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_tH1uec4aDZB8s7GSlEHejjZpnQOihhkyO0Twz9_ugSXkdqvZARnDJYRGOEvngRvlJpAIziup4BkSXHPA=s0-d "ABC")
ao lado é isósceles de vértice ![[;A=20^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uLHPm_jSp6jTUXCPfNnM7DFK2QyBe3G-A3tOpfRAN2ACK16GZ5C13MbMCKziLleFuG0eYAMEAJ3BWPCTFwaXGoMYU4O0u1urZT4zI=s0-d "A=20^{\circ}")
.
Sendo![[;D\hat{B}C=50^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vw8XkbBfF_--N9QOctETBW5Zx2-7CVqcCeVjMykMqIOQMmtRdf8ftNvd-sRi34_J-mMe_X5W0v8nFFnQKhZMH29XLtFjBokeC5WrO5Y9Iv-K6aT2Fw9Cuq4AI=s0-d "D\hat{B}C=50^{\circ}")
e ![[;E\hat{C}B=60^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t9_oBDSNW2eCs3x83VQGgyFCFUVAM5Mgb0E66hZqc4l79THm9H6_ojV0PrcmMdC98HM84q4mtpMK7lE3LWgexgspHNa3yo5_pHNNwurQmy6SVkRJKvDimjUZA=s0-d "E\hat{C}B=60^{\circ}")
calcule ![[;C\hat{E}D;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sWFnSRHgF5fXUFTVncv9jFXvh7vXEaGiQLynCLyTQeo7FNq8WiuRFh-Y-Fs8Hx4pUYssHnG-mDl_Nlb-yDhMSaki-i4EQxKg=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_vVmrXlA0PsOX-cP91rUNhRetdtyXkTRqZxFexQo9G3yPfKXBnh3V2fX6gRNq0NNJ8I7M3HYEeRu76MXdlmqBLf3rzBB4t4MEpf=s0-d "C\hat{E}D=x")
então ![[;C\hat{D}E=160^{\circ} -x;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u6F_F3pTBTCc3MttTW-u2aCeTE-YGp10TzJxhT-blgmR9fSrAvjRuy4Ye04QZfXXyF_iGCiRCAhlj4GpWnovDwIMVi4S-Aka6rKlJWuKmGEvU49SMkVbcuSAZpZG1N3KQ=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_sUD4sDpWNxB6s8-vclmejWSZguxZjQn95qskwClJ8IUmsD8Fj2sYXDHOhneVucFJmZa3JcVuQSGLSzrA=s0-d "CDE")
...
![[;\frac{CE}{CD}=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tLkxIXdOlNtBY7oOYffd-T63FYpRDmbnNl34OQMseUj4WJ4Qib-1Rh2MaTIVlA5ES2rKO0Ty1VSHr3Jhk8iBID2QHbTp6Ha5SPfclE5UsUSA=s0-d "\frac{CE}{CD}=")
![[;\frac{sen(160^{\circ}-x)}{sen (x)};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_stQ_7gN8YKRaphnxzeo_di_DlNX-0oumiAHxjtIb-HjtAleEmvNrftpE6zpQ9iiMQS-2I2FUNp8WKqiO8zC8qWmP5GWCiaNFBFgAV7YgftbssOqlCAXzwjTrnwU45ECxcAnJ4q36dH-o-d6oQlO2bKESoNUlI6=s0-d "\frac{sen(160^{\circ}-x)}{sen (x)}")
(I)
e no triângulo![[;BCE;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sERA3WDDI0nrAgNNmX-Fyr9NwxgeIFZ7u1-gdUWv97KB3Dzym9Roi7w7LqMzvepS3R4xbqUjqq2E7G=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_sxjXRC_wyO1XA2PpPwXlZIOZE_BM8_N_au_5IHoHpP3wRXCE7381h77dUc38aL4fUByWegBcnFVq23fbEfhKWauYpEDKv_pfnhMJ3JUzW2fLUSMoHUxk5wr0HKqDdL_9HSorC8iaIfflEPk6OCZtBuif7o61ywXyzodladxQf08LievHgbWNYBSjVmul00C-uXMbcw9Yves_jJ2Hk3CzrvCkQZHlAjeGFKNNMwfUeJc5EI6duTAJhKkAPMgKgSKB04y6JEQ6OuZwMwNrMHdgQ=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_txxIOt9oJmn0X3AN4C3xaev0PayVy30o-hzM34Z_NxGYVYv2RzmFtVRT8ycJIPoO2yNAClI2YyNLFQLb25gCkegGojbOQBIgM=s0-d "C\hat{B}D=")
