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_smHxUttn9zjQD03BKx6OB8qqSN2oMVXECW7Zg_jyUII90-Lagm2MAUTMT7HraRlDxihlI7INTp1-smGQ=s0-d "ABC")
ao lado é isósceles de vértice ![[;A=20^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vzk209QXWwBMHfW-ql5F6NAY0AkJRfKDFQZxxrHN5ZexhrMDNcErMcrcd-5mPG2O5l-83KN6I1UHwmkIdL6mruIcYGrIZ7g8x94EE=s0-d "A=20^{\circ}")
.
Sendo![[;D\hat{B}C=50^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sSH3ClO76yhGXnL4jKrp6iarYSiMYEblYKNMBryCkc4CrI3hOzkOlb8iwI2GC5MotusNXAquvzW96zLu_NBmlXOmOtub1-eRDz1aoR8XFhYL53faO9DQBtZ14=s0-d "D\hat{B}C=50^{\circ}")
e ![[;E\hat{C}B=60^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vWR0huar4xG4hZZObKmE1V0Od5LbX-rUvjzGF5SP5mz03ccJkgMNTg8aMOpRP0w9ZLBQ4tTx-AWNJEeJm-D96iy760IantPFSmEUJpZuVowSTOCQrkttogYwQ=s0-d "E\hat{C}B=60^{\circ}")
calcule ![[;C\hat{E}D;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sPh4vn0FANuEWX-eGZhBiUeMrXVOpFUdocAoRLOwsPcSfuf6OtqwhSaZsl37BU6L4lgleX1ymC3WNBWv1Iau229bc9i7WgVw=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_s6Q_lv3Vw-rXhY-0Kg6CQbqZ6tY_nGJStjD5ro8URCmN1B8Oer1v3bC7SetVJsVlPanb6WKal7EIFp18pQuUGorXycTeeZZ1BU=s0-d "C\hat{E}D=x")
então ![[;C\hat{D}E=160^{\circ} -x;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uhCYAELYe-tfxWKXOHSKGOH3Bz8u_SbHClRJYjPC4V-Boh6i7-RuQKLS2RqoBdgOvmlOok84-6d-JQwk1JHXJyRTYuzlAbVRZQZRSAymXDtm7qt6LQWISkV6nCjnhCHr8=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_uTD_vZISVbtZY17IZVuI8igFrKgK1zN5Hv_FlTnVdXpvpQbzZ3siu2PF3-QMDu7BAZT1DPtTPrtVUKKA=s0-d "CDE")
...
![[;\frac{CE}{CD}=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v2WvoTrXKL-RHxykJDTj2DVHPDToRE9F4yUXgZqY4iQh47fuFAh88FmFfvzcjio9DbmMGfmVCtu9all_njW0h2yP_0yFhjByhi2ow3Q_kMqQ=s0-d "\frac{CE}{CD}=")
![[;\frac{sen(160^{\circ}-x)}{sen (x)};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tP2K3nzXMqyKblkX2333vce_4VsYybvHtaW2W9tPXxg3t-6V54GkdClAbfCj6HxflTWBmwMGP-iV8c4umcdsMw0TZk51YtrcCsKK26yqxbrkbPwKoL3MQk1R7p54e0iF93nkFGsmTnq6iC39-8HJF_eV1yFeRn=s0-d "\frac{sen(160^{\circ}-x)}{sen (x)}")
(I)
e no triângulo![[;BCE;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vMSagADckmeQYwZzkkVZzqVEXxzg6jPEEdrwTJExEqcY3tZ3ECBMlS_gXFo8dL4YoPgH-vQdR6XCzA=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_vxOEeOkGfW9INepf3otYEPxg1QLJehjrUnAoCWA6c94LOtHAdgWaFehIxGkiHAlL018QnoY97aqOw85EMjICi08BfqLfsZ0pWQx-xCazjFKVLdJ0Gr-TVpO0KwSZAn6i04UGGLby_teS8IJ5QjIKOv9ulIMkFG-jECKkkRCUnMNfzDjGqEO2bAIBaboO1PSvuUsC55YF9yEry240l1QToT4s0K4w47uU9IV4QOOR_qcbYEYPQygQOXozctLw8_o5kNnlbLF6USyh82Ty9XHGs=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_sBBMN3ffF4hk2J1YJj01VRUg0pu-1B7ozBQN783ZNOaFFM7nrhLJ0d0bET95pX1GlPhe7Rn3fKJ7sQ4xLGHEexZVfRDPJmgC4=s0-d "C\hat{B}D=")
