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_uKEv9l0jZE0PCqpRoFqH_xNaTxW29k1gwMi695B2uuQI3JMQjHyIykKrXHatyb9kDDjBi0ar5JD4Ykig=s0-d "ABC")
ao lado é isósceles de vértice ![[;A=20^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tVnBXNjhQTgHegvNbA9DRLJFXRZPFisz3dJYDiG3Z9wnBMuc67c4ChV0sPtNlZSqEoERMaX4QWVmgvsgWg4BuBtDOUDFZrEY2eJaY=s0-d "A=20^{\circ}")
.
Sendo![[;D\hat{B}C=50^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tKUfyfyohr5B-I7ne7OUfCOCr-K1HlWhOPawOpjws14SNi-Gbf_ChTH6EqbVtRQPFJbFDiugvZ8TkxioTRo0QZYi6cr8ER_pRNqUCcc5KGyl6YeTIwwGZZ-ro=s0-d "D\hat{B}C=50^{\circ}")
e ![[;E\hat{C}B=60^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tT-Fc6OikuLVKBPpXZmFg8pmWwhuJYZxC1q7NqcBFNaCzwcNsgt-ZlSMJGajmKsnEE3xVOoMfGamEtOXW_eFgTZjq-9MK1VxS0f-Egn1xcV-8WzU3C2UJtSbw=s0-d "E\hat{C}B=60^{\circ}")
calcule ![[;C\hat{E}D;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uiogIFV4mlkWSwq4xKgdKmbE7ujPdZLLdILqS5vUTtvTGXejGmjiIK9-53o1CijCS84H_vafEz5ULclcPRrdEY2_0xLWI-Lg=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_sewGGfyLp-MuUtA-GQrJwh93oMhb5BAPDQ2MWpgupNjbPZRWAjZOS5iUH0zUXbXWCe7W2UOPh-gDo15fz2IT4zHj9iHN85dkjP=s0-d "C\hat{E}D=x")
então ![[;C\hat{D}E=160^{\circ} -x;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vYrYqBxBT7SSIrGGP7am-IJyPLXec6OudEbqIaABNZEFe7Rb5jTw6TFFOtPh4-II2wKyAieEQmTZ0mmmPtujn7RxTJiErAZdYVrpfLzjHwN8dGp-7rf2HNsVBTD2SVZjw=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_trqBMbfgSZuw4Z0bZ4AD22kwCpWX9sGKuF71vOxogc-ZCLAegT4_Ku3f7df-Lek1wDSERgVdEdl-hLOg=s0-d "CDE")
...
![[;\frac{CE}{CD}=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ugkpzVSCpQAkR2mAb7xfLrLIxFE1wAD-DzBcuDaL3QrTE8xqK6OYZBZw9n41zu2r-Rcco6ryjpQCQUs3RzZ-rOWNOjct6i0ppVOa0ypg1JGw=s0-d "\frac{CE}{CD}=")
![[;\frac{sen(160^{\circ}-x)}{sen (x)};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t92ZF-AAAV-blRlHmYLx4As5Aqcymrc_DFgfndnGRQcrEFIc3AfSaD8MHmN25FkXy_slvue4gcH4OdrGjBjrBVfpTbeGagLrAdZpsoO_ja7CZWIDnM9EkRmZ4MUCFY5RJm8iJ0rFFT5d64eOuGtEUfYMVA5NvA=s0-d "\frac{sen(160^{\circ}-x)}{sen (x)}")
(I)
e no triângulo![[;BCE;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vIE04mNM6dGtDjxKKKDsYJDiYUoj7dxmbjV53plYHUbk45xUpxDPW94IcusIja6JUnOQAR1rrpwVEj=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_trRztzF8QRtGVp_fl1jQrTqXtZCe7VRF9LIUL2aUNtxn5jR8d6GxM5X2H9KbG6KsshOWd4YnLMDLrYObPq4aHC14QeXE6LHxFDwzgFzfQsbyFrORe5LitY9iNy13QsNBmO18J1LITxkG_8x8PUnRfO1sZEacsC8aqjgnYbZYclLkQ2JV2U18TbceHgnewFMArDpIzw4y-TzMeOPcZ1vfqSo3WeMij9OUg9mIATBrqEJGYRyZp9xwJQdYZHQzXeUCC3-c9mCTAblLgFYENCqcU=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_sdiHQT_Oi5bfoZYkrYO_tbqTt6ftF3m6edYhPUFVGdISK3GilV3TyZPTqlLCT8YQWutfn1VJV8PKQvSiZvBdeuq5cLP_VPDLU=s0-d "C\hat{B}D=")
