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_tmF9vKLRU2QGo-6lq9Bu0CZO1cjR69St6st8-IP3R-Hmv5Ph9iu0FxD4L8qyjOgkw3lPXNiNio_oE2TQ=s0-d "ABC")
ao lado é isósceles de vértice ![[;A=20^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uuuwC3IMaMSQtgm7UeIBbvlmNKV3_13vvpnFmQFWeHZRAASiN0LQTvJNpX58jaSbAvwxYUsoPT4VBPUgXJNmQgQsJIvvlOmlecjpI=s0-d "A=20^{\circ}")
.
Sendo![[;D\hat{B}C=50^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vNnndP-6GuKs5okAmaTfoep7BWktB6YjpXPy5STH8NUDQgvafiv4u9HoZPejfYmn6y4MAyqQ6lVtjImrPnuYvtG3TM9a_B57E74W936hYHvhZIwlEp-K7TzIc=s0-d "D\hat{B}C=50^{\circ}")
e ![[;E\hat{C}B=60^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sw8bBLzDLaN8LIa7aQ4tmzm1PStwrlOHQk0S0gKAAzHsa9PmlO7S88u8M3H5RNckJhd1A6z9fF5zUSbdK3rKXi8im1ou7AHzk1_S1XdbfFPUSWTn6cQHwGXQI=s0-d "E\hat{C}B=60^{\circ}")
calcule ![[;C\hat{E}D;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uksNM2IPrRTP_bSI8ITMoOzc4UYWhv8gWolVBNXLqOkXzK1dwJMfDJv4z31Umb6YZfvHlKBLEFGPa_9BKL3IBEfHC8EfRxYg=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_vEcDaqnrOmv1alR-SkILklB6rubnfANigW4Xhr5J5ztpciJ6t_Sy-0z-OZAtbvCXD5qJpjEep4POGRdQ7bFjJkZa7xm6fcjWiI=s0-d "C\hat{E}D=x")
então ![[;C\hat{D}E=160^{\circ} -x;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uoPyjFuuAicABqjema93RcGERC3Fv2a-Xx93-geoskaESfrHuS_vG6g8e53J8ztZol4dv9sKEPo366Q2Ki3vVrYc0FaxuFzCW50HLWiIyUNSfDvlSSBwkZVjFU5ByxdeE=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_skuqppQo5S0YRN8wl03n3AozT1ZxgEoM7nc4Mkc1JLsn3Bxp3ZcVoTYgkBHVJQf6zo9FWTK0wrvscTHw=s0-d "CDE")
...
![[;\frac{CE}{CD}=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_spfEmtTu2LLPgHWRLCXbCnkDycH_eTTKefhJZS9uMFTPxczI_B1OC6PhIt6Pry6mYfFLUr5vPCxeJQY_2Dr_rA26dQ-IcBZAiQzHD8uB91lg=s0-d "\frac{CE}{CD}=")
![[;\frac{sen(160^{\circ}-x)}{sen (x)};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_urLCmCJNrNa3nom8QdS_PcLnREYTJgFV7PM7T2Mtr47J8pQfcjBZ_UCO0_tow05LUcIEkIw1Zsd4BthpATQ7k9i9tJGt2VNsr8VNBv9OVwDtWDyOIc_n2HbXQHTcFqQnWrfb3MqH9YK2717nfkn53z63iFxj7e=s0-d "\frac{sen(160^{\circ}-x)}{sen (x)}")
(I)
e no triângulo![[;BCE;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vjS4uL6S7HtydS88lpx9yWvVi8J72ZgZ2aVU5MnTe1QoPEgC2GdvEiq1LovwdkJiwzvk2OyTn4pmot=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_vLJPX3ut87J9sS5MDvK3hJ5SL6PLh3u2BGPnZmY4zhUVjcazd0FXJBFzu06hvpS8LbmFkJ_cN1Ul-JTj2UpTCsalf0zyTb9VFLbsei9VqOiLU8u6tOGljBoTnksgxWmFK4xFEjpwbaQxqPS8TSFJBKR7y1dHFsrqFT71tsE9dys-IAuWNKuYRkYlaKgWVMX1jqC5KRPn_rgWHvRv93URUFGvvF-TtgpnznBtTvMJHE1tah61fUfmLDi_2mRS36qzW9pxE24dJYlpVk00Rwwzc=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_sYpiGCZgZsyDMPMFmm47IgcLt_NklK9LHEY-p0aA113WeMFeJd0aE4K6B7erndcRoeK3jpTzeOCz6RdOVs8kfuVxxjEVINL34=s0-d "C\hat{B}D=")
