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_tuzZvsiemwebl7Uof_2srAsF_t2P4QvxnhPuERECIl2MkWX6ladz03JDjfSg3N46Oxr_ped0ixh6bFng=s0-d "ABC")
ao lado é isósceles de vértice ![[;A=20^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uMqyGyjj4J1pNyNxjbC-364M3r0dvNdDdRBeIppWN0m-pAyNon55QLuc1qIMRkM8Ef1qAPBQc7ubViystSazqivIn9Vbau48bVxRg=s0-d "A=20^{\circ}")
.
Sendo![[;D\hat{B}C=50^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_smLwTwExVyGhz5EbE1n-arj4bel7fKUVrI_Dqe0QozS22alZ6_bB6TPTm9PAmCxKwjkACzL7sXjHzVD5G_Muq1sdGZXgU1vMYPe1U4jSwIuGBz-Hv36NKZltQ=s0-d "D\hat{B}C=50^{\circ}")
e ![[;E\hat{C}B=60^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t3tn6a1O2Q7SGo452sG4aCghSPV1EffEDuMqRaMcOviIUf5TLlI5pwGoS5-tkv3pam8Erkx2dULZU3DGpi85yKTC5IPp5j3Z-rVt767rkBuh8yHZOjuWmv0ow=s0-d "E\hat{C}B=60^{\circ}")
calcule ![[;C\hat{E}D;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tPHPeZd_cXCGZl4ajcP2JrRgHv1VxydgQ4E7x3aeDxDN_UllqeF8lsLLPXshJ303Ie59laGEzQes2w9qn_gmOvyumQHKMQnw=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_u4Bc0ZMMOZ-40spI96ThmhnIL6P1WPkt8fnybARz4MU2UD-dN_uleZWacJqzs6-u9Fu2xbm2es6Fa4oKew1h5myt8LhjmNdYdN=s0-d "C\hat{E}D=x")
então ![[;C\hat{D}E=160^{\circ} -x;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sAEttefTrISXvypsbMVWXsh3VV1B0h0ZWGYJUSkNhfT1OEPnHctZCCFPtdyOMi8_q28xAi47MNKEO2ZZcpwJCME0wn8L_XI7rlPYhlGacH-OEGEL4oOtmNQdpC881aECc=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_sIr-gydfQJZ8_axsqUFX3dxyTmfBoPVwiqhAO9ykQBkjhkRHfFX4ENimfvnrTwXRTY9MwpLAVWzKIEug=s0-d "CDE")
...
![[;\frac{CE}{CD}=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tQu7z46WL3HOCT24pINtrajPyXhUqmJs2Nv_1FuVnzyREDDWjRWBz0eVid4e6Q5XjW74MJXXGOtlsOo-Id6a4a0LUkOdN-XUdzOi70D50yzQ=s0-d "\frac{CE}{CD}=")
![[;\frac{sen(160^{\circ}-x)}{sen (x)};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u9jsY7n5d1y1ufBM7AKtmy-wP1oEBfUt3ypieX1F9jpB-ZG_jX_GEwoVRC9z7SV_8LKTwkWtfAb9gnZZquIFiktY4RqnZP0zHofitZubz2zoOk2dGriTPfn5j8mMnybLQHHUW9W3-cGdk4JPJAIVlH9hegLPcG=s0-d "\frac{sen(160^{\circ}-x)}{sen (x)}")
(I)
e no triângulo![[;BCE;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vUN4sK19mU7qxG75oI3mOiTnRbRT-QV0FysJEgda3qZE336f1fGAvdi_obGhgEx4LJfKiB-Si8Jzvs=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_uHQj-0wlFA7WFDvB0MOuRpwiVPqCaV2njjKCmwsXRUHULA0xey5TaDGcGf8hIc53I_MCFCGJEXiswuC8J0M6pmZsGGSaP2-Kav4DBtQQvPNkCfbciYkvjANyw_aEI3fXO_qQ0rX6q2C-Bztqt0rhECIRoCgBkNojv8hJaR1ZBE9Gb37eH4eahwMIZyQu5SRL8ht9qoUx9DNyA3lvmR0DJBZ-gXM4eORxA7FT4AobXzXwbSLylMhVDh4AijF2MrLwfyf7XcdmnGxSuAsxt8T-k=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_t8yiCxgT9RA6XDoXwJtl79wOuk9dy4FLBZ0vpNrqKFIMP7caQzNDGgSbktoNlvT5XrnUZzvz3nPKwVBotBOzaEa1Q1es8x5XE=s0-d "C\hat{B}D=")
