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_tTd3xpK2XM3wv0GnicteidNcs9vVIrtYjfCBNOjuABQgXn3npOi-v5gJxwsGwLyJ0Q6-pw2gupxUcfOQ=s0-d "ABC")
ao lado é isósceles de vértice ![[;A=20^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uU5eReFeKWs4gpuHDVBmdz7fD8y-BrkRI20TxSjpVUDBq_TUD0y5ltrayLNT4QDAMFHjTOxo5YuRQfOF9P_ZYml8YjNiEydIfopUc=s0-d "A=20^{\circ}")
.
Sendo![[;D\hat{B}C=50^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tMDxrKi_ZpSZg7I0cB45ovM6HQ1hXZJqcVUNX8wYebBtSuWhJf2a_om-dns2epjQ97ee_TnZaz8XuQqVGoWCF9rGCKcUrorwOCM_bvRutH7fF1ke-K_05YiEk=s0-d "D\hat{B}C=50^{\circ}")
e ![[;E\hat{C}B=60^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ux86x9RoVCYmHulZ4T12G9ZZkL0ZLWT39d9r7zQUDYr_6vDGSTexAZjBJI-dyp5OXiKKGR2SKuB8bF-rz7KsBDLAppxn5yjFM6lKBbIMWR89Mj2Sz45dPJmbE=s0-d "E\hat{C}B=60^{\circ}")
calcule ![[;C\hat{E}D;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_veAlN7wJBmtBwfgSxceJ1Z4a9aSQKdN2UjAq5DJwHftRHBk4exmYX02qt9hLIlJSIccUnvwYoKb456OenIbvx9V9okcVVh4A=s0-d "C\hat{E}D")
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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_vkjVsVfP6udPbNbM6mxlnWJwFGOY8HNofyyfZjYqCXLDFriF_BFymLZdM3Jr8phVLAHC2OD47G1Ecla1lzaITDjLkKgIQyJbhg=s0-d "C\hat{E}D=x")
então ![[;C\hat{D}E=160^{\circ} -x;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u6mPHhDMhJZ2BONQJXjQjeC5xYYFCGxUmfZXRDwb8VmsdeEAdemNrbu3eOHxlWtNrzBpFZM9Hn28lR9ah7nTUPvtZtbgBYBy2c3k8n2AKyCtmI25-vVEbK-AJoL_2a0po=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_uoZhZBMZGZ4ChSUrGmUzoHFq1esTrVt6fZepynbphCRWt17Mnl36v8cym_kx2aVDysxqDN6LVn_LoyGQ=s0-d "CDE")
...
![[;\frac{CE}{CD}=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sY-TLmnYGSJVLqHjfX5y-wVcFCg0cbj5no8RbIqVE93tQBTvqVPTw-tHjP9Lj8dEW5GatgQ6v2zg_b47vFAMqDHOm5BkwP3vN8HecLJ9hDRQ=s0-d "\frac{CE}{CD}=")
![[;\frac{sen(160^{\circ}-x)}{sen (x)};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_slK_a0x7jwikcy-XB6mSvNNRwbmzgrZoOI8g50agxTeIURTVQmJCwjAek6SwgTN5lhNkw27rNeYCMCFeEKtmHrZc9-BUcD-_wjcX9_x42PozcJn3QonZM52jYU0iBffRl0GJoZVrX-dJGXb6Kk5bM6fM_PU68x=s0-d "\frac{sen(160^{\circ}-x)}{sen (x)}")
(I)
e no triângulo![[;BCE;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uss6-CRc4nAtyYlXixJDKXq3iesv4ffstvHjIi-G3ZZstUiuYkTjreyaEofkkfixoLGNPK_0DeUkhJ=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_v4_HPWZr2TkJQBZvRZiVKunNl9kwR2BZHQ3wTvMRHvbZUJPMzgGEMmFj9aU4j6Q6dc9maBrG38b44tpxUMgKu9zPRWHjQa_D91ESgNRch4q-G512N8uPZUolXrrQwBoTfwXKQOQMDkzwEVY12nnd5NQURqxEe1LrVQn-kedJsm7UW9C4Zwyj7GaNhnzcjl4xb77hbRK1nuiI694dzei5n87tF0KlCGIZtggN3uhqzFgj8nKlcIlY0sug7829TSTfmntnjUPoxrwUI-t93vL_c=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_vKaKjRL_BfSxnDxCT8sVD1EDlscc6bM7a5cbxLTmJXDEbvqUErTDIrhy_ME1fn6-14LK3nU-ZNIzWE1V0skNSFGHOM-DstqSM=s0-d "C\hat{B}D=")