![[;C\hat{D}B;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sZjZ9rTPTZotCUvbVqvDw9jStpo7AGWwhvJDumkfMk_A3qBvcEFYiV7tVX4TIp1G3aF0jYOrcqI8Ez-HHreMnPeqnHAxv9Ig=s0-d "C\hat{D}B")
![[;=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t6ILW-LVsWJidm10fvphngbwR9vXyX4L2cQAwAqRkPgydn2i1i9gy52C1NVh9EXpX7aFBw9mt8JcU=s0-d "=")
![[;50^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tOIWduOpDNoqpOBBfVE107nMIB0eYxquFwGgQfnCOCW495pCQtnbSLBmNwiWxUzd6OFexNW4TTFFDzycNrh85x3mB38a6L2KXv=s0-d "50^{\circ}")
então ![[;\triangle{BCD};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tPihzD-ibZk-nc7klLe_bszmfFPS3-Cfk6BKFI-8u_1cZPY8CqK2fdtgF2gy4Tc5xWKZikPnjBycYc3u5bBEPhe6krUniDUWwP0vc=s0-d "\triangle{BCD}")
é isosceles. Assim, ![[;CD=BC;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sgg4FKcJiVtUP-xzjFEBqNNuEGx7kcf0vfZg9s2GBzGceknCEhCFQwzYZ2ySV_ruGvM3Hp4D7Fmj0rygo=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_vtzjP1JfseIwt_lsGbc5EpLoRPXQ-rpfSJ178bpNIdR0pYON66FgWNWV3E3pwY2efcVsU8XC7Xq27r5rLpAAXwxetEg_9-l2VzCbumZWbD9zoB5wMH8iVhYKI6rP0P3Yk5c4CMlLU6IPwgaeDld5jXAEFiwlhBvM91jGkzuAJpqNZoj75qCUTAKf5wzHXFnuejn4zTnfdkDC4DQPuN1os=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_uRgI5jgIkchJL5kvoh6glfZkQYqY0Vf-zJez86SUf8ZqG_MqcEYx3Io_N8Z0OIWMJGKV26YAMw7itaUqWGM1OmXz1qXDNK1wroQ7kl3ch2HloqsmG2tCn5cJsxZSvtozO7DxI2xImh_Ok-WtY=s0-d "sen(180^{\circ} - x) = sen(x)")
Temos
![[;sen(160^{\circ} - x)=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vaU7z7vhRXQO4fjT9jUGXxBnVtKJ9doJjvdRPxdnEC6aF4t6YVKEszUT2yeLvHJpIEn4uZlpjc4hARKy2QQBOOYnGDMDmVTEibjdqnkpgiROwBqWUjh2j8isc6HjE=s0-d "sen(160^{\circ} - x)=")
![[; sen(180^{\circ} - (160^{\circ} - x)) =;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tlxU6KrvRwFuoytAcGvpu1-Q7A6V4aW1CPY4dVL8DHhlTRGRm92gJSaEF6BNJXnncqtc904d6FWqFB1F6Pei6ZR7C7RwIwJFn30YWLq8M49oiN1l64HbQqJbPCib6HX7LKqoOVRc4Z_p7WKDZ4RmhJRfPhK15X_-q8HSGZTAGSTnolMg=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_th8wvmlDGtkFTLM7md3QdroDkh9SP1O6E9TsbF_GrV8bnqebKFtfoYFKpeJ1OSKXMno2fx8PinhpQ2wRqPfMbmi7Jf8rUCZ3RALMZWFzhd989uh8nqWbcV6PfYyKl_YcK3KYnCHO5CJxVfBTK2Bmu8LTI67SD1oO2l0CdcWSEKkxzT-ZzuIVVdFXnLIt90Lo2Xhvoz3lsUh3Yug-fGHeQjRDqp5QvhIII5jgwDmKgZwpwkp0ss3YmyJl-QsNSnG7P1lq6h55JFqKS0Hmhiq5gH3uwKZUe4vNWE19asL3rrNMo2ALL3Nz4wXsEbEosCefBUnk9g=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_vN0M6Wt-bNzlbT8QswXc2wcvFAsXCBmR6PVSgQUP1fNIn9qlzv5CsE5MsOQyMEvZ_yd1LLTHYxwTgUwueepWhC25Qsdrlp_p3BeP1RHTA_kUi8URWByui7bOjxQOZ_gFEEMP01g5E7_WlhRo_06gnsKA=s0-d "sen(20^{\circ}) \cdot cos (x) +")