![[;C\hat{D}B;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sMnZS1dJS832FpDOK-MpmCf1qVVOUZsnfq43K_-80kS9PXztedTyVBPtiyDgk_x_FNJ7ViVNW8p6KgFsuylBKzqB-jLOVXdA=s0-d "C\hat{D}B")
![[;=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ueXj-g2Lhdmu7eiFQ5GX7NcJsBbZ9iIREzSnrY1-v3_8GY6D7FDjwUrKrRbBSXnGO9XeLZsfXAJkw=s0-d "=")
![[;50^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_shhUbWR3HIJr0HzGFvACVyTVq_3g3A1p0KhqcAXjEyYQwSY_GfGCfgRQ03OEKzfXHQgE9A9OPocNoinfK_3VJhZx3_0WAmmWke=s0-d "50^{\circ}")
então ![[;\triangle{BCD};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vzVHstFxbtf7OcoL1DxgqMCf0FfbLrdr8OkLZYVt_Mu-x3Ab42hB4H9lejRu-JXL3YHiiutf9YUNbrg8MwdCUrVylBjTkg2rCLYtI=s0-d "\triangle{BCD}")
é isosceles. Assim, ![[;CD=BC;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tPog1j4RdFvM_ZuJM-oIoxVby4DA7aSlRX8NntU36pRWxBxDCkEYyRSu7XeQ5B6PrjRHGeEeW8yTTPsrA=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_tchYjk9NWG_clTXltEUXL7s9UkNeSD6h6MuOo8hbyHMRcSu8QsZbKywFMq8DNzNk76U__mvoxgOQ16q1NDXWpQQSC3uh4ee4thRyYH6l24Lm-GR0wt8pKxpmZvBUZ1JheIMGo6Nu3BMsW5LBk55jLxxrWuuq5zaoXWSZGtbVHdvOMaKofTSidMjMYyWqrqHM70O19t1XjSsmK2insguo8=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_seGk9HOf2pV-B9BKSh0PHieSplP3VJD4rvzP4_oqWWkkuD81AasjsFDEy4YC9OfWSzAfku7VQOa0eJvxBjlstuBs_lp1xe5d3z7Aqqt5usxtwKSpsO1kaBEbaLOIE_9xSNHPLilzEZXSzwprw=s0-d "sen(180^{\circ} - x) = sen(x)")
Temos
![[;sen(160^{\circ} - x)=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sBOW__yS-LV99pxAyV3ycpdHD57SEAz2Mb0tw6-FF71mm1UBiYWchM0BbYIeZ_52z7mku13fyxHURvWPfdriO927s2H3MVmJXAgNzoBZDhrXwkb8kD_iA2CT4bAnw=s0-d "sen(160^{\circ} - x)=")
![[; sen(180^{\circ} - (160^{\circ} - x)) =;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vlGCF_JGmfFIsGubq8JCELeIS6jVSntJZcydSDasKpSIbK-Az9SJ555R9_0fiw5JieT3739CqasBC-copKh18NmsOY5FsEx7ECya1iq5pMSO-fqhfrlT3caKniGju6L6mOf4I6OKCXVEXL5Uf8_jOiCXi7GLdGBNJPfH9rfdlF-7bw7w=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_uIcUiGH41U3wor3mnJBcICjCh1BT_JuXzwA6LSK5bRj3RGfqW1kcLM5V2RxLEsLM86LKPppKHErNUVSxzlEef4l0LtHTMQBDzEqjdwgehH49551BE5YlcmiEm_H6fPQj64i5gfBp9pDWu5G8a0iAchrbx1IBI0a0Q-Bx17l--Gv63DsnxXLbQqAJfOCerKzXvbIkZs5W0nugidZu0D7eKmr0ulqs0FQLePqEL3nbNSEUGgnH0Zz5P85JTFznF85MPEsR9xSpkeG7AG_k9XSGV0MvlrBbDyfjMhf__qXUkfHenoNhalXLjsSZ9UxMIiOXEJ6b-H=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_tnZqg93A5hqM4VTZvx5d6QELOHHnspEKPYejWd8vZtmA6rPu2Fe5etlYoLhGL18UarYf3R6V2FdaVoESZon5L2zvvxIOlxT2WFHFZqzNGwKnMFOUB9-BSBQWGRDCu73H7LrWmdOs6IwAA0EcUZj2CY2g=s0-d "sen(20^{\circ}) \cdot cos (x) +")