![[;C\hat{D}B;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uIokY6gkZG5imSm19g2bqy1IuakqiP1qnFxv1D581Uh6dIJkR2YsT9IDwPr8avLrerpP0CmPzKXYV-2dM8Cp0TMmfXvO-SAQ=s0-d "C\hat{D}B")
![[;=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tPvJo7Rs1Y9qqC8qq3ayYuxOi0oUenvHUKcIsRdLRqC32g9bN0a4L9ciZeiT9jKtkZDZP5NAL9aw8=s0-d "=")
![[;50^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sFPNyyaupqwV29ABGaG83IC6uMJYM2EpbJhnAUwlRARmXnmWgtrQP94b8xgGcYZSUhFG6pKqTOhX4m4xtfOgCglOgXF-TKYLy2=s0-d "50^{\circ}")
então ![[;\triangle{BCD};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vCXI1t4kyb6M5TjtHK8WTz9jcExE6xCqdNT3WSgbuSr-imaUx7SVJhHROWHNqepFzBWkpyp9AztuPemslGO61UlqOWPiV34DkfTd8=s0-d "\triangle{BCD}")
é isosceles. Assim, ![[;CD=BC;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u_ZvIgAkuPabRE-Wdf-ilh8cDt1W82HSAf10_XVR-XT_OVad9Lf2BdQd13hBCJsxqWF0F3PB236i-qn-A=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_s10EOxkbOO7ZBRDGSZfcjS87_Poz2LG9bjaCiFHSQt2KrVST-a0D_lmEjjEz8avFjzf2IlqAx4ai6nwU8MCrjWjYxNJszae9_OC7TB9g0Cpdg2uV7qMwRgywDLBSCRZtfNTa_i1G6dNxMNoCgP7WpagLCwP4JzNr1Om7urJ2bzHIUJMw0wyd21EgjrASPrKCAQ4ZAdUNjd0HJqrSyo8Wc=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_sLb90LS6Fm5wo5C4gfn5nepvzXxQ9my219tQE2HKDT6v_rRYkp-yQnxx-sBsH1bX2mCFa562P0oGFWik5isz3kGMG-KgLhEWnSfmaR0c7ohvYdUtoncM_QlgluwsPP4E9-jRI0-VmdHP_w7WI=s0-d "sen(180^{\circ} - x) = sen(x)")
Temos
![[;sen(160^{\circ} - x)=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tf0UCdWqQbv-ORenJdIPjAF23ogIbTkfpgv5dh9fy0CZbVTupV7BcG5qZN4CDPUw2RnIbPL_nTnS3GV4gCAe6VacrUFtU2JGjB_cVKdhoRLrjSdHoL5DXXom1ZFmk=s0-d "sen(160^{\circ} - x)=")
![[; sen(180^{\circ} - (160^{\circ} - x)) =;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uG5ko8FGSrCftuTzNXlQ_YMxVYKsVOtMHepetfvGsNi1F7ZWUJTmaoDfHx7rAHZhOtf6q369-JEDBA1JGGw908CHLAScuurgPhHDoRyHEfEESsHaKPrh5g4GIDqlxrdD-SiGbXnPplJvGDuvVaD1LkMGR2Oh2CbiuMDIOHBsW0DEZOwA=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_sSPTC3LQ7PwCnrRRQaYQB3rkDYHo-Rx5U_EomJDCIk1NZZF_tEqJ0tYkkYJD4JAn4V1jGV-sIlK4RHHB5oHoCtMUgFpT0Sjv0-BUEbeEWUC2UDvKI4A6Aa0scUZX_s7_R8WzUDeBHVSPQ2yBb0nfp6QNxPAl0cUDs7hJxoQNPqcpUIN7r9ZLmOa-YaS2ZW5eMxBCiiBHAXghXJHSimFHPy_Qw8M0lm381AMuImazioAH6BaZiR7pq52l0JPMsSKL9XJ4avPg5sEzmZAb5DsGkSTQ90zq5Ixa3fiK8m2nHUnsSaYz-wVEj7Ty0h9ekgGokJQTa7=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_vcC_cIw2M0vupr3imIlSpEGhOiw71a-L09ClomlMG7t8uXNSAyOXZ2vYqk7rCKWPaPOpaDaoJIto-GfN1El3wyXuJ9F-2zZ6f8wVi8WSTJPYEi3TNjKQoyr2DnJYVqgIalmE_zCfZjkw_Zn6pO9zi25Q=s0-d "sen(20^{\circ}) \cdot cos (x) +")