![[;C\hat{D}B;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uIFqrLnUgidV6S-KOW1Ci_pTaEodN-SqoxV2LpwElHxo4tusZDIVp8SsXMK3Q5aC8TDtgy7l22rdKOvsrgglpG_2FkzTaCBg=s0-d "C\hat{D}B")
![[;=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vV97uxHf2JdYwDoot3dTdxJ6_4c9zYI2MQfO76GibP6ny8Pqwz4BhBAVvPP05QpAtu2aSHdk4tGo8=s0-d "=")
![[;50^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uXwFl38cojxpQb32uxSv7vEh3Xj9L6GC60KORQ8PmibpFmQbTWsqlYLahxNMsJ1kyOijcUcccsbHC_pwN8NynPR8D8TE63aGoW=s0-d "50^{\circ}")
então ![[;\triangle{BCD};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vgoKVrmoCjoiR3tTGS5djGhgQugzXmaVx47FNCg01ojS_HPREyLXykAKEn8JXh55ObngqMqXVnw7TMPWMAbNMJ_13VZd9bFUndeek=s0-d "\triangle{BCD}")
é isosceles. Assim, ![[;CD=BC;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uQcMQ5cjCqMS7BjavMiUSwrRuN01aXeIt7IQ_xUv9TWxrfMzzv3JTUzJMLAARK2fsNDUF6TwJ-Rt4Ij7M=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_vw8s1pRDcawZOyhnrAz0xeM8JyHZIiyMP88blrEO9RW5ecVIEYb8wLyegK2K0rtt980KvL4qHttMoOuzkmQz14AhicYUJBoIIFkyWYA78YtQHmd8l27nv9qCr0gzD7t5gvUPeB1Cj25lHIpgyWeVz6GIYbKzrPdZEVFj5KX3RVko_J_8JRYg5ZUz71MTKWNMzkoQnRlYj17cDqgAsiggA=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_v035amdXSYsv1ItUaA0m2tHZiIf81GwkaO0S9ZSCZXOTCALGk2-xdrXFypWKNtBVD47arLcBAPwJmQf3-fA3R8Q2F2lkPViBd8klurl79yXHHl8waiugHgKx26AnrHM9yxfUfYFjICaINNokA=s0-d "sen(180^{\circ} - x) = sen(x)")
Temos
![[;sen(160^{\circ} - x)=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vpn59KI02JptDRM55bZRvWMccj7rGenCNneblr82WwnsVyHAiI-wjYQf31M9qsTnOPnz6rcc9cZC6koH62p5wedeIDugf18owCsoyxOg5e3dDP2fDOlIUGqQKIfTM=s0-d "sen(160^{\circ} - x)=")
![[; sen(180^{\circ} - (160^{\circ} - x)) =;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tlRWsrNxhsuOGTLkDpDSfKR39CMsTn01y5Q9IcG2JRzPhZ0Z48eblhXse0sngcxCnIf8f2_zbNClpZHcnnVhA63uq0XS8jUWXGlqqYOXx9iAouiSX99Nvu08tt-RP_ezSbemxMhZYn-n1yu_YV53G0qGtbcc_pz3KAVCzKyeiDYdwt4A=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_uu7svBU6lZLufbXK7SleTOsAjzfie4978ClWnyc6XfrHmPb3uoJRErQbqTgUPxKjbWKxgkqNVYnz3x_mJiZY7PDlufSW8wFsv7kZKPBksLC66E3C6p27lszRjiqy5l2OnuM56oL1RMg5i9AboqkV8eFWXWgmPP3dF77eL3ymHYLZG_22bQ3tCiAftL56tvCq2EEEn8cU2Sfea6LLHSonGXpBVBIc-G06VC6MhbxsOhKE83Dx-76-pa8MQDHsyhBfDzWSWxJlXeBGDCHKTilUgYBE5KjHdRdbEbdkLCwe5iFIN25FUUWvfNwutfEOG6xkKiKPUN=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_uKM8reMD8opsPtvMNH_hPL-ZJQkQ2NcYoZTkrsLvLpgtVV3jF3D-LpdMweZ7c5y_78GMuKuE-OtLgxwqCVewfxEu79D9jSgNU-3OKGb_NUq9aWTT-wOvyGEdd3rR51VulfTyoegMcihN1rhRxHUxLGvA=s0-d "sen(20^{\circ}) \cdot cos (x) +")
