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_stl35hLfvVXCvRo6gBxp8ItXKlbi4MF4vNKB7gT_GAy0QaWeepaL32xwGxEaHeunVi761hT3yNMLOskg=s0-d "ABC")
ao lado é isósceles de vértice ![[;A=20^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uQFkCimrnALtLzW2zu4mol6rFqNyQ2lLyv3Hbo9GffWsSXattKry5k5_RUBN2Pt7DUdRIzXwIzoUC-iNfpz_0Ru1y0w4yuUXgmlDA=s0-d "A=20^{\circ}")
.
Sendo![[;D\hat{B}C=50^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s_Jrr5OeLuKDoGr1EMz5yziDNXYDbqeIYn_6RntgWBF923d0mi8RhSOp9tdrvME9arHLvPnnOWFug4-A8_s4FEXNktixoDIufEOmShjBc7OJQtj2hDYyAiVsY=s0-d "D\hat{B}C=50^{\circ}")
e ![[;E\hat{C}B=60^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vkG2gmSghy6zBgNeLhifQTrcraQOQdyAMHqrUysEIdzfV82Y3fTOgH85_zIS5dD1iyMXnFFHAO_CtirGyHQRf5kJpGC_PkR4pJOjTDHtRZRKHm1BmzaTSpiBA=s0-d "E\hat{C}B=60^{\circ}")
calcule ![[;C\hat{E}D;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vgkf0ughtpfIiCeiEq7nvlifwLRqKZm_L0U-J-AXcWz45amHxFDvk1LGGI4Jajda0LwbOOdrE7838UfnRbaf70odB_-XMh0A=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_tXQcbR7elM6xjyPWbMWPHd7soVSRemthMCKDWqcLdFff-DruAwME6bEke-UixVGaHTb7qtm0UsOsDZLJoT6BOKbkhEChHy7NuG=s0-d "C\hat{E}D=x")
então ![[;C\hat{D}E=160^{\circ} -x;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vUPC8BZzgTF4QUjVk-spk7HaaV7c3AtTl5nXkXL6757sxebgt9WvGlcB2-TToqskL9G5y6qwSc2reDBam9pyp21jm6ZINF_uDXA2PDUieHKapxfKXtyySdJ1DXTBLVIwM=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_usX5VlpZbljqrrljI_Xs2eKEBpvCwIpENBxTO-TyiHVmsinP-Y2gKS6eIkMWFTTkR9O6ttsbGxYFrQpg=s0-d "CDE")
...
![[;\frac{CE}{CD}=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uDtdbGZxoJg_Lgc26dbd8mke3ufgMG3PXxjcjZJQcLsMOvK9GxLiDTI7VuHcQvhjLJ3U8zdQMy3VygSGNhuw1o5yK3XcJ0q7MPPN2iDQeLcA=s0-d "\frac{CE}{CD}=")
![[;\frac{sen(160^{\circ}-x)}{sen (x)};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_udDQgaaU_uUbZIgfkfyBv9EpMaW3VgK3ZT2T89ow8CYgRYVynp764ymwBVeiMib1AmX_S2AqdsFoGCH8-m7qeTHNRrqmOk-wgYov9UYa4aVCd8QFKV2j334gf5bCLXVFhf-DGLKaSaSWWJpqeRgTXGVd9t-xMR=s0-d "\frac{sen(160^{\circ}-x)}{sen (x)}")
(I)
e no triângulo![[;BCE;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u0T7OkcsRXVpGO44rKwjN1WrWVex7qQOBqdPVu6XugUbCmhCRT62Cw6mDTDazlnPaanvWVxPemwbjZ=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_vWPYUrQNBC1p8Nda24L__zKCPx1xjXsJwGFN6D-c777HglcspFa22U9m5hgvl8TyHjQh4ZXbKzCskn_8u5JnoDfIL69Xme15mv0kVXQYJH4MF2YXYo_hfs91k61e8R-caxs0StANknxVaEzS_CcClxZlNq9XVQzdR8WlLVrV0XupqrgtUr4J0SzTGvpciG7wvu2S_afU9NIVg3jZBNZhFFPn3IJRxYaEZKdwXpdFWoZKeb_nZEHnzRGt1TypsV8GF1noJc49n7nLmuNQdSsK8=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_snIN-OW3QoMkG4goRZuxS6nWsJ-FgjO-ppff6-HR11b4ideWv82w3SH0m4jQw2RO-9F1y_ppCcnPv3f696PoDTD_h89DR9tfI=s0-d "C\hat{B}D=")
