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_uBuduySG0AVHnAQCkawM8Q6UaPgMjbq3Mo2-bNEraXIE0xLEPtip_50hl8ztxk1_2WNxd1fDCqxL4GiQ=s0-d "ABC")
ao lado é isósceles de vértice ![[;A=20^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sTrl3JLDrLaqjfxJllOtwraDvDpuQNYJfLuhLF-NElezcgUEQnklony-HEeUV0lDKwAPRKOJyldhEbIK7ofHyVsB1OHI7SpFXaIxA=s0-d "A=20^{\circ}")
.
Sendo![[;D\hat{B}C=50^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sPYL1C-3FLOZgPKwhit4iFuUN663NFHxQ6-AENj9eF2EBQVU3utQSK5Xk0EVwnj-_d5ry0vmDgaJADZmttoYplHkdTVfmVu-JBzFA9W9yYIWi-OCgsskoM1sY=s0-d "D\hat{B}C=50^{\circ}")
e ![[;E\hat{C}B=60^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t-C7IGUnT1Bj7A1T89usTvBLFWlkdE-GHlMDWy3FYN14G-wu_NyPmQ04mVXtO-gL8-S2Ab-4zUANNlDs4TH7zB3pD4bJge_DGMveMjtcD65F-k66IbJ3tkghI=s0-d "E\hat{C}B=60^{\circ}")
calcule ![[;C\hat{E}D;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uP08KtCsSoQUEfBsydpLmX0eGKS-cql1JZh08mVs8etQ7AGVka8ioJggvOff-FiIMydj78wY0EQg_KNDJFcx0oKHNMjyK9qg=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_tusYe6sZ17P_CQtxvCNyPp-koVyrnc4zQo3eeaoNfM1DHBJoEmQsE2Hva1egsNFu3wbfTCWxlFgtvAe62HmDFsT5WamG89uMUa=s0-d "C\hat{E}D=x")
então ![[;C\hat{D}E=160^{\circ} -x;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tCWW_skwR07bfORmigIMIiH3uw0Je1ALuF_ohE6c6MUaip-FqyqVqqlJSbltw9t2IySelIvqSfL4ZISw8nGPWAnDz5ldGPtfo2rtkRiGWctMx7p5O_ZXPnoBEoWKtr1nw=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_tLkT18GMkWeJoIfVhNrhplf1MNCjrEYskcLOJ2bdvgk_DWFdWLWd8_PiXIFTv34ya_5YEAUnw4DFSOXA=s0-d "CDE")
...
![[;\frac{CE}{CD}=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uIun28VQ9j3vCjrc77GROEcmdihIimNqwCcfJJ86OFgIuIltSRk0C1i6a5rCYS1yRM2bJHO_u0WQFM63DA7xq0CcTp552rwAtEcal3r9t2Pg=s0-d "\frac{CE}{CD}=")
![[;\frac{sen(160^{\circ}-x)}{sen (x)};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sIuXWW5eG9esM487teS5ivQYz7BSqenNoUg94KS3Z8zCcbIopzr_iW_blXkpUexRM23nNwShflU5fZG-LwG5ELNME1Y5dXoUHgqiR6KMltCsGTQtVd5xfl8nxXmRL5fCIfInbnc0Gnu-33MuUI9X5hAW_3YCWu=s0-d "\frac{sen(160^{\circ}-x)}{sen (x)}")
(I)
e no triângulo![[;BCE;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uy-O7A-0dlBlN-7WwPVmoquRDW5nedtiTMQpcHKo2XhAhvpDbcyjpLcVWi6G4oFCIXoHQ272mHAE2x=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_sLKElD8mlDzW9PxF9aQPp6s0ZHIglEvWuvmgGkL_R4t-lpPM3TSR6ZJM_gRAfielmS18OabQnBI5cU78_NKA3GTZkTrybM92BonpTzRsXYPwSfcl4wnCnLq-vL356iOExJPYEe6nxma0K2oMh-L7uUv_d5iyrSSCRpvGWGvsWvcFus-zypu-Tt8_yZi828-ROJfCQY4padfjmNXpIGYzgwNOsmTPCxFSdeHHSsofiD9tMMEzzNGGC9-pTxs29eTcdfsUxCJcxazI_REX2zlkw=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_s98ZhDCEUcMs7bPEA5Deon9J3Biap6DKwcAPI8O2aW2qZgJdb0Wa_VfzwspDt649v4sk6XUfomHuHzNJFlKoy8DZuZ86afOG8=s0-d "C\hat{B}D=")
![[;C\hat{D}B;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sCU05G-mOsCvXWFMOyMrPkLV9kbX5nEqRX8UdfhDj0b4Hck1XW4-2efbScsfnJZrBe_FTld7pImJ4_8pkSZiueeVC_rJWgCg=s0-d "C\hat{D}B")
