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_t4M5RLvd7hqMs-qByIJ9cbXQfzF4QUOW8u3_yJldRwNyfPbFIfyJL6UIIoYGzxrl_haaEe5RLL4leQuQ=s0-d "ABC")
ao lado é isósceles de vértice ![[;A=20^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tizHFD5snyr5YQNTLVYeDey5giv824cIgzzKkCZRrIJjvuy-EGiIZcy-tVnZGMakEWv5Pyy4Pn3TcmFWvYIKKAbRDJ3eD_THYj8i8=s0-d "A=20^{\circ}")
.
Sendo![[;D\hat{B}C=50^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ujFeq_oGZJW8KRgr9EyKA1I-Ki2ghYnycxsVIWOM909exve7ZfsyUnKMUuVnr1E7CBbvQPC-jC4v46rxwUOwXynSJddv3oIeO6Mb3LZ3fwM7UyI5eH4x4JWNE=s0-d "D\hat{B}C=50^{\circ}")
e ![[;E\hat{C}B=60^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s1rHsgic9NBLePGPHFP8AZ-7EArngkcO7hQ5P1ubBbiNDeS2DSnCjgmjNl0jFtaap9CoEzrSuBPSkxTfJ6GlkjqmsZ1j1ceGOEsmJ7f6-YKuQ43XLJwcpe-C8=s0-d "E\hat{C}B=60^{\circ}")
calcule ![[;C\hat{E}D;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_up5SQfyO93ljiMy0c3WGc7Os9PedkqcswfSmBuopPHA6n6JabkvjGFTy49ciq6hTKP9GS01ODZZL26BZuFI6G84UtQwp-KIg=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_tHgC3Kwh-j6xjZEwH_70eD4TXdYIhyKyzR3xGWqk0-_4YXH_ej3npwa0ugJebcSbwOBK-NF6GZaY2fwbs3YyUNLnTb3XRR9_BK=s0-d "C\hat{E}D=x")
então ![[;C\hat{D}E=160^{\circ} -x;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uEmLXW_RQ3YovUP4z25FfwSgdIodWONFNjOhm1Qde6wifl7uEh64w4J7S6_WlhP6mxyyqJjA3dBZ2nceN2nRwmKoZ6mnBqHGiXMAhO6MB3oiCQ-2rY09VnjTwr9AZMXfM=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_skT1ygB0siCbQsjVcHbVZ22YWuYh_FhN8YFQ3GFgTl-B8_L_VYg4ysoMuqSwonFJqSBYPeyBhPV8vwiA=s0-d "CDE")
...
![[;\frac{CE}{CD}=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sR_oVyULh6rjrz31Rf5WKi82_wS4D8y3NUGhvEyS_iHEE5EViE7iv_sTC-nxLpU7AogeCE7jGhAm7DFKheCvYlBXVjkfzRxN_OxLV1ockA-Q=s0-d "\frac{CE}{CD}=")
![[;\frac{sen(160^{\circ}-x)}{sen (x)};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uKCPUjYML1V1rCC4gklCM2C45v5IoawLYb7Le6kOtNLGmATMewm1PJNqU4Wh16kz8wymjUtny9X1QqeU5Fzd4nW6NEGGxjbAUEBuMVJM9cgOz0UIUwGODTQpnUZ3Kg-6n1h3atrLeVNRvgtVq74P851H4NLFa0=s0-d "\frac{sen(160^{\circ}-x)}{sen (x)}")
(I)
e no triângulo![[;BCE;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vYe9bvdSvfiHxj2TFbhyWAQudfdbAQTpy5Ihr4SCtgeo1FsHv9faRiBJPqGLKkeU0gpfH68h0jb4s4=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_veDy6D-fcKucT8pBcIyzr9zN5l1S6oYZr-oZX7K1hiSMrYGIPvq-oI8vKCrhsBmDj4akIlZOrX91hKmGD2Gmcoe4gne0jGqyheKzha7tD6bzzyqR5ewxwQqQxKwnutyxSP_kAE5C-nlXfWTylBGyxtjEDR8krg6BH8xHBQagT08UZmr4e0ZJrHZZa8KYi52kd_petraP4Av5YLsY1R3ohsnsuHKPn-NxIMoObSZi8Md8PO2aRAxDUOq1zBEajXGQVCGzmZ34-Z0nt3cIxYGeg=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_sElL9HcSx9t5SENLo7luwIxSDn_95Y63h4wtCToIvUsx7NJg-rXuJYc3Z-YPGXVjs8YZG7vyxDO2-jh6cKhphypueIYPVbggM=s0-d "C\hat{B}D=")