![[;C\hat{D}B;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vDwGeWknacvMWKRPNiaD07YqWANos-oaftE3d2waUP2XzU_BYzkEN6F2ZiX9zULO_3FueytQpo3p0JsHCOlKrv6dUQtsZySw=s0-d "C\hat{D}B")
![[;=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sayvk3NjkS_dAX4j9ra-NpR64cxiDUHlcqgNk3rZPpneaQ4uO80EsniSLGgekZJjyjFTkBvcGw5g4=s0-d "=")
![[;50^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ssjdrVhipcSHAEdU1NIYMR2389m58wEeF9WhURjulY68PDffV4Xu8w7hs6fAOtwAi8VIfYU29AgwW3spjATrzCz2uFOHGIBzpH=s0-d "50^{\circ}")
então ![[;\triangle{BCD};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tuZOEyNiNF_MtNDTljZnPgb4BQLjIYsWWhbuANNjVDqPvbs68_O2y4RhHn-GXNWh60heqbZlMImb2717ljG9HOlGRL_EF7z8GW7RE=s0-d "\triangle{BCD}")
é isosceles. Assim, ![[;CD=BC;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sK0UcKUBtvr_VzPYdumKnPHHlu6drkyl0J-pzhtx8Y0AEiQCx-uXLAbux4khIJClwve5ttjzK3hDnW1vI=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_smiXt8ZboY3fVqzPiVNBW5WQlEemlFHA6EIZOl03GcclnOVny42m0R45gafybUKCLUp0pwz0DW86r-BIF7XunF51sxfT_8VbOYpqsxBig0XVFu2yMGmPvzDHfH_R2VmyLa99eb73StsFwHQ37uGoY4MyHa9yasJc_GF2NJP7fy7w-qU0kqIZv4UGHLNsB3AnZK-bMfJvSm5SBNsXaFGLM=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_uOq47QidaX4Ewa6_aZ8vRJP3sDLD7lwvlqk0kr5eXxhXZrAu5Bk5zecgPpPIHQ5xN_guJ7xAooAVQKNVSgXNijyWS-NS--DZVoHMDBbk_i2yscPRp_zSxBUJiNZ4l250d0aMUWeWkesUVQfz4=s0-d "sen(180^{\circ} - x) = sen(x)")
Temos
![[;sen(160^{\circ} - x)=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t5tRpj3Sbf1fB5IBQKEZuhoHHOAjncU_2rpsNGRgSTbIWbrcE4Hfy40ofHpDkm4tXZnMr46-gsw5PZM2xPomoxsFV1AWTOOrKFunt66cJNMEzyF3SO9IRnOlwyPOs=s0-d "sen(160^{\circ} - x)=")
![[; sen(180^{\circ} - (160^{\circ} - x)) =;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uWKuP9Ar2szki6WNeLApKC72CDHsPDBb3tzEzINr_Ft9pbzk667tkcAiwWOo9OVTkMnWGli2c27lNRFdn75zOk1lyqriUjrYs5uiJHprjY0mPsuDMGwfomB1G-o80qgX8j7aavSu631TBFMg4RLmANQpurzeQWWtbKR70DaqJp3ntx6w=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_uomhQNCqzxIa5qEDGaCOFCjXfVzfpxMTIWSIIydCrZsP_eX9-DoFFFElJzGXIM8GrMOufLOUNBCaj9Iz3u7UahDj5NDfqk2kQaqppuX6Djq8n9LcZ2v5GCbGz3wEGzZJZBjTGLUdVGDyQlOSlzk-TuqxHiVaXxz_UM211q-So06QM6B6Oz1qUmnBeWRPa16t9OO_c2ETw0m3IG5aULJlQ1SLI0Jmu6lmueERc2oiKyu8FfWyuGo58mWNp9_dBOz9sxkNCNlajHZifQLuCEQdJ7GHDqaRquaINAtoHRSQkiIBQed1AKVeZt_sNj9_mua_U96eyi=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_vEXzP3X1kTxDO5zcQkxB_xPQO9sHfZ-zGHvmP9yRWSc82WCH-ho_xsNRt7divxLyj9vKDURsZpbbq6OMMBP5ub3GbT6FdLm6wqc3N8iXg4nuT_jGKqiUBGeHY_gaYOalUMo0tZ0Qg8wZY-t7XRLIJTIw=s0-d "sen(20^{\circ}) \cdot cos (x) +")
