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_sMfLZGIJDT2OVQe12xAyJ4hn5v9uh_izymMPIsayiQkwfKXxKm4AsN0dnuhgkH36u-lEnZl-YvNyNgJg=s0-d "ABC")
ao lado é isósceles de vértice ![[;A=20^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s4JdQgL7iBWQXyWvRUMydG-SRCXmZDM7ddZBTMcZW9_GHvHmEOZ9HCKKcn6zCzOfihzfEHL-VdRkVIlsN01_AZPlsRP6U6-i10pa0=s0-d "A=20^{\circ}")
.
Sendo![[;D\hat{B}C=50^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vrcJx5W1HhZ3c_hVFcLDD-YV3J1_CbDObsTQ7Igr-ROhiKzioYevJfGuAachCeiUEZXKPe_htrZxS1ydMCkGcNy9jVkZ4yy74E8b1Ah0oyco-bzKqf4GLm6uI=s0-d "D\hat{B}C=50^{\circ}")
e ![[;E\hat{C}B=60^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tvk-e7PLrslSOx4uuI8tRCdhkK4FFTqR_Q12syGaHFh_541U-m0Uw1GmpVrnZod-N99Z4ZHdErxRS4rXi87_FQ-88RuM0F7bzBe8Z8d4upXIqE3BNBkiWzQ1c=s0-d "E\hat{C}B=60^{\circ}")
calcule ![[;C\hat{E}D;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uLAae6MRXmJjaOHhDO7eegEjcFDb_1Fue-N-QZFgk1u-ymOK6xf8TFyjraOQq1v98lj7mYRkQ4MsAzRgk8NSWWxGfYm9uj3w=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_t3lpVny9YN97TVNJCAsPrbgmlt0b4mrYav8uLETnDKsEoRcYYnKdB9LwyNlj3qw5q_N6G3ZGnfxmoFgmQBB8kLJ1N1JmHNCs_H=s0-d "C\hat{E}D=x")
então ![[;C\hat{D}E=160^{\circ} -x;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uH6fH7Jma0w0D9lzzuj9DRmsykycN188sJJd3DqkGM9sotTLoU9bMMRvcaY5_rrXtVghuhmTckjMBVPFgaXbalNW-WS2_9BcP9G_34jlImxrFkf1SFIOqNkWRVU2hptCo=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_tUvUuz3hDSDhWmjhDuBpaz7HS8JgvTXAS6doP61j1CifJiT3-IDiI88IFv7DiCsFQ7PwTg83qFrEJEDw=s0-d "CDE")
...
![[;\frac{CE}{CD}=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tDj6DFnQzZRnBA1bsGxPLxS1VCrMq60OTchifLPNej-1m2FuT8Bm2Ul3VVThxvoSye1FjLV-3ujtnonihio94oOuwJ9kMCTdzSSypmWPDPVw=s0-d "\frac{CE}{CD}=")
![[;\frac{sen(160^{\circ}-x)}{sen (x)};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uehm0SnNISJk44_JUIyZn-YrN6gNdB7o2m1U825ND70X7KaaocJcalMJaPNALraGTLA_UjwHf95hCJmy6NkM8nJ1Wi_5A_TKgwj3XU-Lo5uz24BJQti-PNucbnXf95Mq1lkeU1dl1zPj2vZpgfh0hwFXhdld1d=s0-d "\frac{sen(160^{\circ}-x)}{sen (x)}")
(I)
e no triângulo![[;BCE;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sxBVbYLMDdwMa5t7KBhchMpbITR03dHY61G1TBVZW9bFDavUH3-URTqJrFg1XhCwLDiC0wzx3k_7mr=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_s3PfuRyGITU6ywaWhMPHgbOkDSFxQCq3ir1D1DzYICDad7pvvoOmp9y4LEUJx00bROs7Up9lEeRv7EY4qdKNyCxbp1dwZs5usYF6Qe74ovnUUihAwwfg0WH5au_F2ED72fj3n-nEAXc-jA7PTn7P6ONk1iv-LUWrHYDFlaL-Su4ZJS7326ua6e_76rv-5f1fcB5y_3hymt8Jb-sv8SSCB-RQHfVarEY2zk6RByno3YHD1prQOyDfuS8vakXNFHVGxDpt2kCjc8d8Gr0bZsW4I=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_srYfpHDsjIxrg-JuQPDKiziNrtaEhS1-c6w_QmgAB4tETfrS6wkSZd0oWkt2ooiDw8vBZ_yaURD-s6YvwkNUjiIRjDfBsyyZc=s0-d "C\hat{B}D=")