![[;C\hat{D}B;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tEh2S7_MbtDiG6q6-4mbd_0CURSfVwb9ek4Q11Ch1Y6NAdFMSNvRKnX6RVzPhUUymhcAzFgaiyKaNLPoDy66gDjD1qf2do0w=s0-d "C\hat{D}B")
![[;=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tZguAnhe7-QEV5s-yJwfqVNFE5HY0lvS-Zbeg492e4osvNa9fER0LgXh8Dpq8Z3Zvykk3LyVDA2zY=s0-d "=")
![[;50^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uBvyVYjZfkO4BvOszv9wLNCy98op_pMSRRploP5gV6ahBEdmVR7S0e0eXVw9YoZmB0GuiqTmL9Ys5MzvYB7RO3vOkEtZy5Mxyr=s0-d "50^{\circ}")
então ![[;\triangle{BCD};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_si3WVlemHf0viSil7LHBjohIJQV-ZeP1VP6bOHn7VJyvpTfcxID-OFr54mE_WJWCa49Su-W3rqdfhI5blXyKj36bO2UItNf0nzFgE=s0-d "\triangle{BCD}")
é isosceles. Assim, ![[;CD=BC;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tUji0spl3aUP2nOyrWqXeuOYv-_2D-jtJdXAGlKIDWAgx3rWbJZX7gZii0qk1a2noHpsMMYolHqGnn85s=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_sYjWj5j3IuNssjQ8R0rRSQAvLcoWwg_eWN70wbzVL3WmNgwm2-4BLoMaz0BKgJW22wZl-XUadKkGcvxFAyVBsLa0dk7eFmlfhRTKfOgbKWi68UV7h9261brDDdLFpzRczyeGF3Q8LxUPMwXvqDc3usiajfbSAz4uY4hpxkYQ5p8zsjP2Yx3jhsq9BHkmRE689HFttVGZOPkLARq19ge6w=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_vgdrX65IzbA23jp-QAxikbh1C9QZSYaoX6Ty2d5dLisn51Vq7Bvbx71rbLDdKbICgyh0fnzNx5U5L9J2qxjfL_QJajYOZP01k3GdXdFqNZu8tG-zix9EkXNFHFOU2frTIliiCRKC72T8IdS3o=s0-d "sen(180^{\circ} - x) = sen(x)")
Temos
![[;sen(160^{\circ} - x)=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tJMpShsS9NqcRZzqNH4J7adUv6IJ57reo05zUyLwjO4z5NRXEtommXuovCS4G9iEYt4Qa6-XXHmDdvZBIgXvSrJVBtXssSpT-LKSZ9jTAAPnSfBChJpsig653Qd0k=s0-d "sen(160^{\circ} - x)=")
![[; sen(180^{\circ} - (160^{\circ} - x)) =;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_stRWgSKx-hOvsDHLLd7p0DswfO1QKYV4xtrfoYA1O7WEHjJUpKeOOJWXkUSKbFi-VfzoA-ow5KzS_VwqANmHC0262viCRsXxG8gQaFZas_BNld2Eo9pRnl5CmuCnUsLKGAV-0V3ADZYEFTdM-OYhYTkjHmzFx6U8ypMAHcRhoLLb8fow=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_vvGP0qtfmRoAtKdvnfJA8R8ogTPtROyVzX4mGBWCv9EUTKAkx1ctIP7LRw0FSatFKVt5qqbRkIngF7wZzAZYEPyux9cMNk6hJThBrSWmKucpYxAn7MxbwyNGncN0DVleldmux4Idd3OI-XpvsWpjosR6nlBF95pBq10lKRlNp5-ErabTe5oIgo-oVE73nLxuCdcKCGeEfVKElbtb9M2dBn0l0LvOHjziSo9pOk99MmpvxxWofbUpe5BAAE_v-3yXpKHejhWLjqZE8-vVMRBq5bmesCsZgeHCdw6jSIeAsjoQcto2w2KzkGPzE1RJYuKyTYdb_j=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_u2hVCPLtTyKmb3b8w2qo6PrbHvLRpTSwl96T5dhBkLnT1Gwuw9YpGqenvfWkBmyoN9fglt0-sDq3o_R2ah9AIahm3JbttYVYrw6IrcE9BQzbfjU3nMcp6nKQD6XwwrIdADhhMQuyJweOmi9J2ONag9Xg=s0-d "sen(20^{\circ}) \cdot cos (x) +")