![[;cos(20^{\circ});]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u6njXQlFELWmYGcQFNC8phQ1qvFkAiW9Qk19FiyKIjaxBzmLmz1924bWDyVLEAWPTztxx0BrNflO4OegL6yyFPQNV8D6QCTdoOwE6KBUuBNGVkQg=s0-d "cos(20^{\circ})")
![[;\cdot;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uMLTTVbIFM9F5dBWLdYcxK6aZwnOhb0drrTQsRivj2Z38WPOjJWRa96h8VW0kdzfnvMzVYWJmeYFnRm-ztlxw=s0-d "\cdot")
![[; sen(x)=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_snSf7rcLxCN9l_A7OxgO38y-1mAiSXjqirEZgRSLvQVxTTdtkGjBsJHf1zEaNBE9kDhGiUYRaH-Kk8Sy64pr13wiyqGdTk=s0-d "sen(x)=")
![[; \cdot sen(x);]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vj1hrMWglsgxUzaS2bzoYbhKZNb6K5zR5wlQePluG5SwfTx8uvaSWeOKZYY7CtnDHjdgXRa1VE1UGzfYxJtffSKIZ45C5AAc0WnVhIsog=s0-d "\cdot sen(x)")
![[; + \sqrt{3};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tR-K8ETOkQhMKbFI4Ht594rVzJ6zZXnZ7VRRM8FFZKClOgbPH8Uuufh4HSQ1TRgdwz9j4nmrnv8y0aArIV0NBvz467PiTCM3WlWJP5=s0-d "+ \sqrt{3}")
![[;sen(20^{\circ});]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sI-essHQI5s5C0OKfHEyfWMjnJ_tD6JdDMgyDEmm-B8ZW1tlq6YJMK7L35lIWspHeOrRyW3oKw_RYo8yvy7kpbpG1hhRq9pdJnNYRusPBuq38X8g=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_uLTFcVGcQWxam2ejLHtIyVMfZ7Rg-pVvUiPsGs-JbICpt9spehtrAzBoNiH2Y-rJdryf0bQym7D9bzVMun1jl7VRmkxHwP4IJK9oMJHLMIzOBIqYQkEhYbxWwUKgHP5vdaLRWoVwShp3fMrbcKXlaarLw6xhFNVr8jYFzT5h_zSIjrbw_Z8-5nyW8F3yuBZ3YWjUgrxkbl0aKuZPbRT2TeWyo-G2pu_pSG2z5gyxwtSJCk=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_uLsZihQdTfWSqGQAas0iQZ4YjmbCyf_pHcm363Jb5BJb0nvCe7iDUMT9UgMe-lsksqZL5tW5PH9xaCDJmr2fQqWgC9G7WlNXs5pSYsyMvNMRc=s0-d "ctg(x)=\sqrt{3}")
daonde![[; x=30^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uC4tpyD1LghRUd9XwJwOLKmaqSDnsloedUVWT_vNLf8aYAEmLAsu3oX_P7lkfOR0mNdAnEiJNTNGIF4ERqE85wndQNNS17TehRn_KbywUH=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_tga7vR9gCJ6wINlQqxHCFDHSLOCve9mqFeVOJbG1ClPRVr5tEquwKbBbKxnh8Bah96H28OjQsk1mX-EEo9Vfhs4IG0jKM_-1UDTrSt=s0-d "ED'\parallel BC")
. Seja ![[;P;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sYwwb-0tF2lqqpd7_BnnjmCRyoXtEXoZBZ_bvrx-Q4dekaijxXG9k1fCgwc3KjXeVMAJPVz8cWYl8=s0-d "P")