![[;C\hat{D}B;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vBEDmwQKQSbiT9VCU6HdU0oB2pSOk_lspzm3yTv94yH867mNtfHKIOCEKAbbjoKXuNd-hGDF583iZPjy986v_is5ExrlYRLA=s0-d "C\hat{D}B")
![[;=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uZ1F939B3j1zKCAQ-y1XkeRsmikN0pHBlmzEXSFeMoittlPN8r9lNUlZZXBP0xqhmFvRyRJ1sHYmM=s0-d "=")
![[;50^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uft9isAzbVvvm3AFNtKE6Y5vaSYdHSeIu4QnNmXeY_XHm0_FdaIG87e1r2e9IXrYdCV5pGWgAX3eTdu8kT6qs2ufbhp_Me9dvS=s0-d "50^{\circ}")
então ![[;\triangle{BCD};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_szArhrgnaClhCwjb0N3H7qTsEATLq2L0s4J9cy0cSoZ973AU6T5IbZIDR9e5wzy06gqsxuc1Yzsn9UerLq6EzMuZjLXnsTsv4_r-0=s0-d "\triangle{BCD}")
é isosceles. Assim, ![[;CD=BC;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sQeAGP9bj4OwR06NJj2L70jpBDnIkWc-hs5vMHVV0wkwkQT91J0y5tRLm79-_u-3btxFpaSeTIj7GCwtY=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_s1XvhmlbgKKSCEuwdt0VzLhstN86uJ6vAaGnOOX22xe8J4FAGKmJebB9d1pjj1zfVjbNaV9qklyvCoJW7t-KQLVKvytdom1T5bLxHeZR83oE47bD1OSTlwoX_CBTorjT367dNQAcyjgl8Bec3kUi96RlRYnRciMuLqmjG9kagBrb2MSalZxUkjp5kvxmf9PspOwHG5x3c7QqXNPO0ufpw=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_sBPnMo6C2MLbAofcXmdCaHWhiGRRqLqH8lXgcEMBdXIHCnu1ofiyig_6PBFTM8sP2--M5-f9IRibBOw0kfrFr3-gFqrmTNxc_FYS87eIcFxrci8rG0Q2mNN8kJOzNm3xGBJbEOpjesCm_UaQA=s0-d "sen(180^{\circ} - x) = sen(x)")
Temos
![[;sen(160^{\circ} - x)=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ve9GugEKGkQTz5--cN3JBvP4pR2EG7fpGpdojEFt2ccN8_9LJckuzennf72h00h1c6Bv29hwcLorxd2baeCxXReddeAC-XhfyGT9DT-85rE5cR3CSt7uhXU0sHHWE=s0-d "sen(160^{\circ} - x)=")
![[; sen(180^{\circ} - (160^{\circ} - x)) =;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u406YsDBtP0l6vKyzT3lVzLoEKMoUaYlzZ7Xy2bsM-7Mr52FShLRhpbWz79OmkfrLqQDzA-aZd3fx7K9HAU6DAMzFrYAzl9t_iLm0jr6HPhzyWcMhUZcRNz90TqZP7bJ0Ul_ED_8p7oKQbqOoqUvhVtugyjB19rT8DbrF65bqH8l-CNg=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_t8e4crcaSg0jC_Gqn6AcaY4n_tEISe022eFHxvkzvM63Y3GwFZBUyt6mcPMrXvdbMrfkOtV0OAnLab5CNNl4PZP91nDB9pFFyelgI10MRkEewYF5h56g9cYidLoBZd_Hho-zNPY9fdOKldJTXwgANatuLwoM9WYx_ferdrhkEilSagzdFOjzI9f2Lya5ojBCAURktc3FGK1BrfLPyjce3_6FaKWvUSCXBzHh7EQkEEtYYoU5hyqGI3zoRU_gzFCIRwCXh0v_hHk2R-H1voP6iF2ugRm30_cpYKbNO8V-4tcCOs3x50jbRxeWA5UgJsEwRfmCPF=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_ut2glfyWTUw1dFHRqALgTgl94bbx-fHY6rJwa_nKd1hBpenPFyYKoAoHw3y5GLHQWuSXqbdNnBlPaPdrNaHLJLlAuggRcsyD_HIE8tJIADkvT4iOWVq1au66Y9r5CGvEs8ZWnTv3npfeGAKKnFAEE1Wg=s0-d "sen(20^{\circ}) \cdot cos (x) +")
