que![[;\triangle{EPD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s7sVq7YLuP76x4R6EquhYlnhC_F8wDEATJ4IBN3f2LNzn967L5MRK-e-pUhJe2s-6AeyEBYhZgAVdorEAgzdDorN5xLCN81-PJyyYt3kDD=s0-d "\triangle{EPD'}")
é equilatero. Temos ![[;ED'=EP;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sCO80CmSxSCabmRMiGF_UmUnox32CCw5ni5FE2sMAao6Qp7-2LbMsk0_A10zQ8DZiXOy6f782OeLx7LRZ-xCS-=s0-d "ED'=EP")
e ![[;DP=DD';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uwH5JQ9AIS819T7_2ijI6Dcjg2UlkuZuixiw8FLAcVqO2D3Wf4IniYkquNYVTvnQmosThlkGisa-RweFlD0_ts=s0-d "DP=DD'")
assim, pelo caso LLL, ![[;\triangle{EPD}=\triangle{EDD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_stMb7Fz_rdP6Rh9-qUORq7ZM2_uO12mYRrcXoVXDbAcVLUqhKy7Tc42hW67IdBkYo0d_HGPdLyoOruj_b7OVzNX1b8dADtAPaDApqXBfRE7kODR6HT1YHaoCx_SdbCyc7z1bKf=s0-d "\triangle{EPD}=\triangle{EDD'}")
. Por isso, ![[;ED;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vqCDEKNGVpbD3OYEPgiC_ldmtUXnDPB9LGoet_7QABri5kY9Ie3UwnorErQiUmPYmXK-CFotoTAehL=s0-d "ED")
é bissetriz de ![[;P\hat{E}D';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vyv96NOx5hwaWVYTnFP-APMydFTWgKvL_yesqp6WIdL-OFN2s2CTQLdoruh5dDCPT6lTlhQkgpNMvzkNIDFaIR0ECmQH9VgOuyMQ=s0-d "P\hat{E}D'")
, logo, ![[;C\hat{E}D=30^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v6Pmtgti6XjtlfLqMtjFMO25LOV79lf1n5owbOr464BVLer6GzMRKq_dFxPL3haGAOyX78T0gY39wadfkUZSGrF45giVfcboj91UnAZa41ayL2Aj9zrMSNqg=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