o ponto de encontro de![[;BD';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tCqUNvwl8iD8BOG5QjiWd5LZ4dXKyNY917gbdi_i8Ww8MPvMvQdbLvAmpy4j7N38to_8Yzg0qGM4BIzh_Z=s0-d "BD'")
com![[;CE;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uNMp4ptM56u4kVIHBXfenqWKnPH-TelojfMFUcrMx3JQS3XuHaqZGyKI0yFZubKCT5G1U0PqWQ09X6=s0-d "CE")
. Assim, ![[; \triangle{BCP};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vGwPNlpkjpvgUdN6TVtCHHP1JR1lIkZ_6BB7nEfkWJEy7NURscyQ7UcbvCtRAwguAB_3BzngNTiULWD9tbh5EqeuVYD6uqzRz62e3H2ACf=s0-d "\triangle{BCP}")
e ![[;\triangle{ED'P};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u72EjJl5lR59XfzUC1BiJ11DFOBjd1NW1K8UWTDM9i0j5AFjVZ4wPK9YRno69SWyQbs5rpc1vBg47AQrc4bUPd9S43Uy54JqIzhfDgMrcF=s0-d "\triangle{ED'P}")
são triângulos equiláteros. Ou seja, ![[;CP=BC;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sL4-zoIdvWJUbzbn86P00NTCJxKZ3n0sFbbQQnqWVuZ07Jx-Fhs_Y8mPargI6QMbXLQJ42dRRoJr1FUc51=s0-d "CP=BC")
como é isosceles, então ![[;\triangle{PCD};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tcAFr4DKr1jVPOTRvC-cDq8tH9mKhXXxDo_pVrqUB73AIAmh47HllL5Ee-NgYI6-_PEdl7N88_cGrIebOIxt40dq-Vk-ZFdVgpKj0=s0-d "\triangle{PCD}")
é isosceles com ângulo do vértice ![[;\hat{C}=20^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u0HeUXkg8urfywrlydzQwu6eD-1bfkQIcmhdYXIZ7EAhq9G6Q0Syi5HVBBnYDmnRKy3u5W7RI3eBNsvjx7aZ1mkElixWx3AmUitK2cVYnX6xbTJOQDiw0o=s0-d "\hat{C}=20^{\circ}")
, assim![[;C\hat{P}D=80^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tMw0g5uygprBMQJunZyBJbymS2WZpPUVl3BePkApccWtcmCgis0F04P_8uJ8U-T0pwecZeDwFn3_Xf5VyvtTzUNdwbgU5ExGhxssd6zx1o-yhI1cEZz8xsbOQ=s0-d "C\hat{P}D=80^{\circ}")
como ![[;E\hat{P}D'=60^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tgYo7hUUp4T-49Dm08nIo9dvKp874bNtgYMuE9nEPMB8-Y8aR2NpH26YoaCSAZbGMRlf8OeyeSUvEcp2UAViwpDNyLqGZpYrvcBxghBo2tavfjoGUBZXDVvvqOP1E=s0-d "E\hat{P}D'=60^{\circ}")
e ![[;E,P;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_swgpA7iGAkHpmI3rMAE8q2CpSXl4ax3d9ye2XSlqnCju3YsUjt-5-gx5W2GVJV64UoMh4IQPD4kFFwyQ=s0-d "E,P")
e ![[;C;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sj9IR1yVre8b2ZL8WEXSViX5EFHc2jIbsj8UCmlgUHMciJd9wbjPZIPS2exCDJkKqAYpsqUjjQaA=s0-d "C")
são colineares, ![[;D'\hat{P}C=40^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tj5ZAWSMMmXRq8dLcmWSv2EvwTuVDNgHRXqJuQ4vZauatO9KJJuMVAJbIIr5oYumXqMmrfoWmEM7O4K7u1xOX82QBBUfM484EBsaAcqHD8teISQjVTlyVLUnx8wfM=s0-d "D'\hat{P}C=40^{\circ}")
.