![[;cos(20^{\circ});]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s9ISHe_ndxVCnnAFzRG7vRlgOfedL5MJi3UUQMSpCbZOiW4NYJ1S4WuUgpS90diABut65wvnEJewPv8xIavKdWZ5DsCNV59AexgYzog9xHXElbNw=s0-d "cos(20^{\circ})")
![[;\cdot;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s6Rwjw92mztm9Xu49oeTg8eLIGcf32dE_M610Ph2dXVXwgAlSEu3VctSi9e7mT12VL0VTkA0-hodei48r13p4=s0-d "\cdot")
![[; sen(x)=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_shaVZEmIba0AUSbIMn5gmRI0MzmSaZdxCVXZfdyvvsVD4oKO6-qWu_VfnAJ9QeOSFt9DB__SCh7eiW9mqjrCOrD9k5Tf2S=s0-d "sen(x)=")
![[; \cdot sen(x);]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vliC1WhSGTo5ILvans4ABy4ygyDJzTt1pVV9eRd5o2dGOHcjyiqdhVghk2DKm8fOjU3cMaE3JH1EzLCdE-W6VcmiHI7IAdKj7R6YtfP0Q=s0-d "\cdot sen(x)")
![[; + \sqrt{3};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uRHNHTqyKWZvir28LNHQbnDcaKV4t4J1OOKsX6PEobD2HZ-jHu_qyJ4Huyjrsty50N3fUN0bOobIQA2epJahm3d4pF8dsN3pAg4cpu=s0-d "+ \sqrt{3}")
![[;sen(20^{\circ});]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sl2ANsLLoxovoLOD4qHWZnzqVaIThKSDUmf0lM9x7akjgfVKKo93Mirb-AhdjZRbBq4Lgs_zlQ5CjIVfjAlA6asXiDus6la9mGMCrjAiiQ1pvRHw=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_ticj-QEEchpp0cPd9sNxgN-KGVjxqP0Fc7McFnLJBmJ6neBA0N__izz7xaJ-dFVP4yM18OSw468dzBU_xg7lroaWZmvLgRtYBggw2zIXSrCksWFUEVguBOw9WajNtblI9dRM_Gjy8szNJW0LAfkUEUotfKKBDwdLkb9xJ4vC9j9n9Y_toHKkQiO0NTUVh-hvJt6C4K2mQcPBy6ocypwkYA6CB2dkyWV4jcU2JpSI2UzhQR=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_sgRu_Dl2Jq0yob0-3n1u1HZArMOC7fiO3Q_-m4AF8fQKwGOb5eKGXlGkVSNAGVy8ilPdrNUtyPlzNyjAGsG2HP5LpxHzSk11dRpZW1yF1NQ7A=s0-d "ctg(x)=\sqrt{3}")
daonde![[; x=30^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tla80GoVMe_TpONGvBnpf3B1uGBM-8YT--KlBcIuWJjV_rmIX8-K0X-lWqWn4qoyt6Gbwv3nT9Riw3mzmbaQTm9SMEcPxcqIDjJA3YEksX=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_vU1LD557VAp4VNf2JVFyBuherM9lXWaGK8awe3d3uafs7Gynu-WdJ1_WiWDxUyoNJ3JBbKxzxsbuTE8DgMQGMyv_OtoT7xzW5WcYlL=s0-d "ED'\parallel BC")
. Seja ![[;P;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sHl0vurdeUuSNGSuAZmE1hGqmIsaM0xRCjq3o3n4wbR181Ro99lI6NjoepMQhZYwCbGvH4WEDSciw=s0-d "P")