![[;cos(20^{\circ});]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uqclcL_Ep-57epmJ2lF3jGixxicqHY6QTkxfnhtAOMMQetL_FFsKaMwJZcxLDJfA06R9ioqus9_9C-RsmjctdPftdRuUAKk1ckeZwQlcBw74DJhg=s0-d "cos(20^{\circ})")
![[;\cdot;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tWLfyIJbo0xu_EjOTlvAUoZIUThPZ30f87LAFnh1l54nzjo4kDQVpWDYCWrCp3YCcdfhSW52QNroDHi9lL2UA=s0-d "\cdot")
![[; sen(x)=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uqs8vrlt5q_187tiI_RGZlnuTudBKkbihVGCSaybkAx3cSWyZARAL_c7buO1AylM8fyAe0cX9waarLA9LY46jw3-2eqaOJ=s0-d "sen(x)=")
![[; \cdot sen(x);]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tedvm1jtpwjMF4nTs9tH4PZ6VKB4cprNgjZUk1-68NYZoXOXPYe-dzlwPyqaoaShQ4BSrnH3fkOxYfWOAzMw9i3cYscybkFaXd5bl-SpQ=s0-d "\cdot sen(x)")
![[; + \sqrt{3};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sDG91g_GYjQzLN6kSxmCtcPIUTwGHYEY94Dr7GD4CBxsTPNCjHu7Hc48vtvGWHbOrjM02bZ2uPOelAAt9UU7BOqIFQ6_cxMhgDSWuB=s0-d "+ \sqrt{3}")
![[;sen(20^{\circ});]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uh_kel2NNvgSNeWRA5uA_M2QngmT3wNcr7E7Lsn9pAExYDDJR8Y5gcvi1rT26uo0JnHyaBD4uTbvytkxKetMEzOpeBnaOAjq0de0q7_rwpCsJ37Q=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_v9WP6ij41aXQ_wSqqzjFQl2burwALDVvNSWfap5tIxmqsJipxSmJF6J0-gq7CK-QACXybiKxO8I-t_Z99gTaXR-OCt4VXFPBWNTgb-2qcpUeFgcG2XgwlMLUVgePk056SjE4AGnJTOR3pySO6nLJw3i3HNdpVB04zLDwgbWaFvyJjFRw5HUq6SkjAK-JOZ4waucI_MmpXIYl-czyWPJubby-YPmWvz2ESX3ttj-4ZJ6t9p=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_uIoNUjb_9Oocw2ZZUkiDT7vW3FPOOCc-jWWFhDbOTszmwh3vkobvKbQySJwn5pn9QOGNu1uFL2X2aOmqVmO6cvU15eMtYvRg0IujQdAgJ1uYg=s0-d "ctg(x)=\sqrt{3}")
daonde![[; x=30^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sseWZIZyIiDVpB37JBN0NeIdOwgn_acOejXo6nw0cKNuDdTe8BTDJv0NmwY-q_Tu0RJYqBAKmMptgP1gljXrpA495w0AEWVgiIpNGCYZhO=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_vyceKadNaUagH0wJTY2SQGQE3xxVgU-ZZWG1LyZQnqwf29CiV-ii_vLAaDj4hqVpOiSnuuqivHx3MXi7ym0qBkcUTxSm7ylRHzHOxz=s0-d "ED'\parallel BC")
. Seja ![[;P;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t0oGO2vuCf-udVkDcNXqvVa8bTyZY4gjIreZAsZC6--BTgyNTS1k8kMP91YPEbBK8d0K-dsOD5LKw=s0-d "P")
