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_vXflJAmCGA4exCRnuFrvRQAUSEBzHmiHKmRd3TYFYk1Ka-TASVA4x5TQIGp-VM4iM_cQDtjnrzXwi-5Q=s0-d "ABC")
ao lado é isósceles de vértice ![[;A=20^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_smU_4Fhr8zLX9jY4Ulm_XTSRDfQqgQC1fKMe0MYjXRFxd0CgeZv-bU_6kRbd6Akep7lffgXKtlmWU6MFD0gOpD0iSRsTuO9FujgcE=s0-d "A=20^{\circ}")
.
Sendo![[;D\hat{B}C=50^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v1XYoMVMIKLCwDKcThLPh88N_luBODcMc_-G_DjGKlHVS9pq7yoIX5PCpNRbae0s2KqUxsmafp1ac4gm75zALqWZVqDCcVpJqr_uoUZXQDE0rFM5I1k6AzLNs=s0-d "D\hat{B}C=50^{\circ}")
e ![[;E\hat{C}B=60^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tSYXsjsq5jfiEK4qcjrr2VqeN1LjBumpf2fKVbZL8_lgGAIbnbM7TC3HUo3gWuf00ye97pijF8awdsQzak4ihf-9K9pgfZ5qKWJOO0jCz3dwdhNbkRwnOlmNA=s0-d "E\hat{C}B=60^{\circ}")
calcule ![[;C\hat{E}D;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uPqaPbDj0w7dC3mXVbYgltuEQq_qkP1vTKi2nhQhO6WBaH8u4nSKf6kz4DPGe5l3DMnSap-dg06X06jxHjAcW0xmAEQ2_wwg=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_tdB8Dbpmt2nYsKmdIagFsuSRk0El7UW99xy3XeE4ytHbm38W3K86gAfITyLqH74q13IIMCA4FeoGZ0eX3g6_4XzuOsPD0AmO1H=s0-d "C\hat{E}D=x")
então ![[;C\hat{D}E=160^{\circ} -x;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sgcgpdYU1B0joKFHu75ITw4qipYLOvld54nLN_G2IL4JDaVBXFzpnkvNDmQl3ECLgOg3iYfxPqD9LwZwaMDBT6IRadC8kdLsVLiFVj9oZF5JTPvJjw_7iKX5MFo-usi54=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_sW0LFk1P9WicjWfxnc6OTVTKTuehOXq9_QD059Ef3TzC3a79czhCwjoEMDSP-HFIiZNQZuThPuybdoIQ=s0-d "CDE")
...
![[;\frac{CE}{CD}=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s90R7mqhnSyzo2PgQdWTihsEFzXocP9vrdyW_lfGOOULgKz_U4-oTBXQopLqic38YkhsTTg96Kym-ypbyaQMa8hzoyqKZsZ-FEJj8fPSObcQ=s0-d "\frac{CE}{CD}=")
![[;\frac{sen(160^{\circ}-x)}{sen (x)};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uw84Of5Y6PeHCgQDQNQf803cUgY0WtAEFAdz3y63RbrGDKzo8IMI7FJ47FruxsW8E7GOuCToc13LHdUY-Ge2y8jxPHPj6yONqmI7nuXY1zazJZq9sBnHfNKDfZsASmcsti_Df5X-EXzeufyzw76jHp7Tgba6DN=s0-d "\frac{sen(160^{\circ}-x)}{sen (x)}")
(I)
e no triângulo![[;BCE;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vjlKNuIiZL5eT_1fDueHagf72CcdtYn_RVc3e5gNEZLQh7U2tatx3RurqidAbnKdz9PXJsrWpPinag=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_sBILTwkaHSZyKCZk4UcxoLwxDLa9fCXDVom0N-zC5s_PbGJAHswYWQTWm_e-qoSWk-ycJ_lQOr810szYVXQpireyLWMDBqTdUFS79G2CperHrTza9yzdFPzHiBKoH_Y3uxyaH_-7ROyLZ4ToDel6bPzw5oTo0kUTop2cJ-Y-3l45kIQ9xlNZYhC3O6h-cB2JrJx0Fsdd335dnrzMFs-vcFkzC90nVDTEliMNNH4b2VrcZFY5NUJpIEYOXbKGAKRCxz2TnTA0kAcwaaDvMFJ0U=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_vEMXIlAJZy8vBDW-fFkhugwGzU-rgDopgwGXQItxnb9NygW7p_pogungOHgd1Z6C8yEXf_SJIV6Gs9qiFNMhrUN4uTkeON2uo=s0-d "C\hat{B}D=")