o ponto de encontro de![[;BD';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vJ_Vz8b9CUUlPztIMZMPmNzylQ6F2NPDEWsXGADVczGcY2uRulnaTtL57okEVRvwh83jEpuJeZ1mjASSmW=s0-d "BD'")
com![[;CE;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uXdInnh8VwrE4Efpe2TL64kzjCPEqDc3vT_1WVY2A86pAJoFvPF5p0973wuSO7ocos-6pch5QiaamX=s0-d "CE")
. Assim, ![[; \triangle{BCP};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ub9O_RPWDK7_xE8t6g5g96AOZnanH4eFURqspJj_Z3I1OUz_FV8pbsviVdFea3R_V8NSVjIecm5egOco3h1Cr6WEcOKRcaLoQwviwnAMED=s0-d "\triangle{BCP}")
e ![[;\triangle{ED'P};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t4_D6BD0xIxuM0PaZwtoIO2suLxkC3HkNE-qtKsv-YfYAeTSox8TiPaogAJ5UAqfNyW9GzSgQ0_0dMLEpNc69nFObvSLF7Y2myt3zMOxkS=s0-d "\triangle{ED'P}")
são triângulos equiláteros. Ou seja, ![[;CP=BC;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sEkfFYVeL-EmMfuyybkYUcaM8NiDpqQiXVZewt1KnM7ykKM7YDQgb8l8Tm6XQu_QS0qlaq8JtVnwdhCZV1=s0-d "CP=BC")
como é isosceles, então ![[;\triangle{PCD};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uHoxeY3217VXpjTLyEHxYw6NrNzdcgZezl9JPgUnvCFR-KKk261pCfH3vPcPT9M77nwDGL8rJNpe-wFFgyHkgAxDk8JPAKk6Qvxnc=s0-d "\triangle{PCD}")
é isosceles com ângulo do vértice ![[;\hat{C}=20^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tdjHAjA4HSVifuTdM47TSkfTcG8GCT076vGf2MsXZtKRJCK4bi5pIt8Ewdu1tzBdflf6kEHzx3c0Jb8nd9MhGkrsutTRd4m66NTNRiUE8yxaNlxgISzbl7=s0-d "\hat{C}=20^{\circ}")
, assim![[;C\hat{P}D=80^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sJlbtXFGiFvSQzDgSBxaXPcJpvuuffEwLuz76aYDZkil2PN9aQjJkOh-BzJ_A1cWHGc27eXhpIqojNy1P6jpVBIgAk8DsEjdd1OXJ8SUf9gfc4ktl1UboC1d4=s0-d "C\hat{P}D=80^{\circ}")
como ![[;E\hat{P}D'=60^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v_aHHAXDAOyEoOwKUY4lMsWgeu787KlMebF3Naf8dQmQAt26mR7iCTLNJgAkE9nZ1YQvbIvYxAk_Yv9MrCMunTv5L1BlDcZBoxpVE7OAV80gO1KSB5CzJlKcQ2q5M=s0-d "E\hat{P}D'=60^{\circ}")
e ![[;E,P;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vwBBB1veeVB3_qQOj0JyTfjx7iQgtqKaQ6i7KCkWOKQEt1Id13sUBFgBx9B3C0IK2k2Ver64lAN1elNQ=s0-d "E,P")
e ![[;C;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s0iLFUb4fkegzOqPxsy5oXQ9kNB0saYquPBG8O6ARiuF6g7spOKjReCFLHM99MyZul_UOSAqOQyw=s0-d "C")
são colineares, ![[;D'\hat{P}C=40^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_siZXfLAZ4bacD3wqCaqVvEshv50DsrqsP7hAGv2SADlNJnAsPIhdfVl4bEc2tjq6fYC1KznkHfMH2Hm0ke3aJPWRKdO1pK5709awiQ5aSflu-7o3KLYi_KLqvK7eg=s0-d "D'\hat{P}C=40^{\circ}")
.