![[;cos(20^{\circ});]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v5WbWo8vVIahqb1xK8wDn2-vHUtN9jQzSvSZfHS5e_ue4vtDZY36RVU3k8ffjHIxjCwCs15lI0JpJJXTiukKRKoSHfIrSYKdGy83t_jk2JfNGt2A=s0-d "cos(20^{\circ})")
![[;\cdot;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uTaWcv-DuR5AZG3xIrG4pwI2lsXgBPf893rDmnCFTmPoHhASdnjDNB3vs3UzmdDB8gaWGj_BFI88ChCf-Mcd8=s0-d "\cdot")
![[; sen(x)=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uZIvlSJaYdeThVybt1vJxqiI3dE0m0AgqXEez7K9T1TIi7zKzLEPTIi_MpYtYGfwazBeO-StrpPlY-_jqV1gGz9hWYlFfJ=s0-d "sen(x)=")
![[; \cdot sen(x);]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vb8VPogSEDW70eZkj8SiGERG_FtLJdN-A00VqCbsrgi8L0vAsuP29CtptasjbNaVglYZ3RBChTFXU4OeRsbvzFL2bmRXzr-f5OmnkaFZg=s0-d "\cdot sen(x)")
![[; + \sqrt{3};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uqjpKd6Bu9Cs6GtaKblL_EPACeoswGvCa--Iz9vjxvRoRtltFbNQIsdLN0o1En3EX6fQuQpCQuf8ZjBb-oJwbLh0-fbmm2uVahYWLP=s0-d "+ \sqrt{3}")
![[;sen(20^{\circ});]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tJ4RW6j1_R5YLPd8Rm9xf-iIx0PTk_LoE1jTkIM-oGqueUzplQ8HX38gd7rtd5fDHCB91k6frBY4CoqJGns0j2ddo66nwZmHDtPs1ErNSmzr7b8Q=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_tHyS3PMcBu_bsX_4Fymp46JKKjPH88mPa8oZhuSaIF2GuaL7O0JsL7j6Wb2--w-6DECdAMuRd_7LblFZ9RR5j1GfAC5sNWCyDaDzjlOcC503GfvIJSm6519qHpCHw1STDLluAspjdTUIFAkknSY12qQ1dzEPI-n4msb3-CWatWBE5ulTwNXsoODNltSsYMM0bmteBtz41iwfrSZHTrxMO69PJRcKgOGwVH6xuOxMfWn4Nq=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_vw2WS-HvWrNBAQ9tZ0O5_-HNDPqliOAol-hbwOWWyE_wwJoD2gAFMF1c4K7KQVMh3TSTnmtbgYdHt5UdqetFkpih7umrsK1mvAyDxgBx4H2Vw=s0-d "ctg(x)=\sqrt{3}")
daonde![[; x=30^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tYLsNI4qqIn0o_-oK1HCXa75AZuVzVIQXkz5mtQWfa3i9SSb5Av09_Joo_rQ_iIkJD17GOKFKIOvOUqQbB-mE_nQQZ8bXrY0H7wS7X-3N-=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_tZpfamVCT-sfeIST-UyLUfCBPDeZGwuFY085Pgr2k0eqSIfeM1Bt6agOBZ9ZmCv652Xcn0vW6H--DwyrbgPBM3pWVwA6m6vmgZSNuw=s0-d "ED'\parallel BC")
. Seja ![[;P;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sNDEUmUvXP2pU-gMRcVv43o8ljBddIL9azfkt8rbrFTGeRnZDfkhB5UX7sJ6Z09FuMPJEI9_fj4jU=s0-d "P")