![[;C\hat{D}B;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vNkqozTYfNNGbQwupoG0z9WXIrf9-mFzh7Q1doZnm-cxsQU6UKGqACjQzmlnxNX5rRQZQD0KBEjZIo5jvp3XLAplZEjrPXwg=s0-d "C\hat{D}B")
![[;=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uRQfcbKhwFN6xjEhdkK2dsRnmURHC7FR2PPrT1LfvEVcrylpp0rrDkAMNOdkpxXmPTMYTYDZspL_w=s0-d "=")
![[;50^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vQUAL9FxIPDpVbI4GZ4hHRo6daf2w7dgXARxY1Ow0VlTxZSAIw84svMsZvDzCTLcgCYQM9iQ8xkYgfs7GwI9Ir-siSIL5dlKNc=s0-d "50^{\circ}")
então ![[;\triangle{BCD};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uJ4cndsUjDdFP-dBnEKhrA4Kf1fmI7QUaEwB3kclf5bkkng5gR5ehPFxSDE5Qjdru2qKOtx-VFm6F7_0negjdfwnP6MPDLuiDdGRU=s0-d "\triangle{BCD}")
é isosceles. Assim, ![[;CD=BC;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sRfQDy5WmZgsagSKu3k8Bg_UMW6dk0gRiM_g2tW3h9t7oRRhxjoVbB9QamguIWyM1yNeZ_qOoglKWZQKw=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_uCeJw_IK3Rq1kHEb3A0-dpRZZBhDVmm6bCgn3h-vve36cO4V_iQv7zI04q3C76F3tpz-2TdW65ACrYCp4mInqddDLmm7yW-a6yZODeWE5QBuEYZjiwMEBs4GUf7U47QYTZPG7mL29UIakRJOrSSuFWMxMgj0WOO9EqBnH-XvYkL7i0rd3IiYgrH0XE8384UPttYXk1gcOhnWqhSp1aR1o=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_tlu3NinN3z3P15HgcI3yOgv1kGV099Cemj1i4ka2TuB9ExOnAdpgbtLwxu72WVr8NIl21LXFaZLAf8NmIwfBnJhhci4C7g4Od9-R8bSJ7U0yNdJnWgLUSqTGL52fMQ_MWTVHLWxvmrMEOCLQU=s0-d "sen(180^{\circ} - x) = sen(x)")
Temos
![[;sen(160^{\circ} - x)=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vLbgOHxnftW20zh13bsDo-FZUdvSnP7fxh2D6NyzIJZCPRJBJAtWlCb_vK1nFZh_J4p-d207p2gr-tSXmf0DfbnhcA1RJrL7GMEJS6f83IgB_aZZOQ1WpS6dWPM_Y=s0-d "sen(160^{\circ} - x)=")
![[; sen(180^{\circ} - (160^{\circ} - x)) =;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v8hbGkdkC_Hu7f12sXrtWzxQZ4qJunI1u74knHJsDhhpefl8rYAGNYMjBkLsah8OUGqJQqgHq6tpgBQ1-jV2DB2vNqxVrSMAOFEmF1BN2jauVgB19qdVKD7TAm0oJkKbq5tKdR62ndXsYZeFQC4KmNP81213awIGsl_10fzkeyAGuZqQ=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_tkF706LsX1v4kSYYtYbk7_gUwc14RUxh2zLTsFwJJ-l1sZ1y2CnC4rM3BYjHJD045a481Nrk-RflO9NWFAfVVPRzw7Qau8B4lE3Qmya1zEdOfkL4U7LDK-poV4rK-BX19jumN-rOqTAp9EgQuDLbpDgxln6EqccL9Pw3AnF2FSSGX35NrkIB2JOt1FFoqAIXApfW9EYdWPv_b14bEhcgVWLHXTgXrkxVJnsW3s9l6irDznNRwNMjiKkHrOP8okgejQq2mrVKEcrbuUp7X6N5pyTsBRVz9B5S-hPpbn1A_hlrach9VGzzau5jRxyQ_gnRRGhfP5=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_si40srwV8FGD-CDNgskAT4Ykgy-0fWxLz4_lc8sN6bjKjOW3fHNJmeUoRujLhNBB9k5jmc7je_tnNFEbVU1vKM_2irChAAOh7c8jeUr7Ey8ZjBwgYjcQfIgvobeMk0sh9fFDulxEqK_SZwcF-5uSYYbA=s0-d "sen(20^{\circ}) \cdot cos (x) +")