Como e ![[;B\hat{D'}C=40^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vEpmqeYF-5MWLTuOOUPQHf_bmeYMn_CGSTg8pH-W6BV08v2w3Bz1V2EGRjs0ywrYnMBnTOGdkU5m2mhNjNOaHOh3PL8vsFc0Im4FPBLSn1kUFttx8vI0eQd1yrVdg=s0-d "B\hat{D'}C=40^{\circ}")
![[;\triangle{PDD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s0lrrlHFeL_1q1kTckZ1So__PwEmgKBvpYzvv0VRWNmN4gWq9myUcHlSvlzxdp3jiaRQMdFhSCGMuCET3IbdyTS1041Wj32jmr68ojLXvL=s0-d "\triangle{PDD'}")
é isosceles com angulo do vétice ![[; \hat{D}=100^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vZ8y3aUgW615Wt_MOfHbqDJG8Hp0VVK3VEdyPNJfKVCYBjKf2Z7JE-2TNwCpeE82vr08Y6IWuoeqFNzmMvXUsfMZf5TkficgNmVuwgwyGRKnvn-a-4LWj05VqbSA=s0-d "\hat{D}=100^{\circ}")
.
Lembrando que![[;\triangle{EPD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_to5uyYRlXlqqG4wWYk9BH9mI5Wpjz8MyU6Ga7a0qZIb0EVZMK-D2sfI9pENK-T7oAgnNUK-4mtxwYcv03hs_Q0ta8qlxXbf7kB9BpD1vdb=s0-d "\triangle{EPD'}")
é equilatero. Temos ![[;ED'=EP;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vOoc2Ng3Sb9_sbaSuPlFbjzbCbVZg_rtOV3UMGArkH9d9R-9XvF4KhkWBopMNr_1A8_dc8ypaqoDuKqYfCzhYi=s0-d "ED'=EP")
e ![[;DP=DD';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uWYSrO5zGPgxiilcpXXt-954JxLM9YnGc0CWMWxlX9sZl7xyF7YUcZvRXV3SLYX1vf-5s-tb6DDhFKKpdALxJ7=s0-d "DP=DD'")
assim, pelo caso LLL, ![[;\triangle{EPD}=\triangle{EDD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vpMJTuZ-XsHi7ox3w_3hYEPDRZzNWfmTxvvfJ0eq-FN-v6rje6lc081eTLQkNxEw20odb5ftNbTxRhdiLytmaA3f3r4eAFG89myjGl4hHRdi55fwiSE_OubAkiMbppbP4tL9Xs=s0-d "\triangle{EPD}=\triangle{EDD'}")
. Por isso, ![[;ED;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vN5u_sJe6rQW856mTmD4i1hmY5cg6Sgd1xcqa9es2eTL78HCDNm15358fQ5EsoaITBGPqOXU4ceMu8=s0-d "ED")
é bissetriz de ![[;P\hat{E}D';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vpysw9vfNeY8fCb7X74ztp3B_b7SnjtgvQyOcg3LTNLg53ForDTcnEHkMZ_yUQaRrqaBHYaKTavY8TIo1t_1AY0J7miyFLGmkJmA=s0-d "P\hat{E}D'")
, logo, ![[;C\hat{E}D=30^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t8Chm1CTfTpn-1f6xM8Vjo9LXCLo922EeO8F5vDaa-2WgO2EEUr3-nadGCjEYopxHVT8H8yf8vOAp1JKaCkA-KzLQV_UweEaMvMyukmMc8TY91bUCOJVipKg=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