![[;cos(20^{\circ});]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_te7nIzmynmxgwhPcimPUDJdfaUlW6tCFUgfNKpfctOnyJS0TkNZt_935EgjKG5Praeb0gJ1cMHvlOKWGmKv5Gx1mOK8VYcMo4ne4NJY-QbWl1T4w=s0-d "cos(20^{\circ})")
![[;\cdot;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uCZRqZlWV6KTCoAJEvVr-vWE_3NmG0n1xvvkUmnHnVpVsfnnYFc8Ejb-Mn3Xj5CBUcloLVh04Dp__aFNxWytQ=s0-d "\cdot")
![[; sen(x)=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tg6wFovTS2g5a7OwISAeDII_9YQD-4Obn2VQ1Nv92ihHWAn_c8iB7VGpaF_y5OFUSH_b12RzrfZJR5oyLO0F_gOHKDI_95=s0-d "sen(x)=")
![[; \cdot sen(x);]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vaqyIFfIx1OMZORQSYvbznWYDpjcNMybyqCSXB5OTV1NH8L4jEXzTTVRLWGDdw8KPTzUysPVb2zMrclETCe9Bb1Hoy8gu76NMVFgKqqAo=s0-d "\cdot sen(x)")
![[; + \sqrt{3};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tpKEIoOaW3bDHdEMtyEE1zxXKa2dUASo6-9cTeWXSEDF9V8itnvAHA_YEuPIdhtVjvJnvnuHWODqWIo20J_PthQd9Q-AwzJsZ7GvL8=s0-d "+ \sqrt{3}")
![[;sen(20^{\circ});]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s_absNLaUx0NNMN8XOovqSROUk0u-bGCiRbIxBjTybFNSrmG948qNP6Zg-imKZK7loG0yJFbDL_2Is-E7K76JC685_P3nW0KPLsjyhZXpUUAPUXA=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_s_8HVX_b8jpcUJlFeKuP6MsL8Guh7Eym1G-C7jYyH6IO5X-u0DGLl_yGADDmWdouMT8AFN17MCtUxeN_MrBj9kv6WRPB_LW-GFrCb08_w0hVcdt5LqM6H83ZZbt-2JRRk19A5VzrSw4LBOGEWWkXTcLVWj86XLfE7BVQ9tlhaUuP-V2BQRTA_87EGoc2TK8wfQSuuxMb0TsjL5KlqeszgVPXsiaGyJ4xbN1XzEz3xnbIO-=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_scOUucuoLfAbHPnRvVWqdesUMsSXRWk0tq3C31RZnYNV0qI2vzj755NFtZkXSa2DRCv6jO1iYEIxqJFWa-sRl9aHA7Wht10XGfRfEt2otHVRQ=s0-d "ctg(x)=\sqrt{3}")
daonde![[; x=30^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t7lpR-QjoFgcT1nqcV3sHNFG388vXoluxzD57v3D3jmRdfS1RhuSPZGZFJe-XoiSUAxggSOLLs8Rk_IDKKtxNgVYCqPu6L02LKNNt-T919=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_uqqYYEwwIayoSgFNQxKH1-XswFPIQe9tSeONZm9U1aANs0RO4ts9n7OQ67BxQ3h_rjXn_hcJGWaWq7zenlaFOiSCd3MsGaf9UPlsXE=s0-d "ED'\parallel BC")
. Seja ![[;P;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s1GmnOxVb5iZQSmwGRfvsstcHq9O2LDb8THqP9wFGwRhialUCZ1FsKX3gFzh3mA-_lJjnNiBzB9-A=s0-d "P")