![[;cos(20^{\circ});]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tnrGFckxeawj9yOrC682mHg4zgV1Ks571ZSWvX1j7KDPIgubWfrWDxEn2ycELnG1R6pvTQWHcbzig6qYUgnR8OP4eLiu-OLQ55L6aoOUETx1ENpw=s0-d "cos(20^{\circ})")
![[;\cdot;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ulhxhKPWhqLlOfHzOWmL59WNVKTrCbAEIQNIkWp6JtYcwFl1IWWCkfMhSnOlUEXTZ28gdPZstqZxNUHu8wvyk=s0-d "\cdot")
![[; sen(x)=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vrqq4Epli6lbe4MlUSSULY7nrtZjYdSZ4eoAgt9htvf43m_tOf4cCcADmvQliL1f2obntUiL3V73f-PdWyh7GeU08nOkiv=s0-d "sen(x)=")
![[; \cdot sen(x);]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vgEB9Kd0tHtm2ay4MmIP1YyjSoa9kMnwfXxyT9Bshnue8WmFi9azVu6IsWmHY2jldgHH87aAOm7LIP8rfo2CFV6sB7OPGeDhixEHeVGBs=s0-d "\cdot sen(x)")
![[; + \sqrt{3};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_srYev-9ICdOtBYVMQ9aNafFyNxtYogi6B82iZjxRomrObKL5ODHi6c7wBHJLhPSb_lMrdlZaZG9PdRX2WeCHOTTNRTYrDosv-d0amL=s0-d "+ \sqrt{3}")
![[;sen(20^{\circ});]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u0Up3rQz1cJHYlH4r6A8J48W38lSUdJwWPmfQj66xxtdsavdEdO152YDBfVqISIhO_hYeujU-JG2lTG56ba3MsvobNVhHV-EchUf6XZPkGcAV-8w=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_vTedHgJimuUeVp732VU_C44w2kxW408ve8b0oqjsdzdNzl9U9JvSemzJZDCYr46XocxQcdnUm7Uj6fYz9pcDonm1v9w1vz0LHMxy-IqhIzEWCtHJw-ChoE_A-5vhYEdJMLuGIJH3bv5b_D83HE4is-PE6t_mffo644z3H_zJx5qjvln7b8OtlFHO6-avjBOLaqGKykyzcfyO1askd8_Wz8TuIR5HRDkkWUj8iUsPWWTRx1=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_sMpT2db42uOEOhvvDVLTC3BoBehONm-vSrUHC3k-MTMcj4iF0XIGyT41myv90Cgch2e9RtGLmVkn64RAOJtC37KN-xTrBVxZtPBcjKVsHNae4=s0-d "ctg(x)=\sqrt{3}")
daonde![[; x=30^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t9Iopy_NxxRmhy-6Wb4CNxbHNL9rpSnIVXy8ObEf3ZJr7MY5uXPFXRYrFDEjZnCoxTH2tKsgfk9Af1iLkv4M2tx7lilGz7xKykEnFut5J-=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_v3xGU7eHKvXpozDuKSffqVz8tica1Vudm8P6dsEutDGjQ6kZZCXnDJA-Gl6_0DwLEXhhPNvytwPppwm0k3VPiAHCYjWPhrWUzlBSNX=s0-d "ED'\parallel BC")
. Seja ![[;P;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s8tHQ3E1vZk-4m5QwmoOAk4WhFksTONPvBTQ-eGuZ1D_nO7-V1CKVdz9vvN9LE1uvJ263zzVbG-7M=s0-d "P")