![[;C\hat{D}B;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sCDOvpHVpKTBIMDIABlU2p-Z1GbDhHLiSKkd2d49lifk58v-tmFAwmW6Z4HOZX6JvwgJ5cQmaBGE4lvUit41yvqPiHQx12Aw=s0-d "C\hat{D}B")
![[;=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tYB1fppqEXyPvksvd889lfhekiwffUpU9OquePCieYuySlbJZVhOocmTnFsFTMaC-82toaKjc15tA=s0-d "=")
![[;50^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tNpqJAf6JzWaQZY18copu7wK9IfifapBremex0y6ZHleu57Euw43rM4f5jk3cHK_JpM8oE-OL-2xyU1B-DT_VHzIwvoOY9Lx_2=s0-d "50^{\circ}")
então ![[;\triangle{BCD};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ub0JIpkmauSk7GFnMUxaMoK03Q41JNoBqdKKCKWS6VOfI1PuCvRaWMVZA5TnPNxYTYpEW5CaqiZkIUBdEcABf5UdcRRM1P6j4krNo=s0-d "\triangle{BCD}")
é isosceles. Assim, ![[;CD=BC;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sC4CGaKVia9k6BO8HLmy5qq1H_3oi07eezPLdC0GCPEgKnyfdPaqQ4GMNkZo3IctE0yA_ph9Ppzw-v8hI=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_tildOOl0lTFohiKe30fGhnPiWQSgsqePPldFvxpSMQT0wHwGQeOU33_RNEsu5hoCOod1Xfyypk4D1itfDVBpDtJFNpFRrpy2Y2Lr0xzmL7Ak7CKm3CVtmG7AYpJ6T1qKsK7sqnsY4aPp2mS6tJvyD766DtI-Qx3dpK5u2j3navXCRXxFFZIlDLHgePhh1vv0HenbJL7J_W_n4YSW7R980=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_v53FnaHloj0aSuYfu_dvGBmQS7RShXAnREx8Mwo2IzRi5c8-gL_udSITiz-myLp0kytmo4UGBPYCHYY0g9DlsRm9DjzozAoPJOw1OEXnHfCPzDnPUMKqUCcTn-7ZKWepgdOybLwuB2x1MvHh0=s0-d "sen(180^{\circ} - x) = sen(x)")
Temos
![[;sen(160^{\circ} - x)=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tzi4JXzyAEhBcld-kOoAdaNL06TOx2c0EBGvq15HPiqLw9jxdhfKSFTU3ApDm1NcYqz6v1gVuCClQGkpdO-C8XrmMMJ8jzKha4p2Z3mbI4puWDqEZQ9ST5SXY9Q_o=s0-d "sen(160^{\circ} - x)=")
![[; sen(180^{\circ} - (160^{\circ} - x)) =;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t_-HWxethh3GcsCzx2Zl-9PLxuEZPCxFEg47gzmLWE_3Li3JEDm4JSQxfjpnm4Cql7Jo8THuWC1TP9-5ZYHHwpLWG82XQL0sSNokVZ2J6ty7YS-9ecJD1uh2szo5sd62skQOT2Z4cUFX8TA9M9MOze0lP-ECqaUGnpUtYDuWAvgj1kEA=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_tpSGtX_9cj4BGYJIKRoO2surdft-poPrw65j7P3RE4OTjLRWyrXAHajvhJbeAly3s4rg-86zBV7XodDXXxSz_oTiHaGD_FkonSpwjSree517ZvjRbLH7H663l5pnxmw9ydj_TEI6h6YE9zxan-FPZXH5sL3QOVGJs32aEkqIQiW6-siR6wMCWIMEhi0EIz0dBPB_WWjxq8zNvdro3QgdbewQEW9OHs_399Ya5hgUYN2wBGFnVM3DOLuxEE_Bx1lZrBFXvADcZq2l3sWBj8SVUQBlr2gtnwJPne3LudGBZ6-2sLQ9DBa5VVSWJFPOZ9-rIcqOu1=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_uwI-oaGEr7uC_NKTxp0tGqWsodtVIk94W13jMClBRv1eTzRxbZygYakBAypK7GLhl704faavqas-5p7NIaGYnJgSBYbGtj3z5YX3vOC2XDkLrbrBeQgbCOetgHGeUFrneAmO2MEWcAXLRFDsFnT79Ffw=s0-d "sen(20^{\circ}) \cdot cos (x) +")