![[;=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tnCyfeyAei-fLTRAgSTk_grbleQn1YlLUrkQ8Vkz1iywwBKrGr91dAajPMMAyeG8S3esWZVhheuDk=s0-d "=")
![[;50^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vfgGObKSEak42MQMTKeeoQs9EvnJ1A4PG175L2s9EjhLXqKe8KxkNiEiG7e5Ty60T-GgkDs55gF8VHuVjYdCLi2xfdNtbEBo8d=s0-d "50^{\circ}")
então ![[;\triangle{BCD};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sPRuZRnyPmiBcotoJbbBqTF7--xOh8yA2HY9wy8d54o9KGofIsWa0moFGiQq7sfRRMXSZakq_ELDDl0CPp-DD7csBUPizWMI46--o=s0-d "\triangle{BCD}")
é isosceles. Assim, ![[;CD=BC;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ttMz0nCHF6p9-oiTWlfRAVwfUGQImPfymUkJuSTD6mwYwqpAamAPVrdNwASv0uLprWCDUgj5bYXBw-f7Y=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_vo9V7K9D1iroudVK8Wp-fi6yrW11p2YegCbSCTzZD1yutn_zsmaHVC-tLiFcX5AcrR7bRnozUDCOUCb7Cy3xzCx6y0pOOiH_Eto7Fes8-ZvG5xuPiKo7LxlTE8m80agpu-0t65LouJ8q_z1VJM6QfO1Xfe7hsAMzqde5Pv5GELhpgGq7H9dg4njwqJ_ZcXtgcY1JLR09wNCJK_Em3SyUU=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_vzjYKIS878Cl952MqvHvUBIToSsxawsieAkWQH-x38P_b4jJr-c-d7KVI-BFJNUjKLocEMwtTaK7iKDfHzeux-C3aM2yaUtdifa68DQ0pJONuqRd3vbLMZr1qsQRsCnQ7y4yWzp4b7kr5Ejts=s0-d "sen(180^{\circ} - x) = sen(x)")
Temos
![[;sen(160^{\circ} - x)=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uZ1DUL1Uyvqb-rGbQjZbLCxsGrblMP5kxS5hTiPJTtAUkxdk2vx9eZcpUm_wKsDbmNZRygOVYXA4W54iUgoxaBnWdiDj9st5J_uzW5E768PRG-iQglTrqNqOIuRBk=s0-d "sen(160^{\circ} - x)=")
![[; sen(180^{\circ} - (160^{\circ} - x)) =;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vtFtUVnWN-d9LwVoSNP2irp_t8kFfPhED50RvJfjxOo5EabliC9FfVk1lbvGDa90jPm3UzJf3B55G_4lXO37Og7Qwpfgpr51nu3kClAN9bWSJtMNMvh0TTdtIzSi-tOvQ_48tUUhO0ELw6qZistlWvu34Afh1dPF4UmZhZDK5CTVVMlA=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_syCfW_YUqacZi0REyqxsQkiHVACPHY6OOtjntmK1dKoq6-ddY56V872R7CquPCIUWrZ3QFH5Sf7LGaSKHhMK3tnlDgN2FEjWPHHxghRtZC-ep-O8YMowRXdJIQ9oIgFm3rUO2o1bIUC0ckkrvsGMkgfQlL3qEJq5bdobuurxRM3ndoPZuuM1KehRtuBgFxqoTG_TvFoCrI0PqQLJ_gqswn2XX58QW1CYy1DJBJxEUc3UV44NkwYMx7sQizG5ikH87_fArdl89wO5NBfCHAKY89akJHcT0hJRYMHhyxpmFqnzaCpCNqbiUZjTwTZDxO73UgUGV7=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_tZFymWHMpkIfDtwfQEAUvn4WJqMUwjxnrGtV4Ph8-J4Sq9uOnkx9tJSJWckbCfc9tZGMqkPR3UXPO9KMZwAssv5xnbjdXc-eVbWH0V2lz3lqssKJFo8KV_Xn34u9GnMN-q_EvZ5J7pNnTMCGicpnnAUw=s0-d "sen(20^{\circ}) \cdot cos (x) +")
![[;cos(20^{\circ});]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vGXp-ddlsGBGQNn_BtDKauCcnpBwJwuN-fO7QUUN6qdGgz-sOd5u4k5GYScRB6cHa_G8ffCvWwjfa6Y1hFgWSJ94Ai8SJWTUbEXxZTRLgc8iZR4Q=s0-d "cos(20^{\circ})")
![[;\cdot;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uYFXJjqD_19KOqDbnaygZ7QwIkMsoQVzj_hRDC7VwDC1CGDqc_1t0HfZxdQvCcYOzb82NKYHrufUuAD_JjAwQ=s0-d "\cdot")
![[; sen(x)=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uTbdz1TL2T6Na-cFcKEyjGOcQXhhS8txRurXz_kj9SoA-fk2tssCUzIp0LYxCRzIfEvjSzivGeT9lLa8aIu0klGFhspUP0=s0-d "sen(x)=")