![[;cos(20^{\circ});]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sB9tL2Mn7kefmFP_kEBwh_1ztK93i9YNShpN_6yaud3g5Uj9tueEkGCmm2qOrv3NG8H6tdSWKGF22U5VG6nk7DgGbwBUqvTNrJhAZcpSJMs-0lgg=s0-d "cos(20^{\circ})")
![[;\cdot;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vi_7yu9iHW4ZB4nbySQdpWRyZVikel7YwoLDLA7jTRtpuPeGtyUitNSPLvNC1qOPKuO0tD2oPwwz_schEUGUs=s0-d "\cdot")
![[; sen(x)=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vjoDR2db23dosBH7-eWg-4KDVtHvTs8rxOorVu9wrUdBjuUrnAFoQf1Oo7zJF4CUcsEupGCAoPM-ZN3P96-TVEYaC7Bv79=s0-d "sen(x)=")
![[; \cdot sen(x);]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uta_2slKFLfOisXApjgzE52AsdFlpjdJ_L5JNVytFkwnqjiWEQh-AKDCaB3noyBk5h48bRtviGqbLhbv0G8dBUyKospRlXKZkMNvn0h1o=s0-d "\cdot sen(x)")
![[; + \sqrt{3};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t_gi2ltRXC7NclTne123ZtxKErbb77zm40gdNvu1wj1FtBd6-7Z-bL4MxqZ7dVJ2VGYj0dcf3c9nSZlCZL08yux23nadfmSYHB47w3=s0-d "+ \sqrt{3}")
![[;sen(20^{\circ});]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uUnNY7kjE8AFgYCJVJWe8IkVK-U3RodhtTBhUa01ctbJKfbHdYBEdZ9lk11LbDNItJnA_Oq4ZunFC1zo9Lx2FrOeZNXVWj0dE6_aCFlb-zZSnoYQ=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_sQ1Iiee6h3KnQybNHCLndgyDxk9TqTvSLDEjeg8e-sHMbjC3rN40NGKCySsekKOUeq1kQW6jYSNdm2EtUpo2d7nAGhO-irKZFLM2or6DhtxcWe7OHZZNDw0oJrjKTzLKOduOds2IsTSCNv-uG-cKBbCG_O3xz63BTNyKZgXVnX-Km9qLYpE3ZYtU6eNnclHnJROEXpv7dYG0JitECAvXwjYXjjcb-kYAgEZECiATOqmERp=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_ukrHt74aIaBKo8y2i-cLSz5goAW8OBMKAwoZNYt5GAMO55oTnyj-dyNj9xGQ-mUgJrqRPNhEfp6D7EjvRK7863SQ9U4urMXCdxVeGRtAzf6dI=s0-d "ctg(x)=\sqrt{3}")
daonde![[; x=30^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sLHdw7W3oMCRck4-Mb_QrpK6Ck0dxlHJfDIYD7OmKnE9YlDoMo4oU-yM0oGhwQmSMxBAMBY7Gpji-enp6_sHg1gXpN4yXFx57u3FcMlWFE=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_sBhgWjtzJXpMFO9VjM16aCAeR0saBzZJtnPe893uZC_mkNVQgAbLA_Y2mYsf-3Z_jyFKUitJc8XAZwfULk4V-UWnVFwx5cGEBYYj9-=s0-d "ED'\parallel BC")
. Seja ![[;P;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_teLEzaE88UeynS-2g-NHRNJIpRLuDifDZV8xHLw_hum5YOgp88sfwcSFguLnV_G1sRR357crZ1EZ4=s0-d "P")