![[;cos(20^{\circ});]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v4g1lc30wIJn-0qBaBZdXV8LaGh2ZiNcl5mGhQhjjh6fONcpYPFoav5wBWNN6oJQ4qFSPzFNJt_QbDss0WvIM4vgT4ZBpFd_OIKouWmDWksv1spw=s0-d "cos(20^{\circ})")
![[;\cdot;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_syIn8oX0WaY6RW5Umk7EETBw02skEn5luAZoUcD9eP08RkMZE6Gg9ISOiazWXeHAZvnwTnrz_wWelMBYaxQgY=s0-d "\cdot")
![[; sen(x)=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u6vSr3ql6nM4lxVh_NEtGhR4q_jkUHBkh6aOCR4lEbHUA7xvb4fSXxNuGjm8GzhwvbP1vH51CrdPMXpyPVP57qCVjCW0y_=s0-d "sen(x)=")
![[; \cdot sen(x);]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s_mPtn8d1-cjOL5iV8853_P3rZZJ6C5UnkY_r9BkWz-d9baoa47bKQC-TjzGCnfU_zFCAH73dIMn_EK_X22yvXRIUzcmlzCRQS8LdX82A=s0-d "\cdot sen(x)")
![[; + \sqrt{3};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v_u9dgIZZwZKVOUMOFwU1kNTUKA8tO3Wlo4n8sSHfvrMhAex0thBWmF7b5jfMnpzNamtn3Mzu2aoQcBcniT0BXv9vZUacDYXVPI41t=s0-d "+ \sqrt{3}")
![[;sen(20^{\circ});]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u10KgtS3KeYw-0BaBjsLjvutASmBtxQTMR_R-D2yglJ08QTLLzd70WzshLJpt3p8IE88fQWcIuNa8HVSRaZ0p6GBakHicz5-Z7u9_WDe3A8Zh_2w=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_sTGPIGuU0T2320Y9mjOAGA3L-IkspDBjjIo9OPD0WY31KI7jjtVw9LTplDpJ2XtLIjrO4hOXuJhgAfs96V74aAPD8BfMyG2N5Y3Wm3HMsKtULnAaxZmrSLMuPGztitP2nBnbfhLnmPBuBVFhXptLjA0A1YzaBSXGdvHSm3_wxcn5-lNj-vYWv3pWynMM1fpjHI225XRbnrjwVNVWRbrM5MAuYe2hmcSOW5MQlVvy0cybUS=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_tcMikA-oKhKbaeYPM577qaWjjPqSbt1EZ2fnNLCacdny2aFEiH9KM4t7gq0e-a6fxY8D68fyOYtoIpz_GRRCW3cRngTE2A9VV1kXzHiDZkD8U=s0-d "ctg(x)=\sqrt{3}")
daonde![[; x=30^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t26uQpAembA3hB95PbfuvVF_wW5TTon2_Y3JDyOHW79Pc7jYIj9T19Ja_FOCO-hshY8878pNiM4N_yJhCFmkRwGvd2RZfuUEnO0Q26r0-q=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_u_J8aZzi2WKTRr3tVRGFlDqZOmpjwobY0PN1TmCtArCpyX6YgppNoh8tr7M9tDwsnl8wdudzc--p_Ax8HAFZ3hH8eHCfsF9VEeQwVM=s0-d "ED'\parallel BC")
. Seja ![[;P;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uK7Eiz4dhYluhYgc8ffwzzv34x_dWCo1Gw-wuH6SjRO0BG_-utxS6eHYml0o8A55Eu93ZF_Y6AyC8=s0-d "P")