![[;cos(20^{\circ});]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t7LV-AhiIxVvDX7qtPspyqsbg4LYoIBC3DoKSWFvnwCe7ZLxF4_yzNn1uNAPxbgGmDNkk3cIrTL9xtFPrvJyBSIOF-9qbTYLGsr72-rG1fIyONrQ=s0-d "cos(20^{\circ})")
![[;\cdot;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tFswNNLFtPRytv2XFsdGaFNMfoJd7NFNCQKHGCAZRwZKmfkr_1rCGD1sO-h9dreLNapamaci4xhZj1qYGfgr0=s0-d "\cdot")
![[; sen(x)=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_txNN35WneVacMaGK8lHp9SlzsRgVV5fkz5xn-G5-PrsU_vkQzY4ep1KVsvTSSdoC0lJQpBedt8lEyAUaQRTt-1bDYn1vLz=s0-d "sen(x)=")
![[; \cdot sen(x);]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vatl9qsiYv9cdBT25qTjfe5SxeLbwKPVktUAtPcPAsQVWQSn7EfzAjoWOORB_rkjn30x67T0rLG3odrNVeT0Bz_gEvPBAZFN8UnZ7PNQU=s0-d "\cdot sen(x)")
![[; + \sqrt{3};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tjvbbxHIdK3TGBYpQcXr-aF-_jltuKPmMiWI2Picllxh-UbMqC8GfHnGldT2_9LIq3gybmaPOTqyVVWc2lfS8Nbx1Ys3C_GnWlGEqg=s0-d "+ \sqrt{3}")
![[;sen(20^{\circ});]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vCvAnI36OcyWLJ-lIIlZvbIFo4TQn6C_vZrX085HFg3aGqP_hzLB2e9tCnHC5CIe6CJrJEO08juLYT038HYORxxBcIL1Eoh1yfPmqMr5mvLxHN7Q=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_tkWllXY6ZYoDnmGQcyYJyDorRVJ29d9tq-8HGRFzg0Xmv14Q26AQNAorVshJ3c0bnrRyhFgCE4-Nux3vrZaylafPwnBoEOmLv3aLsT7ESwXb6AFySFjqxfpH7ZWc5WbgHlK4oPysCeSCzVBOE43jG2yp1KGK_EIXrznGe_8zyrudYEg3W8LbzSJv3CH3xtfd4d_PzLPG4GetnYRM8-B8hXO3p6BHxYFS0J7sE9PglLksCb=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_sYiygowqyfQ4eie5JSqLmv8VMuLnUPb_6Cd7-FRODM3OEOOYBz1fz2dip5QAQC7P0uHVlEhzflN-IJseRYoOrpFIq5ixqO2mYC7s8fZD20IT0=s0-d "ctg(x)=\sqrt{3}")
daonde![[; x=30^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vbVexWhdPo4QLFlE-IDq52cuvwJuGKfFUoctLluz7T0z97310EbSo1_pfuJydKyHA-LUhGQ2_KuI1fDwF2L47yWGXhKBnVOaPUbr24MMxk=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_sfKNb8A07Re-bbaJhy2kAMAk3GZyBoJlzThPczI7UIjB3XIjq4dlZ8lB-F_e4F9IXLuCtu4xtGawiUtD-xhM_PRZvafhR3CaOjkhl9=s0-d "ED'\parallel BC")
. Seja ![[;P;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sKqNPfIEqJePYOhKER-RPNZ-tuaaabCDCDgsSM5Rd4UVTJ6tpqA3akcgNEv0A2E0xPLqF81MrK6PM=s0-d "P")