o ponto de encontro de![[;BD';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uBbkZKINJeeg20JHbB4jEeH_Tt_aTL4NQv3eItVJWfl3lx8GfDm_6sUv0uGF-DTvbUo_POE6WGX3gEjgwh=s0-d "BD'")
com![[;CE;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sItk8YM0d3M-P-ZlZPLnCcjy02NVvHzQikBnPdn624vi485wfiQt9w1MetOMCpIU-7DOKg8WRnQnpO=s0-d "CE")
. Assim, ![[; \triangle{BCP};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uFKaWQjAyoTPyqHerMtx4YbU8Cqori940x845R40gWeQCYOyO1pWyOMGiuhn5KBZJhdIvqPs8aKzseESQ-sEC6N0zVXk3rvGaDIGcYf57h=s0-d "\triangle{BCP}")
e ![[;\triangle{ED'P};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uekKvQLh-lqw2RS_wEZLGxgGGuQAx7IZgAj1UrsRSqXHF2fRFJK0pq4hrzopGy6LlW2L-5sYPKUgxf0XUnq02nGoG-HqD6qFZ8e9MlHeGf=s0-d "\triangle{ED'P}")
são triângulos equiláteros. Ou seja, ![[;CP=BC;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u2KvXmfAMFHP6wETnB2-8oarOqqZgWvkwGThEVEXxSAFvBfpNAXZvy_p1O26FuLLTAfLKMRWe-jewKHCtT=s0-d "CP=BC")
como é isosceles, então ![[;\triangle{PCD};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vseI109VE_K1LBHX-qrxnHB-I-1GwnoVm98twJYfmUFnlawDokm_BriPyAfxnEbDgabqn9avE80ohb0oDMYMILzvwUEPGPCQg_8RU=s0-d "\triangle{PCD}")
é isosceles com ângulo do vértice ![[;\hat{C}=20^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vX8gCeskL7jjmUpXpeWnNFyuGYNtZvmVQkvsz-pHHiL5G-rn3JFx6f3yf08vJaP_I-jLDYdVw29CpeYK6fngWKKPrkptwYSMKKOZ5hOx9rVcurCXPVbBZE=s0-d "\hat{C}=20^{\circ}")
, assim![[;C\hat{P}D=80^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tEUF-x88MjSX1ivY9vCbwPNWVDhXh8einf8nyBI7LvAwrcjFgPyn4QVARohrauJkB2ygFs-SXVdBK8z9gkDAcPT-9FXwBrirSLR2be9pMtnUbqsaHF6frnMa0=s0-d "C\hat{P}D=80^{\circ}")
como ![[;E\hat{P}D'=60^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s3YPvJs-gSs9HIO5BGUvm_TeFhK0w2cNNSplF6_Lq7KPzn_yMf4UzmwLl0k3dECZQ8kV19OEydDZPukLZ0Fgke-HAXLvULr0y5RHHY_DwfcNpf26zjdG3adxB988M=s0-d "E\hat{P}D'=60^{\circ}")
e ![[;E,P;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tZUVQrKorCf1PabGEsH5seklFLlkK34_RD2tO36r-plPfaLjY22pHaZIVkSJRhOzhQa5YXG_5TjuJa4Q=s0-d "E,P")
e ![[;C;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_siJoyQrWoqnxgawid7Q19CcehWDvCJ-VIsG23AMnptHupGfGGdwWDVD6Mye5kB_VutSyYcldEOtA=s0-d "C")
são colineares, ![[;D'\hat{P}C=40^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t50EuNzXDmXoR79h2fR-bXPNJSlNk_UEhpBIVkK4nbk41zurk8-jpWmJE-C8wOhm7MLjxPKVPAOcEhPXj8V-fnv70_ISyMEGIrSq8c_aF0uMrgc-zL0KHcY4ERcQk=s0-d "D'\hat{P}C=40^{\circ}")
.