o ponto de encontro de![[;BD';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tRD_-KZvP75FXF66c3OmFobeZr7qxvgodXuB4b-wGNMbTM9JIpLh3hV-5HDJzP8UAsXkHJEsbzavfw0RlG=s0-d "BD'")
com![[;CE;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ulXAgksrt6K7c5s6qZZ0bylMIFB30a2zEjMgBB1b8d3RRsFwah0vQlMVvJDK38G1vQ5DLkRMbhaKUU=s0-d "CE")
. Assim, ![[; \triangle{BCP};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ssjjynn47IcX89SP3vNDWQ7BOMY_riD8mVYBeHQ1mlGL5ImSfDlNGWZKrXwwTP3hl8PLf1P83LeUoZv7EfaanlgGXOQHGEeC7hOmWeWyMq=s0-d "\triangle{BCP}")
e ![[;\triangle{ED'P};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s2bijDdcb_ceaaGGn5uSD-aRuDgFP3WRCZeVB8MmZ_Dt1QZox19rxNpjbjYM0pmdnLCc2ElNuRBvXXPMZ_jQm2ensBhRwG4aKwW_E4v8T2=s0-d "\triangle{ED'P}")
são triângulos equiláteros. Ou seja, ![[;CP=BC;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t5S_17O8Me4J6295Al7tHwg99LdT4gthB2lAHJuUTdQ2NA-Zqd-xkmvHXp4C9pB9DsKNf1BHdGnj5tyN6z=s0-d "CP=BC")
como é isosceles, então ![[;\triangle{PCD};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sYJLqROeVo4Qaj76e30oAjtFUd_8HHiVuGX-1jbhzpVeesmyDoFVzi9VBTzAjwt9TfJq6MyTxBtSIJn4ewHV_ixWKu-k01OXDfwGo=s0-d "\triangle{PCD}")
é isosceles com ângulo do vértice ![[;\hat{C}=20^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ujpGfzo4Gen1WNZHXkyM0_k7GN_x8F7S8WaT9FYAnF7FOkwsoBLYHfcFkJJXnBp3Htp9WqKrcFDOu7w1j-UKCoLCuoR4ya8eBsx-v0S83YrHTaJcnTZ_74=s0-d "\hat{C}=20^{\circ}")
, assim![[;C\hat{P}D=80^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_srVuo8M35GuLncNBGy87tXvZkmclF9fnJRZ28XipLd7pqqsdS-RHlDSBDiskAq4DUQ9ClKQo86-P69QnD-Be5lJf4naVdPb4ju-omiUOJvJuqyWD9_du2azfw=s0-d "C\hat{P}D=80^{\circ}")
como ![[;E\hat{P}D'=60^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sr4JJR7GcmEOJjnT_KEHx3N5uoVkcruB0zokR-nC70e_cERyOmAMonIxFfUkKfYDz6PzUt5xmUpWaDudMVFuYNE7aO0ZcT1sWSTp7ICTi2ihpr7Mffyd-NpIg2lWI=s0-d "E\hat{P}D'=60^{\circ}")
e ![[;E,P;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_upZRtt3vmvchLxyAcmmzwKfaSiOQvnT58ZJNOngckcQyH3j6jNURtPCsvOOmknnOft_WjCjz5iuvjRgw=s0-d "E,P")
e ![[;C;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uFGjubKx-inm48oa8jzXUgIf3ceGSRUPWgtuQhC-K0LvOVneVzIjw3BCKwGT-fkaE8NuVwtryBdg=s0-d "C")
são colineares, ![[;D'\hat{P}C=40^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_saxY29YUvcMgJurXqOnhSec_M_ST76dhzk1xcn1Ei-LKWIC4ckZ-e9oLlEwj3B3DsUmcR-Cn5Jo3vWA19XsIM_aMf-VeKAOS4Cmu4fTKJqisMD_FWj0rCHtjdtCHw=s0-d "D'\hat{P}C=40^{\circ}")
.