o ponto de encontro de![[;BD';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uEZG0bxEBSzhoOJcspLqsSPKpR4hIGi67AYYguE3L55wdxYoFIklslw--oKEvoJkRl7fCjEVTSxY0e-s7B=s0-d "BD'")
com![[;CE;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s3ZM-XzDBM8y7zOAvzlabbfIVK1xM363HtbwhrfAY30kn2xOkCjmEa1GAZ39RnZ1QXTI4Fg4m1mdEo=s0-d "CE")
. Assim, ![[; \triangle{BCP};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t43QaKuqgUv4ihN9Ug73-dAkxjAhZPFj9duWjk_hfqrPUwWjI7XOycMyr5g3NB81cHP6DxntIBhcxQN7GNTXhMWuGo8eXOoMZrkRPw0_dW=s0-d "\triangle{BCP}")
e ![[;\triangle{ED'P};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vg-OZMhPg7SW-POzQ6DYHKm-rf7I6U0WHsIeHaj67fyM92GiX-KWj83BEHeCAKel59YisYmoWJss_MGps1DrUlemnrI8vlzj8GteqvNZTI=s0-d "\triangle{ED'P}")
são triângulos equiláteros. Ou seja, ![[;CP=BC;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_va6wOdQmU1GSHhAolQqvlrncMXU9JWdw5e4dKJXblEes1XS6CQxBFmYB8pl2vIGGeGfx0VHyII65Kfjsfq=s0-d "CP=BC")
como é isosceles, então ![[;\triangle{PCD};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_svexjKluK95KLsUY6BmQJkqYyIq_OlMidv_46KXX66Q03B9UGdW2PPC54PPjXMYfV7q3qP__R10U-JZAhnv5nBqUw1AWxQ_4UdRXg=s0-d "\triangle{PCD}")
é isosceles com ângulo do vértice ![[;\hat{C}=20^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sD2EflSYR4g_G5_9KUmIFa-dydCTcSs9pk3taI8TgmLOosPDhHQm7orKu6sMzafQIy0h-SI8XbLNSqwnIf9Ovx9L9FR1XM3-is5yfTZ_WtwCEtzWnui9py=s0-d "\hat{C}=20^{\circ}")
, assim![[;C\hat{P}D=80^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uV0AjAVBf0dpT8KviGgp3BOb9uF9MX9n-XRH1LQeyVbftpLfyXfs0oGQafykXtseZOPApum02Ieedqyd7uzlzrYcd6ZEn_ZNrK1PggPEOR-sPsr7N_SEn-Xw8=s0-d "C\hat{P}D=80^{\circ}")
como ![[;E\hat{P}D'=60^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vHorDBJAwp7PJ-JcOaIMTD7N9yXp5vmMTwqw9MxiiNftUhcQ7JaMbDKnEPo70SCF1WZlQgwPGyO_-txFmaNjzn3Fgh_3X85KKSudHYQUALp9sqBWBqiGtkCDS2gZ4=s0-d "E\hat{P}D'=60^{\circ}")
e ![[;E,P;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vJp-O8wR0yjbxKdtfqM434_vlEh0dDVEQou07X141QAZ7zU5p2uoEXYpG3GfiC7AXH_ahG5His6pn2Ig=s0-d "E,P")
e ![[;C;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tvL-FZD2zeanjetIh5RhKees4Qi2GHOv8BaotU6a2ixfwOBFAcdUQA-WVgACeFes5wpOzKq1W1-Q=s0-d "C")
são colineares, ![[;D'\hat{P}C=40^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_twVctYm5vedtfVVtu6rCnZeSCL5hnMOe552MpxEAcsthx3k-tze5xra278JZ_n4sSM0TxSohM-2Y6q3LzBvmclGPIcKCf4Emej83ZjO2nn1M0BcsBx-LK6HZvC6Cc=s0-d "D'\hat{P}C=40^{\circ}")
.