Como e ![[;B\hat{D'}C=40^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s3g3MZ-HXs-3rU8FJb8LLCX-VHRHcJamrQJ83kTs3susioa6PWnaJWtcnrfEFaHayG5_Wpp4lU1NgPnUmFMububjmyS1_YgRdql6To40tpqbh7lI0XKuEWWASCOPo=s0-d "B\hat{D'}C=40^{\circ}")
![[;\triangle{PDD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v3N8C9nyg9M-utFHpUI1nzba7CtZ1ZxJECLz51Oydoc6XTwTAe07khsxKV_c-o0kJZMnkixcPkbWOx114AGxLm2lLDTmClUuuvRIQhyMg8=s0-d "\triangle{PDD'}")
é isosceles com angulo do vétice ![[; \hat{D}=100^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sFQXCYFLXuW8JE0zWEC0mcqXGPhD5IvIhlona8EBfH07RRHd2aDq26_gnD81qdSOhf-jcZbvNL7qGstWC1ixQWOfbAA5vBBjWkEmOcuCj0DzwkPRAGAuSrJWaP-Q=s0-d "\hat{D}=100^{\circ}")
.
Lembrando que![[;\triangle{EPD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uNB3IbXqNOFXs5772guTXxPkhz61ffqOuHFSdgTv6J_wFi1_JyVx9jzVslT4lrBcHbVeiEADXqlc6gQ-ePmnvSI4nBAXjegb8OkWHZkUGM=s0-d "\triangle{EPD'}")
é equilatero. Temos ![[;ED'=EP;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v9yuq0xsY0gOrNicu5MVSRNIofxUKD9SVp6SBXb3IdA1j6TKjpQq7OZpN_es6OhD1BPeKTTQ0blvmTCxyHTX3d=s0-d "ED'=EP")
e ![[;DP=DD';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vaZNtvAhV2xt4ByF1WMQzttjemdBU79uyFUBjVhIl5Obi_mx69EyrX_n1MQ9JkNb5Y7c8ki2iLLirilD3RyNN9=s0-d "DP=DD'")
assim, pelo caso LLL, ![[;\triangle{EPD}=\triangle{EDD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sZo3caq5Vhqqsp1hbDgveTwyZDTtYeEvX0dD-3R6sLH4t92-XBFW1_pcDhTqv81lvowsbt4tFl4xIUodGutARi0HgjgnYb_-2J6xWSJlMoRlnSPFCLe6dq5VSmhuwY7OhHnhlp=s0-d "\triangle{EPD}=\triangle{EDD'}")
. Por isso, ![[;ED;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tl9ydInLWnGYcZoCpX3OBLTm8UbCQLWtOm5Y2RCBTGnjRlRaSc4-Dfk1YQ98-9zNE3HslXLEapFda9=s0-d "ED")
é bissetriz de ![[;P\hat{E}D';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vhNm4tTo__qu-LOYooNEv7ISi2iyTqrEdr0zTCEcokt4yYutMtHB5iV5umn0Gy4r2gdbbX94Zb8ZCUmihy4kyrxhblr6voyRr8vg=s0-d "P\hat{E}D'")
, logo, ![[;C\hat{E}D=30^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t8nzDbZolYWrcarz7OSPM9UCvcigAGx4b466xdhnCenVOPH87DPR21YSds83nKlXM3ptbnG8gABatiuSxcUQS9L2Ya-mVF7y1EvA_3bV6_yZlaHrJ7B4-u-g=s0-d "C\hat{E}D=30^{\circ}")
.
Há muitas outras soluções para esse problema. Por que você não tenta a sua?
Lembre-se: Para melhorar a qualidade de nossas postagens avalie-as logo abaixo. Se você gostou do blog recomende aos seus amigos e se inscreva por e-mail para receber nossas atualizações.
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?
Lembre-se: Para melhorar a qualidade de nossas postagens avalie-as logo abaixo. Se você gostou do blog recomende aos seus amigos e se inscreva por e-mail para receber nossas atualizações.
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