![[;C\hat{D}B;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u1OHVd5sJPiWjGZjm5C8Nd0PwAJ6qFpK0E2eZayXk9JFssLZ50cvYWZ3cK9cVtoKUhJ1s6UWhZEGnlnYzEiFrJIsoQIt4C7A=s0-d "C\hat{D}B")
![[;=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tmhzrYRXP-UjV9GQ-GG8-M3EjaGjGSiNXGHUajXyGF15SAWboF6fsmznEaWmVPRUI4T7SKpZ8TMP8=s0-d "=")
![[;50^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tHa8G27JoJTV3KW4S1ci5S6rRS1a8vd1Pz70ufxYMSpIqMzF4vQJPpFCrSm6VzbOymjcFq5sJI24dLy1mvgxDuXi5IDyntCl_I=s0-d "50^{\circ}")
então ![[;\triangle{BCD};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vJXYUA3DzZl17lsN0UJY619_7KVMGKivfdSS_pFPIfhc17Ri-baiWLU0V-4nqj6ZIlIm2fbN0FSFlZcqahIcUOp69HLd40GIq21lU=s0-d "\triangle{BCD}")
é isosceles. Assim, ![[;CD=BC;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uA4A4wSUiKWSwZSZHFiO4Lg_dXuk64Eg8cJA4JDm0GnrYcKAj_6cLAlSmX9zDXkvm54KjmTlzsxoqfatc=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_uXuiTqIC_RTk8GFVvE3VeXxuOT3Ijrz7QwdR4UlM-wOz-qqsAN1uEKXJnOdnVBZwcgy-iW6cm3hxAjdYhfRZ_s1923JId2gOx62RUYSoM5G0CRV92UkGnjESWu76Vyc-mceB-uMahrSiAuiseOrykRzIZopWzkmh5wvonDbteHTfm-fhK5l0psjIdp5qyKTMXgFwwZCnpqDwXjhvX2yto=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_uCIyLAAW7KN-B-4tpmx40eEgjW1UNMBfpLZx_SCHFJ3K6tqn0MP6Rg8D9pYaMdBmMJUL45LCNegLq5z27TPMYXyDWpnFhWnf5qVrS8UYi8WWm-53gL4beZV5_MC-C8I5Dc3D2O0R2QkyLWo_E=s0-d "sen(180^{\circ} - x) = sen(x)")
Temos
![[;sen(160^{\circ} - x)=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u9_aSANCPAXzy4EouIB4U7gyeZc2rPYNpIBunqmn28utpKWJCRebLUWDhrovkxa5w8R8ADEYiuMBUhjJjH3kAVac9XC4B3n8K0qdK2Pa28A4Cvx4icU4kYHoKSrcg=s0-d "sen(160^{\circ} - x)=")
![[; sen(180^{\circ} - (160^{\circ} - x)) =;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vKiCf4XvOAnK-YtRAqfnfevb99FYWv3Duo7zroGCdokZaTY4MMYDbrYoGCpNZZnfo6yASpn9nm8Q2Ql0916idTTysJdY7gTQ2sHRyqW2YikJc9UtLw9x3CsM2a_sn5Z1tH__Di1LLqeiyubal-8O0UZCWqqYy-igY0qWIb5uxweuuYXQ=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_t02CgMHSXvK31jgijWhTopsAizkGDFQMi3TbQAm0jLiaFf9M3L8kCe_mw6_ALHJvpftLHLInl8fh0dcQiuU8cn2nIQfBUOZuai-dLrg-Qn8VJlMMrPLPR-EvPSfFfkvC_8PD7qYaczyx9DS3-qqn_scJrSonJZpNSWa7iE8-GKLpOGEdKmDra2iSWEOX0nQuxil1F4tQe0ndJbvQkIy8DblXxwE2QimYrUvrDiHii1nXnWw8wSYWz2uBbLFz8NByNzJu8kxHYxS_jmf1OmXFcs1jBwsh4mrErvKD-NDkwLWYQCw8hY_dEdB6nSOQ1opzIZKzVe=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_u6Cp3GGVNKdHYqpAaaqicb_jNY4TGuqfgkM4FIY1IvfJfdo5_NDp8I5qmyzG1DfDnTjfD0irCpRXaV3R4CS4qYDVF_cb3NsY46yB1DsaIm-EIw6kIeWt11nlUEDt09AbNNMjyIhlBW_E16RwSDm4QrYw=s0-d "sen(20^{\circ}) \cdot cos (x) +")