Como e ![[;B\hat{D'}C=40^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t0vhEDmO0oh8m964IuTVVE8kgzKs0H7rSBMyoyRLneldwYKZMBRuKvKpxjpZdk3RjfGW7zITTURSfkC0PSIyUv4CopPaZCpQrR1leV05fHrbxNCUtPMNtqP8TE8sQ=s0-d "B\hat{D'}C=40^{\circ}")
![[;\triangle{PDD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tS3-3moHppWWr_AwMf9tRvieYmE3pmMKcpp8uyzk6-Nzy_2VEYBwDkQdmezZUFbtJBSQ_o-OrQR1JtitQZArVjZrJzEl-oi6B1uSJCjuWd=s0-d "\triangle{PDD'}")
é isosceles com angulo do vétice ![[; \hat{D}=100^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v2STh9c1fH22wr9iL2_ch7Le1xVHEc84_5HmjJf_9R5sUVYWeXBH4oPphK-5e1agSY89FcAmC7jgdQAT_oCmKEqZsiFYSvRgPaJ5U9dOfSKxMqH41xsKZkrIWBYg=s0-d "\hat{D}=100^{\circ}")
.
Lembrando que![[;\triangle{EPD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uBBQDQ3LONpDve592KK5h00MSlVx8xn_jx-bUjDMt8h2njITI_wZvz904vD8SyN--iT6t8i1pIVCe8zU_37XH-M2sbOL5usM6xtISaKOvk=s0-d "\triangle{EPD'}")
é equilatero. Temos ![[;ED'=EP;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ts2nmwHdsFQPiLr1tUYQvjtb7ADQjc4ETuUKNN7VMjA0m_GS-gD6DU2sE0LYP1Im59fdhx0O8ngiqrxu1U92Md=s0-d "ED'=EP")
e ![[;DP=DD';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vzg9MwnFF2ysHR_AIj7_z_3r4FURZ9URXjkvZhQ0vT7y0oW8F90eMt2RxFcs3anmYJ3ndvWqdAn17m9VNFmHVo=s0-d "DP=DD'")
assim, pelo caso LLL, ![[;\triangle{EPD}=\triangle{EDD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u35efKpxUcE9wSGSYtWu0DySzVXkkik7u8C8oGBcOlK7BIUCsAlizFdUfT4qeMFZ2CAUzIY5jwEoWZeUhmAJTriQOFxuArH1jE96fSAIdYx_pb9ojkmqYI0iwGtPsHgVg3XBFb=s0-d "\triangle{EPD}=\triangle{EDD'}")
. Por isso, ![[;ED;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vspT9ha8UsjrERVMmtMcXwkNsmvqbaTSLqjdme9KAaN4kp7ve5W5pXGBpF9UoX6a5jXST0Q-CyY35k=s0-d "ED")
é bissetriz de ![[;P\hat{E}D';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uuURyD3VGXvKsvVuWJQkVLqLrg82PriQfv2pExdsuaAQ-zDEuWzVwmt42Umhv4iL8XJ8N9pn-4NV8njZrDZ72oIN-G03RBB0EFmw=s0-d "P\hat{E}D'")
, logo, ![[;C\hat{E}D=30^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t1MAeDcfjab5Hrz-AaSuP_1fKlUlBZdI01a4B5DTLjiBlOwoWnZMQFiy4rM8FmHAtbDtUER0vpy5vVP7L_9vSqAhsFHxfIC9ol9dpTvA6eCKwxjPdONXuB2g=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