o ponto de encontro de![[;BD';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uF5t8wER4l-pBDjLlDEIvsLSeTvQM4xWBp55imjyO3XLLiLgSfbnACM6zTgeCWI79Uys3ZljJikJ83bkcT=s0-d "BD'")
com![[;CE;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vciODfxfgYE-WnCMM1Enu7XiHhJBrM-EPQM_ETmiSRd35e6187vYzcK4qWTwu61-KDjIg5l3iEijvf=s0-d "CE")
. Assim, ![[; \triangle{BCP};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uzbpP1OJVPf7PpCDHLxugllJv_cLylRlDrsB-CmP_b_KvxN_IUL6PpSy28YV7W77ym5RKt5EIc7QuhL8XRTjDm8UqLJZ3rv1P0PK2uxibj=s0-d "\triangle{BCP}")
e ![[;\triangle{ED'P};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u_-h6h5QEQpKvYBd0cR5t2wDWVv0Gy1uV-oF4l32tarDNyFqlchUAZHXANmCnBWriQwhhUSpUVqyFkkp0W_cr4g1cIztGzM_J7A3T36YoY=s0-d "\triangle{ED'P}")
são triângulos equiláteros. Ou seja, ![[;CP=BC;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sbsElmdSeyoslKXx1Wgied3C5Vs4j6XD4FUg2C6JTH0l6fqzlT6dpbsqQK_GUeAehbTisriGjwZwDgJL7k=s0-d "CP=BC")
como é isosceles, então ![[;\triangle{PCD};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u2HIugnBbG83kS-TSSVqS4HSaRoGtte23G6Nft770qWmjE8hUtzkMDU5bNAI3ImxOOzRx1faKoGivTh25nuHsJY_hKpgViZbk3KiE=s0-d "\triangle{PCD}")
é isosceles com ângulo do vértice ![[;\hat{C}=20^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s8O7sg2VTbeilF0kswFocmioCp1m3XGEh_Grw9fPs2OTIN6nKKns-p4N7VNpQ8d48ezXPvmqeEWu9tgTIYL5MLmJMhAYAq_w742ba8lABP85GP97oktwhf=s0-d "\hat{C}=20^{\circ}")
, assim![[;C\hat{P}D=80^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uwgEXmoRMbJEVgVdkJwkpWYIJQRXVuxCA7TOYLv9MOyOUNcyq_KzRIWjU6OCGrvtZNMOpgyeswL4YZGiPFgvdMiiwLf83znwVqhtnSg_cRmi85B8bAbnf8ZDE=s0-d "C\hat{P}D=80^{\circ}")
como ![[;E\hat{P}D'=60^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sO6ccGAyiOw2qIpek7dTvvdoTXnD7OE2hLvUrPCawsg2DDgryWeYrLZb9lUd6ON4412Nd47ovwMUqa60476qPSc3HEzI0IOjyW2CbE49dPObk5ryvxOEj6KcKLou0=s0-d "E\hat{P}D'=60^{\circ}")
e ![[;E,P;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v_TLZMlZz37BTN3cAbh3THAb_A1ZqVKBXPoSJrOi9IhXvX10Z5KVx1IVevpX-GEbtq1MjzxMyeFvi9Ww=s0-d "E,P")
e ![[;C;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uCzpT4KvKtlPqK_YOSQmK3JvC-UN0r157KX8C-pAoLap9fRBUlwpfP5VnIsgF_FPjbUWrfyie7eQ=s0-d "C")
são colineares, ![[;D'\hat{P}C=40^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tABJ92E0ynKDwUwSoSHOsR06sqxSIH4Ps5A8lX8IY8-_xTbTJI5bEljXchgwqNFB_oDMdci0a6bMRPadb90Q4j1DUfgD-2rYuzBQ91Y1WQp_0YYUN8MVpoCY8kPnQ=s0-d "D'\hat{P}C=40^{\circ}")
.