Como e ![[;B\hat{D'}C=40^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tpXRIRCZhUmee9TNgblmqCLOkdkgArIJhcOVIHyk3-JGBie5ymtJJfG4Swvz8VGyzCivGyLctxk6_CtR2p0SKhdRSoT8bhIJ-4lG6cu95h2Ekyptq1z6qmDXfKwMk=s0-d "B\hat{D'}C=40^{\circ}")
![[;\triangle{PDD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_up08EbmcMmvLxlf3CTYFYeXblLP9C2IlK1TmZrJCTCVHsQMiqwlUweKL67NfR-SHy65euvQF4gI9gDxopm3ssWkbUUE1aUS5q4dc_yeU1q=s0-d "\triangle{PDD'}")
é isosceles com angulo do vétice ![[; \hat{D}=100^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vDsN0GfcaLEvfUjEEQNT2pR81iVsXmQoYYoLX_bOlLyCPnpFt1Gokx7YNjLaEp55yLCJ5ViMfJicu1oh-H1zxwnHNVl36SRb9P074zK7U6zRvFFscmoB9Yi-PkaQ=s0-d "\hat{D}=100^{\circ}")
.
Lembrando que![[;\triangle{EPD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tq8AN6U0a2ISt6jwrHoT3kfAia5tasq1yRU8B0h5Oosvvwy_G2pxWxbg80Esa3qSe4-Fmf3yCySia9Vx37IWWRwNc1rQLQxItZrtgWky24=s0-d "\triangle{EPD'}")
é equilatero. Temos ![[;ED'=EP;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v2u8Xo7Rcj-2_K9Y9WXPL9CLrHANnxm1MCW8G_fiMGbBhdnwSTNVCTbZilrOj1L1Bfj6euk0ZSOwNQ_BR0afPG=s0-d "ED'=EP")
e ![[;DP=DD';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vu_B3T4HKzM8fxZY-HSY_5vSD8_prusd0SVFaWS6aoSoJS356Ggd9s4ksohDdSUsr2BMJrYSZaqEmHARm4dBEM=s0-d "DP=DD'")
assim, pelo caso LLL, ![[;\triangle{EPD}=\triangle{EDD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sVWQuMY7BtfUC3KLwd-Y_D4W_ABvU0mDYbDo-EnIZSqhn6vXtGuaN6y6zMkhV6VxAry9QTUHvaccO4ttFBoNa74Gu5WwxxZ3HY2AqaArHM-5OPCljXV7hmNShVs_hlQepNpWOu=s0-d "\triangle{EPD}=\triangle{EDD'}")
. Por isso, ![[;ED;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tG1rokJZkwnn93NtDdJm7exYG0JNPlUYmr1K6SinmUB8EMEujPic0gK86HqSHAxEz8-LVlXXlGHvof=s0-d "ED")
é bissetriz de ![[;P\hat{E}D';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tRVP2472qLL0YWCoOnuP0K5wahnx6ebYdesLPUilMpcICNBeOI8lLSALcf77FaAYbcMW_J_kNyJLVeFzLXj7sJ8oHN2raTDHnBYQ=s0-d "P\hat{E}D'")
, logo, ![[;C\hat{E}D=30^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v77n2IE9NaP19L_J6vPWj4cpr1p1GwKgEicbRLaw99Bxh6pcuOnpde8SYd8rBbod0AuONvjj_5iCMP74tq5WR0YcmzSBYBI8qq3kVcu181FzDduPYRLdIneA=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