o ponto de encontro de![[;BD';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uFnMGlC1gOjPYTvDT_uGX7zyAXSQFHe6IlhfGGwldruDzO95gimEDQZNzg5PZyYBXxxjm5-Mqu_g6ny6RZ=s0-d "BD'")
com![[;CE;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vQGcYYm73uLnIeGMcILSPPtw_4hrTotVatuAkKKV4ObLTazXDhcwQcp8bz5-REskzdXUyAw5eQ6Olv=s0-d "CE")
. Assim, ![[; \triangle{BCP};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uRTLhCC8n_O2wwUes09Qfhuh4DpbIYid6X_IgIv1K1JaowODAbyx1XRJ3a9wb4Z9NiQAdY9wrfy1dLuRjIbc9uM8XbPC1hJCKSs8KzPlMT=s0-d "\triangle{BCP}")
e ![[;\triangle{ED'P};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vGVlstqfapBZQF3SQP-IAI6bujt94sIvIT4lz4j3cwnvbCCyHWX0toZesNNBe0K3MjzkyozET9DL2GmSUJvP4ZNAQUrFHredU7Nj3DRS8b=s0-d "\triangle{ED'P}")
são triângulos equiláteros. Ou seja, ![[;CP=BC;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vjlTeTPw835MB4U33H269ADV74GmNeFefcqB2etyQ_qCsLEcndCxG5dTJgLQGWWT2MMifayQSb58Bhrud-=s0-d "CP=BC")
como é isosceles, então ![[;\triangle{PCD};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uwonjdZxfmlDiggY-cvQ2q8QzmMpBgkcLbyLYVcKcNiDIq8WObU-JVVIKMyHJAmtqf6vX_jDVnP2xkZC9lJr-623Hlqt4l-qpk2hg=s0-d "\triangle{PCD}")
é isosceles com ângulo do vértice ![[;\hat{C}=20^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tGWnVvmEBgQxkJeMTE7n7oI9Xa4iFQzczGD0U9_euT8ydjBRKFb9jd_aAV4_6A2FQz9CsTeDfxmf5On2T2jkBK2Wr6D2TDI81w7Y0npHosRUvRkn3YMlcW=s0-d "\hat{C}=20^{\circ}")
, assim![[;C\hat{P}D=80^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vDG_j19_wD1hDIjPDsnFcu7I6dElNKLMu94I98YUyRAwDkMqQIOdlpDzkGkM-bwM7sXz3UDvav_JtJE1Le2K2JKCI0XWi3i0F9c8QJBTvk06V6tZdulkNZB9Y=s0-d "C\hat{P}D=80^{\circ}")
como ![[;E\hat{P}D'=60^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sxjTjzslEjMSl15Luw6odHXrcRwxDBAS17ffLfTNEz4DR5JlAsbfpXqO2C9qKQApGKuCkk40jArtpbtZMR6OtwdksIonJVZuzguQUAfQHQdP7ib-ZPXq3LYkuZDuM=s0-d "E\hat{P}D'=60^{\circ}")
e ![[;E,P;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uq_SH_EluQLad8q3ddR3YAu8Cs-VEwNUqLLyfSTuA3OomL6sIAKF0rBvVwjVhq7-x5D2HJ6gxeJ3N2xg=s0-d "E,P")
e ![[;C;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sWzwdAFxsKUrShT-1PGoqfuHrDSCAZeFY9kCUD9ljRXS8a52YErEyqFtDMB3SaTzOn7H0dIToVng=s0-d "C")
são colineares, ![[;D'\hat{P}C=40^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ujX2Nl5r1KnSN2nVBnakbUfOf5xZOB7BF_dDkLCk-lPVEimJzkFza1eeCqyRR2zbzh4-2BrONaqz3XBusGOarKIFVevwxv8W5Pg2sjBWTcHSBZYmUORvP4j8AlozA=s0-d "D'\hat{P}C=40^{\circ}")
.