![[;cos(20^{\circ});]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_urFIK1u7FyEZzkc2HtYThGjAVfja2Mz8lLTtIAWz104zFCA6ph_qflv-qZg3FriGRhJVrPEgkFB83DJyu_AHvUBqcdbApfarVTRf_wIKp9DQHmRw=s0-d "cos(20^{\circ})")
![[;\cdot;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uUZNFENf9G2QrN8EDtru6XxLwt9V2hoQ27o3_ECL50WrHkZRnvlg-RnrAVDBMZMii_cLupCZeSbW8JX7Lz7bM=s0-d "\cdot")
![[; sen(x)=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s56sRjZfcw8qPfc-AxZl2Rb_dKu8hJJfBJDACTj2ujMU-Md4k60j1IqfWil4mZ7B-s-LWZc9xL5jJQdIyXty14goqErsRk=s0-d "sen(x)=")
![[; \cdot sen(x);]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tK5eS6VVI5z6BnCLqbcwTxBJC1o48HaxIU3iiXjPRN2dptEJk2Uz9SSebvhqdX5T3dTVZBMsbH3dtDZ9ADVXL01J07B91RxZUeV9C4GY0=s0-d "\cdot sen(x)")
![[; + \sqrt{3};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uW2Kpj-dJ92B9Qxjse6KPnXYQdp9DWE96vKyRhs-nZqsBnkGkK9TAdH1y5wxfpx6IQWNjJtry_LBCNbyMcFOJZul2t8K1HUWgn4gEm=s0-d "+ \sqrt{3}")
![[;sen(20^{\circ});]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tcObcwzSUPARBzoof_cTYP05kSHKp9-HTKr5Xn2a9lz45Io5uGRNKvtezuWB8EjvKf0d80rZAkfRAyRZEX5EVsAVUl53DCtUm9rBeB3JxkjjUGUA=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_tgTxQxQmElaK_xJj1Y5onPsRUxuGJYta3VqJTYIAzLnBxwe0En2tYuijkZXi83x9G0aKTyALnBCist5FAI-YgHXOQLRPpBu3LqZ-kFimSX2KSvCQBDisujC8_tLGnaTUU5TxlzOX6mxJ95ZFZm2r53h0sdNiNn1Xwhw9UO7QdGX_npu_rK4Dof2evn94Y5XVWoDj49Tkj9CAjSqmY4mxyrJhubxKzTqdYJUG1Le3d-dJi8=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_uOJCGKgoDSBSAOlMuynsXH8vhr8XVg8JG_xVAOXcS6z80UTDyP5NUykdd2P463tYSjyx2TElzgbXP7_v32xPrKskP7XoWE0bYZOl4NhSMfNxs=s0-d "ctg(x)=\sqrt{3}")
daonde![[; x=30^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sFIxVuaaBi-Va8DVJrRyFJ94W78QEY2WA-cPFIlYe_aB-8fyadD9q4YTP34baeRAmxExGxGInx5akQ9hDO69ghoOX3UQxT8nUkzkrIll-m=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_vS5eNW72lWoutzeMNI2VskygIIaJUf5h6K0IR3SOuqwnqVXbc3p4mTK0zdH9mBcsjmW0y_y8aRAr3TDoG4eFLtrQE6vMVu9OkvUJth=s0-d "ED'\parallel BC")
. Seja ![[;P;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vi-Re-86IL8J-03UjRtQMGnM-sDFQNmpvfyV9fpQbE1my43SNbyziUfbh8DxQ3g9StZYngLAZyan8=s0-d "P")