o ponto de encontro de![[;BD';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tmZUqQmeLHdIl4ww9jwyhQFjBCOkhLMX8zE113B3irvI4xLznHHAIyeVAO8CLccw1uMp9824Wj94V-AZUn=s0-d "BD'")
com![[;CE;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sXl7EOpEUU-Eaw1ttdHmTxyBy3xTa_T7fenKI-P7f6wX9-t_BroodxXesx2xoRrF3m7FdbBMZlEIz4=s0-d "CE")
. Assim, ![[; \triangle{BCP};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vnSVUN5PDa8jaHoOpECvuvK3M8Ai22THvMu4J6lPkzbrXEdKx365uszTFzUqdHfugBh9hoHmn8KnQGITbWbTUO7OeRc7ulnZPtC40MZ-iJ=s0-d "\triangle{BCP}")
e ![[;\triangle{ED'P};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tvVv-PfO7HCP-MVGk9f33Zrp1OBdTDpIylbUwq_VH3aQpwSOpmvMYYVukn5UgLF_XAFoikYzkrlrIzJBoGJP27skrkBqo93Vv_uKOX2Gaz=s0-d "\triangle{ED'P}")
são triângulos equiláteros. Ou seja, ![[;CP=BC;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u4TTfb-j0BR-Hg1pixXDdh89pXAofUq7PsYv3Xr8O_u3c_aM85aZ2YIQGwcYDfTgafpUZrJLzroXwyICOx=s0-d "CP=BC")
como é isosceles, então ![[;\triangle{PCD};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u4Mr94iYoNIobaFkQljnUzN3FvrBBSVzq3P9NRrWfWpCjnUww43mbFHfdtKYXk4MZynaJYIF2nNErSj62gs0Iqm6QOoj6GV1LnXuo=s0-d "\triangle{PCD}")
é isosceles com ângulo do vértice ![[;\hat{C}=20^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uzLDYjDpKV7IW9kFopYOHdkx67bmVzh1LwVmJ3VDEP0ddgLMXYTbt0ERv7q2zoLb59AwHnNOUncbhaq0HjWOXSwLRq14vmuZGHz_1x39clpjPakrGJBJk-=s0-d "\hat{C}=20^{\circ}")
, assim![[;C\hat{P}D=80^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sxFU0jnQVSBFblw1R9K4qd5Uh2MQPvSt3qVsE4zgzHOdPQzybOxZIX37PVjj0gvMLBfzpU1T7s7Uke7kqUmf7zhNPPPAAVIvH5vz-o-iwOC5WMF9WSCH9My4A=s0-d "C\hat{P}D=80^{\circ}")
como ![[;E\hat{P}D'=60^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vv8dt5coYNzh2DL8KhQAogUw18fLIIVM_sUqUVYKiRWynp_vkliNtPfvSIolLS_9WBCOdBfN2QwjexPzWnRkF0UrLpFBeNKYTPZ3XGFAzNf2gHtuKIb45S_tZCHo0=s0-d "E\hat{P}D'=60^{\circ}")
e ![[;E,P;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t7PIMB2YcL_n1O8pLrBUbowNeTfVSA62XEiE1sl9ZF6BQFrAL3aEblAOUypzc0IllXVUgNsjL36dW0iQ=s0-d "E,P")
e ![[;C;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vpuLhG8JqfeqQ95_Sk3LFzLqE5yJaA4ZrsA6zZyi2c0_sS-K85ri9S-8S7fBKcby4hfCjVhVHFsg=s0-d "C")
são colineares, ![[;D'\hat{P}C=40^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vR6MC-6hI9EVg1oZpb_QjiEOu2D-pqawg4T-WPpCMMZCjWIrhcJGXAF1N2276WzN5KK0GT_DuLfhzeX7gEG1WT4DaRD8dX4tjn7nG7UC4lpqRPlHBEihP6MzDai94=s0-d "D'\hat{P}C=40^{\circ}")
.