Como e ![[;B\hat{D'}C=40^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sfPrhvA3tgBeDeQZOPrdgdnsL9PuPJw3SxiW2mIMPquAvbFjJMWI5MjrLb8VQ4emytsICiyRS1wuwCUE1mAjnpCB-7HG2pyDHzAgot1dMKHqISXxhGq44MQPRsIsQ=s0-d "B\hat{D'}C=40^{\circ}")
![[;\triangle{PDD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v-9UVCfaXoCt-vE6OhN1FvMs_efcLOP4Jon9Uo5xFns01kkEjlpiM3QQJo0LKItzlmoPFCHn9CnhU9kXJHxQj6MnwaaP3g7zWn0vQkifSD=s0-d "\triangle{PDD'}")
é isosceles com angulo do vétice ![[; \hat{D}=100^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vK5CaBF-8pxb0H73IRvBwsT7sXgnNsx6bk-ixQyMRIekZ7nRApspovb3iMwHz7XHAmLJCOMkY96SuueGduTDqd0brYmON00DUcqXWNfqKtNd83jEtvCCRm17x8Yg=s0-d "\hat{D}=100^{\circ}")
.
Lembrando que![[;\triangle{EPD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_shLfjVUWgRdm1vN_tT6Hbv0m5eqh_O0_5V_cBkhFUSQ1BGycPSxFHYpcTMd_w49wuv-p9wqUa9kkv51iECW7cDQ6VTzvfxfZhMQcls8Meh=s0-d "\triangle{EPD'}")
é equilatero. Temos ![[;ED'=EP;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sVekcIylYbFbIXqhNOeI0CTJBu4BLqp6sKsb6d4bvldpFtnnZxnSIPuLTzLxrEC2hipAT_c3jYEbAny-DPrmtw=s0-d "ED'=EP")
e ![[;DP=DD';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uvxm2phGn7W3nrVR9uhZ69ifCX6FDLlwkp1TY31_5ZVdbG77MgyCyYJMvFk370waAXeT8ATKKSBUpFTeyclokZ=s0-d "DP=DD'")
assim, pelo caso LLL, ![[;\triangle{EPD}=\triangle{EDD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uSOFKEgwwkMgLvqW_IZXZkrtxuWUenFZQ7gxDTPABBFKrvkTszIQ5-kABtct9a9hqmDxXt2yzM-wQPbgNodQUIIeXaXT8gmMlnoysYeu9HgQj0mEXYG4iNqlGrStnijWgFzjOR=s0-d "\triangle{EPD}=\triangle{EDD'}")
. Por isso, ![[;ED;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_syC3xwOzrDkFoiRsLlN0__YIxPBNZraF1T5nu_YO1b38zWNBeLdm_Y6flnVqwrZAU5aH4g9r-A5RhS=s0-d "ED")
é bissetriz de ![[;P\hat{E}D';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tHohknzdGAEE931AtR5AjLY1ukoJ3SPpKOqnA-yIO25W9EME1klDhCvJGtqjbDvHVvTEsEaq_ST_F4zFEjfSVYzvzTI8Hg0xNCeA=s0-d "P\hat{E}D'")
, logo, ![[;C\hat{E}D=30^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tsF1QyAunJs9Ja3NAHDIHl6qiRqycR1HvBzAiSCSQ1-xte8IS9WvvcrthbyXqLzeO0oQF8Sh9HEsnfntIcZeySrnAiqMK_h9d3zQWJ8BZzjQ7wvgr0J20EFQ=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