Como e ![[;B\hat{D'}C=40^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tl-jzsZ0cHfLDI0pXo5z2NX1DWg8GnC5OERFWjAJv2x-0dnGpDxWYC2QWfhOQkmYg2BkUlZ2ADBrCVvSvKIJ8R8JHE6S9zK4YyXtm716SUE8E6nC6NnMAYF8wFE_Y=s0-d "B\hat{D'}C=40^{\circ}")
![[;\triangle{PDD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vFHOeUqE71rksYGRTEg3hOLO5cJVwRbq7gF77xcdWo9ftQKuWVQ0mS_EQdb690IhJjOleK_zPnReLvvGayxiHNfQAmetQo6c3mB-59uLBd=s0-d "\triangle{PDD'}")
é isosceles com angulo do vétice ![[; \hat{D}=100^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uqANzDAkxuO9NDDz4pSr_J5eF7g_LkI17hrwmgi9AuNlYmiNp94sCNAh7yotlYH06xIvRdUGjDQ4s10eDgmmRKAsTGR8_5ogKFEKVnKmXIraRWA3BTNtMBE59Gcw=s0-d "\hat{D}=100^{\circ}")
.
Lembrando que![[;\triangle{EPD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tQp3xN0kUniWG9yDKmEtzn4coYTo-X0uBcTC-c_dYBFNTq7kSOxfxMiXwEKNUKuvmElq_kJMt2elgF17sGOpz7kYObkySHkv9pyALJwjFs=s0-d "\triangle{EPD'}")
é equilatero. Temos ![[;ED'=EP;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uXR8XankUglDBoV__QqjoZpEkRH3aV37Ew9Y_MFjBX9whFsuscjX3Dc6UzMyphgudE54e-rUbAhHehzHhG3lwW=s0-d "ED'=EP")
e ![[;DP=DD';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vH9VKPwP_c94ZDpIWNsnD8uIbWaOMx3Kst7Rj6cVB_JvrmJTfgl6rFws2WqJWcUN4ScTdrKLkBUB-_WIYEnMlI=s0-d "DP=DD'")
assim, pelo caso LLL, ![[;\triangle{EPD}=\triangle{EDD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sxurfc4PG5exbPgYFXTlmPczlTSSoziSq_HsEMsoRomQZ3EAoqwgR9ICNaBRtbqe9vMNRNgzDtYSErCzbonMVOFWq1rIq1tOpzlmlFZc8fgikh8SD17UnYRknRf9QX1lM46YBj=s0-d "\triangle{EPD}=\triangle{EDD'}")
. Por isso, ![[;ED;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vtvqGDk86nNzsB3lJc4Gyto7fcS3lVdcsPsC5vxaN5BSlmQSOQY_a45py-JfnC0wsCuVFl-Jc8KNLb=s0-d "ED")
é bissetriz de ![[;P\hat{E}D';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uF8Sdpop-l3gmsobMIFGnSZp29rmsz-nBuYzCsYKLnV1GcKWps9aZgrzQkVOQG9u_m_A5-Evawtz9uHVMDiVsXGA1GyjShMlu0Gg=s0-d "P\hat{E}D'")
, logo, ![[;C\hat{E}D=30^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vOoWo-OzK9Zy_XoZEZJmWGO3sO-tS5BlHx_cVileGRmhTgVoTY4qKiGgYGuqeUpe1sLXc70LuVGIHzfgozcixkvUNnN200gMtuaSndxtsuZczDvaDpKzA-8A=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