Como e ![[;B\hat{D'}C=40^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ukYxIVXoTeF3rGFUAaBLVNiqxmYYo9QCSQ-eR2KO8cfbSypCydeotJ2i7CQ_v4w9vZXI5soMf6Pc0Thn60bBLX6MrIBsc0vTgjbM0PBJcPNu8oSYhgF_4xoh0mfmA=s0-d "B\hat{D'}C=40^{\circ}")
![[;\triangle{PDD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tNpHLJGRMisj9OGpzgswmiPCqWBNAPHLkEYMtIeFedq8W2g4EyLM29S0rPNMVUEcespD9LuuftvubRqYpb4OEpsdNTNyCOQ_cntvhC4RY2=s0-d "\triangle{PDD'}")
é isosceles com angulo do vétice ![[; \hat{D}=100^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v-nODRp41WHW1DSUiFKoaWX-e42cUmpIUHcE0F7yZeHT8l3pLBvz_1k0ZE6yR-O7fRZXL8XW22hPoWPUaDDyALg4gabZ5YhzEPdRs9I5dQe5FnSXN9FjWFcdZJbw=s0-d "\hat{D}=100^{\circ}")
.
Lembrando que![[;\triangle{EPD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sh97-T8SGmSZfHBhyKGcNRGk8H9CuuIxZGta7EIkT9e_fHnfu9BV1R7nuBfP2ipXN8xCPj-6QzTy4QQ7-SI3jELRKR7S3HF5boTsgUv8YX=s0-d "\triangle{EPD'}")
é equilatero. Temos ![[;ED'=EP;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uvjNsg4Rpo9LW8sdTJzTYDHJUCchcnwwwZOrFkRs4QDFtbRva6FcTGOweGK2VHGNTBf3sezZImClE7vOGm8TWw=s0-d "ED'=EP")
e ![[;DP=DD';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sjIQnTvj3wr2QMvT1wj3ehimaf-v_ldlCbB1DFvkVXacnvFU0ICZSgjm_8WgNJe3BSy81EhRYKv_-aLsa8xNDl=s0-d "DP=DD'")
assim, pelo caso LLL, ![[;\triangle{EPD}=\triangle{EDD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t4g_UIaQLu9xjy1JZwu1NKHXvXhUUvkhOH9Z44gIlKvf7HHicMKb0CzQX9H_QyeCNSgwOatEVuw6m1gU0RvkorVByGfFOKeWCXG3M_voWdmkUPDXTGdbvDvEKACQzTc4DFvdQn=s0-d "\triangle{EPD}=\triangle{EDD'}")
. Por isso, ![[;ED;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sHAzvVmxHO6Wany8qTmAsSADak4de6Lp6CmPMxFj7m6ZLFVw6JJkKy3Y5EfL8DeLHAv9u1h2V4IOYg=s0-d "ED")
é bissetriz de ![[;P\hat{E}D';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sKy5tfBh3WRYudprz3b9TKPNI-4pjEG7bO-Wy_aZjl0JuLy3wmTAxiUmggXnUMXw3MBjOeoCWt32tuMYsxGVtNV4Q-A9ED7u9cdg=s0-d "P\hat{E}D'")
, logo, ![[;C\hat{E}D=30^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tuqUOq7HGxFWWV-LCH8ecbxrqdUtNy_yYvFO229Ky2Js1iZLS6y_Fs9UWJXmeS_kGj-GDXRWYZn41uM1HTXPNuSWVuclsMistX9UClwrZEdH7_4vfoJsq-cw=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