o ponto de encontro de![[;BD';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uFVKRSavGSccrDkJHr9Wa1fy5YjTUKHnA0hvnTZZxAVOaqSBYC8xDHZiVicXKhdUvBVbbuRJ_vofA-wxsF=s0-d "BD'")
com![[;CE;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s7mxq1NNntDCCnPSOfv5Hnne6255ccPlfDj85TlLx2uGLFUUfArn805CukTUaGrE8Zs4yEX8Z_IV9Q=s0-d "CE")
. Assim, ![[; \triangle{BCP};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u6ZboQA6z10sSvtFLviODTD9HNZdJdbM9M2MV_tKjbcdK2KNeTxgKCrvR-SmlRTg5p9M9UT34huAfM5HAZtHBoAlP5gzjvbsQCsR7FcE-o=s0-d "\triangle{BCP}")
e ![[;\triangle{ED'P};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uZpDOvAtmbTuiFqBSZ94B_vLpxVqSTbbEo6a8021TuaBiHlTv77ul0TY96s2astXDuAdfWgLSQAqjMBxT8K8Nv5xX5rPsPA9qecXMmzrxR=s0-d "\triangle{ED'P}")
são triângulos equiláteros. Ou seja, ![[;CP=BC;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sxENzVCx70OMjJNWLa7fiuqu2FKtFv87Z8qLNSSquRxY9z1KBACTR1CoRskA72htJo68ZDgbbO4_P8vADD=s0-d "CP=BC")
como é isosceles, então ![[;\triangle{PCD};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sUoDs1Mk9qLfyG8FiuamYbagoRAM-iHdVzhW_LRQ7pnLlkM_NhCwl_LwqdnaxS1sxOT1KPTI2py066j33rHHZcELYnMmW7MwtEE3w=s0-d "\triangle{PCD}")
é isosceles com ângulo do vértice ![[;\hat{C}=20^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v_SeC4U--2MWT-DIbq-4ykAHs4zHG5x5u2OXtiFbKQk9P_S9kziIy8_iaLGDYMMxRa8O5qA6Crn4tb-wapXwsFjlAhVTRCfpSr9l8EL0MNS0g7LMUQoZ3O=s0-d "\hat{C}=20^{\circ}")
, assim![[;C\hat{P}D=80^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vNiT0ePErz0nP_aPjVjIegpaZE3AQ2v8zV5XYDJyxWH51CC7pJw_xlX1aM4XSFJdM70yMbWHAinmXTZ0fTsCONJ4PjzOwKK9qY_oZ-7uZUPObyNKUyjOb_bJc=s0-d "C\hat{P}D=80^{\circ}")
como ![[;E\hat{P}D'=60^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vxLLVDWUf7xDU0XRpGLGZtO-GUvjyZo_geP2EeF7qyxuW3xsadCsnt6CFd6rJRhp8w08ZTjx3e95Oo8A37nf7japcBewzS59ci24u5rlhidGu1lMV_rRcdrlGvw-4=s0-d "E\hat{P}D'=60^{\circ}")
e ![[;E,P;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v5aDQVwhLmsaSV-jplDUWwyMwGKkTmo26Z4kJwylSWE3uq8TMG9z9AmGvUu8ZOYlpNKRiWmp6Xmc01FQ=s0-d "E,P")
e ![[;C;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tD4NDUcHG4u1T-huRh5wISJmaVcQ8CJUATyuC-tzmSic0FGE_1mT0L0wnZteUc3i5aeDXvmVuysQ=s0-d "C")
são colineares, ![[;D'\hat{P}C=40^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ufyJd13ll4vBVNI-JZFvbQSPQsiVCM9TM_ucP8qbdNB_dZGNhP_xvUCl0XGIfj5n9zSvvS9x5HhokEtInvZFxyXYRgeopRH7Kd8HHJbSRKXpskSXgwh6CQyiXrMNo=s0-d "D'\hat{P}C=40^{\circ}")
.