Como e ![[;B\hat{D'}C=40^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sZSsrifAewIsQsnfwfFCDszYwCVPjkc0c5QhY31Ul976pkVKAAiAYIdW3G1yVMGLcjyAFkTxdF4UNpo_n43DCBgBhaCdUnQRpBO1MfTNuvDlXRr5_RQcgAfRAuiH4=s0-d "B\hat{D'}C=40^{\circ}")
![[;\triangle{PDD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vwLQFBb52nlPqLkedSX5AUD6hhspkGjfW7br-mgT1DTNK8-35l2BMHmunKt7R8cX9TdV64WGappkUgGRKmET1c4DBPannhft3ZBjzL5XEs=s0-d "\triangle{PDD'}")
é isosceles com angulo do vétice ![[; \hat{D}=100^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sNaA7TLC5kFjlN02ypZby3dbYMx45FKd78Jq_ob7rVquRzNI4M_KO-N5vbTPgx-Kupu8mScaB2KC3kDW50F4FlTebl17pVY0vuDccd-w0Lq67vQ3K92Sr6-iU0Jw=s0-d "\hat{D}=100^{\circ}")
.
Lembrando que![[;\triangle{EPD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tJwpy9XFthWTbBfcSsjt1uk12E4bqHE27vxuYcCIj6jUqWwvuKERO6sWrbhmJkffU1zBzRyj2DFqloipzGadKHLBkGzE5tdx0P3FTwENwy=s0-d "\triangle{EPD'}")
é equilatero. Temos ![[;ED'=EP;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v6r5Jba-ERL8qgqb1nne68Tk_LMV1RXtBPyU_ez4JMl9kK4EI7Vo-_fHVaPVapgapOzX3ldZH8Zfz6DfXqyxrL=s0-d "ED'=EP")
e ![[;DP=DD';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tPFHD7Nal-QWOEIjLBHb1xaE0dCiQZ_hL1Uyr1HCBDWK_NhGec-J0wD5zLTJqdd57_K328xh1_3NLrxpF4QJ8Y=s0-d "DP=DD'")
assim, pelo caso LLL, ![[;\triangle{EPD}=\triangle{EDD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_szSVFTqlcHsmMK320VHxg7KfxnJHByjHCbpD6ItgqlzAH6mHDY5uWQQrugv_v3RSbbEgFy3inCqcie_Ta3H3HUFKTOz-rfXoLA96wHNOV9YSYPE2fcou1TqDtCmpRsLWe7hsXB=s0-d "\triangle{EPD}=\triangle{EDD'}")
. Por isso, ![[;ED;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uQJTPmxTg-GfSsedVZgit5uP07PeE5ee4c8VypBj7tNX6vI3p025tDssutz1h9_mta5Vd0WaXb8Vug=s0-d "ED")
é bissetriz de ![[;P\hat{E}D';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ttDsQM2OZj277UVTPRtORf-d3ZeI9wLHDSZ465phExto19RvPb2OjbAVvyOlgkXGHX654fqiDmHI1V-kPyetKMnBnUn-q4gqDX1Q=s0-d "P\hat{E}D'")
, logo, ![[;C\hat{E}D=30^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ufCCDU-882KxUbXteNr-u-68cbRsw5nPHM-yTTu3d0yY4OrOitckOLS2IeV5WDJSeNB2N0SuvZBdCwUUlEWGdm_vX1q3po49_wOnlg7QP6t5-ivze34jReiw=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