o ponto de encontro de![[;BD';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tjquue0f1WHR4HC7NuZb_ae3qm8MhHIatOEBiPJlfId6J9Zs3A1r_q4MvXWDdLfFyTlQVEiI8Ld_PCLyr9=s0-d "BD'")
com![[;CE;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uScDW5CrI5ifLOQahLMzKfpjQc06H-TYp-gPbCi-RrFHBMjoYmy9ozGrO-NTNBMC8Z2WTsbzXExoPN=s0-d "CE")
. Assim, ![[; \triangle{BCP};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v2lnk4goFDCwmdOgLOa_1V7U79BreCXaZY0fqtIE4r-D9VvNHfmCPIF8VFR3MwH7InMONszDW-Pme8dWJbmdjVkdD-BxrXaTKQ6IZ22f4R=s0-d "\triangle{BCP}")
e ![[;\triangle{ED'P};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s17Z3pO90ba0bwJg5Y7wOrgVVJbCIi-0LS8CqiOMZbHM8YLhbYrL9yXuZbhpdP1T97N35rZ3hoe0MHDuhjp8PLAicjvi4VH0w9HmuIiwZd=s0-d "\triangle{ED'P}")
são triângulos equiláteros. Ou seja, ![[;CP=BC;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sOKE3gk-Q75pcPJIQ5bWj8jhj124aQVEfOAzzQQWnMmncN602yd4Jq4tEwzQR8oKiSAct4hJcA1rWPJhyp=s0-d "CP=BC")
como é isosceles, então ![[;\triangle{PCD};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tkBu1kPP7SjQyQw0wS6Ti_ImSLjmq8SMBsPE8cVMUZfqS4t1stMkrksB3RUcFTPnA6UO7HqkVFWLfrF8QJTZCkriA_33zHLZEy408=s0-d "\triangle{PCD}")
é isosceles com ângulo do vértice ![[;\hat{C}=20^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vMiH7TndprwirZapukXFR1YiHSzsmptJyXaNy7qktu8uAnsKf0jS9VFWHB-WYU0rd0cFzTWhJ_CzRxFT-5kXg2lUsKAQnI46rnIEKMwsYhVzgeYPlw6WMs=s0-d "\hat{C}=20^{\circ}")
, assim![[;C\hat{P}D=80^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uVu8xqrgKSzr-hJhfwUI_FSV5PbRSVoaphtjXHPGrYJVzpYmohTN4ma239P8naoHasXIrZqcgHzH_dULq2B0R7TphfcJbr-bQtSI2E3c1pmzpBlXthwJvC-oQ=s0-d "C\hat{P}D=80^{\circ}")
como ![[;E\hat{P}D'=60^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t4OwIa_VadZUH6oDVbOa0fZeJ9GhstGv0VggTMEYRlPjuURw2PzVx677b0pH83eo55ohCER-il4P16VvubYUhexnnI6rGm5VI8ISB2nBuSnePBcRvzy8zQvM1SP8o=s0-d "E\hat{P}D'=60^{\circ}")
e ![[;E,P;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t3QQjN6JAlH32JWuQXjONJ4Cd5Kk216MoIwZl5t8_AiY8tm-OFI0ccqvRqLi_a8m_blhWvh14xa4vL5Q=s0-d "E,P")
e ![[;C;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vdMoEUMCyQDIi5552laDZm81WtGrXzqmlvKPxTOya-bRigvAvqeFETZd_wQIaSlIjhRbz1BSEZgA=s0-d "C")
são colineares, ![[;D'\hat{P}C=40^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tizm7--esLCbyQsZGxTa5mcUSonWNBDPfhX3BZN1CGtjl5WkeqMkgUJKnZyHg-fmwnkbR4BcOh5Q1ISstSWW9opM5cxr5olsxiEKEYJ-Gt6eM2MGitDGCckxTx12o=s0-d "D'\hat{P}C=40^{\circ}")
.