Como e ![[;B\hat{D'}C=40^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uQY3bexgR6aFZ4MO5f9Cot43JbA2VVrucloBgGYYpWgMfR8-JRliyOeQ7u6NlYTonVuSODSoQps22SpOik0xOnAQ40lZ4k1d2j1_MBW9DCj7qhzZP8bds5AaR3uWs=s0-d "B\hat{D'}C=40^{\circ}")
![[;\triangle{PDD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vguG6LnJ2OrrTzBCKA_efVHS58mw40q2mypU8-33QeQArI4VzVF6IjVIRqABYd9UkmzPFjbUqW_AM1-0dwZ05Wfk2Clgy1Q0eyQWOjg8gd=s0-d "\triangle{PDD'}")
é isosceles com angulo do vétice ![[; \hat{D}=100^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sOXzHJB64ryLOwSPTVVhSw71CYl1abU_jZe6gA3bMN2loLvfQ4sU0sDcS7Sst8rKStz0zYwi1NPXusFOzqL_2-rJZ-PQiPZDRICcoUuHOkgJTQBbFH2z1o-Z78oA=s0-d "\hat{D}=100^{\circ}")
.
Lembrando que![[;\triangle{EPD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vbYuo0b7Or0X4F5rX4eA0DXnUrCoAyMdgHPqRsPqBtf8bqvTPPX0vjl68e9tD9sR-HcD47DgAka8QabpBNL2hNthO6P0rDJEigyjf9XJxh=s0-d "\triangle{EPD'}")
é equilatero. Temos ![[;ED'=EP;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vWG_IQO1VKaSeDXjWj1zkN_5YiAOsSFEfTvDRvwsGEctKutBglCQQKf9e4pnSI5xgJxv6a6ATc6nK_wvt3M4mA=s0-d "ED'=EP")
e ![[;DP=DD';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_szBhYLubOXP9yhKAHjW2puSlRJXlaJTUQmxccwFR9qhE0qlbdcM4I228DDqrW4UJWwo5qnXsGLklABKY3UgkTR=s0-d "DP=DD'")
assim, pelo caso LLL, ![[;\triangle{EPD}=\triangle{EDD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_smfX7Kz4ZA1H8cVg8_SBzEZIeQEeKfi-xW33e3gAJa8figCmAW3BC6V4IInldOufgBC6SDMxDuf9pcnVfBCvDXSg9wNlgkzKwm3id1X5vW8XC3GR3WbpT0u-F_qTWBLtOy9FG-=s0-d "\triangle{EPD}=\triangle{EDD'}")
. Por isso, ![[;ED;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ujvYqyacUcuqq58K_HNphSUnCYUQYbeKfJh85GyNYp353DeUuZ6PSCK-06XZiQrKv4drvGntCgPCwd=s0-d "ED")
é bissetriz de ![[;P\hat{E}D';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s9T8MuOrPzzuufMo85sTHQQRzIy036H4bPJxcJ7m__oIWYPowVmkqfieKh7ybW4eNVxkSeJ00cA-kwUVjInBCKGtalZKl4M5S42A=s0-d "P\hat{E}D'")
, logo, ![[;C\hat{E}D=30^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t0lBaX1Jb3fyYMlI5qUbZo7-7-q-wIjaIVhJA1h2Xe2Mf2-IGOd8-1wMaYlH0AghKZ_K7FwLwshPTKCJEp5zGyuDbgiddOvynREvhzmPuhVjpJPlDOoxSeig=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