Como e ![[;B\hat{D'}C=40^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uXejozgfHN9ereOfRgYUCz6MZGoOc8-WYAHrSzTDsxCrCFjJ2yXcGTT-QU6szMjOkolCOZsSryyf90103QzkwhGa4K0d4gGWcPPJzA474IZSgxiaceqdfIP0e857w=s0-d "B\hat{D'}C=40^{\circ}")
![[;\triangle{PDD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s3ATGbpdGTzehybmEQ8ZS_laOgOUkK-ojlJYWhW1zqnsuHnT6mnLul04AyeT2sTiOnFXKkxUopDAHmTjeOsiUqVDB3MARiQsg9Hbx_IGB3=s0-d "\triangle{PDD'}")
é isosceles com angulo do vétice ![[; \hat{D}=100^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t5Kq1ybNP7TeKbpsLkOnqxF7IN2xWWvxK544HYXfGxqNjg20W7j7UfeHKc7Hy-UmDOl3zHOYTNEBNs2tn4A3WCoMIMTh0fX293XE-RG-2G2y79LvscNgQtHOhswA=s0-d "\hat{D}=100^{\circ}")
.
Lembrando que![[;\triangle{EPD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tnrd35pJ4oADbhrfXqoct-7BoBhxll21tdjyK0TvigJWRdVvhMGTiGYtvn2V6KgRcytR4Sv4DTdZCC9Aee1lQQFdJYxtge71INpldv3EQq=s0-d "\triangle{EPD'}")
é equilatero. Temos ![[;ED'=EP;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vAs9XSKhA6v3HdFaU4kRnCzwZv_YvWVgqwqVXgBZJC_yHqaW4ziGHMG6nUal2Ss4CO1d6ENGGjz5vKm4vxDfez=s0-d "ED'=EP")
e ![[;DP=DD';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v910pNUA7zprs2v-BOSQR2H-F2OFZ8gE3pm1U2I0B8Ohff8dZdJOWw4DeREseSAA6iqykuul-KHCy8HjWN49Ub=s0-d "DP=DD'")
assim, pelo caso LLL, ![[;\triangle{EPD}=\triangle{EDD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tNsW5jOhkOQpqbC3a1UL17bTroJL77WnJS0Q5EjCguKwsHzXn-HiEE40blP_yKdE-hlW6QsqVX4zHppJF9PQULHVU3xs6dbtkzE1lcYeVepqfoIXpn6bHXjnoGig03y9oOjwTa=s0-d "\triangle{EPD}=\triangle{EDD'}")
. Por isso, ![[;ED;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sd_BrgRqdk-6L4znXyQFzbpYzw8W2PxNA8_qa7hIGgz_qFBC2A4yPlNPO3sVQyJb6Sle-yZvUZoNAh=s0-d "ED")
é bissetriz de ![[;P\hat{E}D';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uOCYsSf7CUwXMOBIigyqF8IsGDtw8YACBa2LNxgjo6ZSYkMRi06wqI_PAwADTpfauk8290dTc-4fuUlMNJLh9q1f1iNPO_RUnptQ=s0-d "P\hat{E}D'")
, logo, ![[;C\hat{E}D=30^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uAleXij7gVDs1G2B7_gPUEugUIYelwkYJ_CNhT1qZ0XQLVYf-12fORC28Xa54Zhes8UKG-vhxRmxclfxW42W2inttP3z9Ys1P1WctusxEkIEC1MN_0UyaWOA=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