![[; \cdot sen(x);]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vaF6IKONWrScb6QZm7CDoodOc7T2KmD6_JHCnk0hyHNj7yMNMLdBw9Lqx8JeSBGt7jwBXhdh41BVsIzWsEmDbL3X2UAViHGxazfJVKo5Q=s0-d "\cdot sen(x)")
![[; + \sqrt{3};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_szzhzGRFjG9vBc0sMw7MEZbmw6TSllHlpdWlUQKVzF9TVy6RSbK07oSvUjNXU4UL8NUlylS4WVpUSKeohJ8BgpCJh4-F2pgpfxFcbv=s0-d "+ \sqrt{3}")
![[;sen(20^{\circ});]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t-7UoYTGD1IQJXQ1YmXxko7BRwtfysnqBiz1HY0WMOpXVvuMUfB0Kx_f8D_uCBDgQJKVvvgBNHXrpDv2P_Jb-wasgpIBif_9u_Nov8QaEWYCd37w=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_tItPXs2jwxX2MQzmnz8Fn633lkWq4b69xau-2mym2xvemf3876yPMrGLT6rZItRIvQb3PB7-Ht7Q__4zFDXcyhSZyNdqTEYRQv2UwLB1SzYCWpYE-0OMHr6X-64YVIpZI527Ad2MUf87FMtjIewnFIGRmPr1mLrLOsuuw4s4-6EGlKlzCV_p8YJbe7Lxs782kMKlkwtaWt7gNOYAYg6FxHM-OIeFihfe17m8Tg9L7ZYB3f=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_vIZLbJ2y_6baH-BhMA6E5G8G7u712dlwv5LOtYUWQSUmSnhNEMHDVAtT42Gpe3d6pMtt3rVZIAr6VRuQV0Yw-kCNGgZlGhXcSUccj4R7gd68M=s0-d "ctg(x)=\sqrt{3}")
daonde![[; x=30^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uqhgTIb6xGFEv8NE-3Jmd6xcNaznK8xOmg2u3kmnIpqRjrBrB2UzHiqEwa7YXGHHqVTmSLaIvms2gYGrmdpnTvxyhxNQX6lghbSmL1HCUb=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_vU4QF3DEHCt8tjYIZGyimwMzIGTK7uk6C9Inhwxy2BLGOtP-3otD-5mN2IUzbig0cn5l3NBRrTRpiQUDea7Lqfz9bkKIVRlAn8807a=s0-d "ED'\parallel BC")
. Seja ![[;P;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_svmBe7CTl8OoyCuwniABXzumSi7fQ6muER2gtiDKIlF7ljQM3aY5L_7Robojq8ol5yXkRgjJNXWFI=s0-d "P")
o ponto de encontro de![[;BD';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tTmilUqZFClFFVQYxE-9LIXct__pvsUYgnr4Pkqd5vsvvKD6HZdNWQu3lFO5zUpIQBhZGOPTM9spYmDKXD=s0-d "BD'")
com![[;CE;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uyf2pHPy9ouWuNygNLeTKHgYVC60XkMvcWHj8KBCR3Y0YxFKz7AhJ2TfF5Y0qLvs_Ll7gvDZU1bo3N=s0-d "CE")
. Assim, ![[; \triangle{BCP};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t4VpT0zYaI04d5XuC4ZTN85o_tih130b_hGrJb4Qah0-36FIMsaR7cGOQLXfqAHqaFaFoYDbr997Yfbqnmo1AkiehHxb5UydXPe-339Oxr=s0-d "\triangle{BCP}")
e ![[;\triangle{ED'P};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vSjfqEyduAmhcPqhfBJUoGam924c2eAk-CNL9mVSPbe_rwYTbUnjanWSAX5XsxIQv13vuHly8iHqB3yDbok82vU3hgvHTMU--6h9ZzNRLo=s0-d "\triangle{ED'P}")
são triângulos equiláteros. Ou seja, ![[;CP=BC;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_shuBskl6dGl__mQilO5uch2kyyNZ30bVfzb2niKmTdhYKaFcSQqgRKt90hLJ9jS5kTV2D7Pxn5KQZYVELF=s0-d "CP=BC")
como é isosceles, então ![[;\triangle{PCD};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vKOHVqno79wWtNtFmrhEH-2T8la8MNkarvIMjpAldI9sGvOOkhMeXBeZ3x28MhZgzjKifDze1lnbodDFY1v28BMhLbXB6mOFHjIYw=s0-d "\triangle{PCD}")
é isosceles com ângulo do vértice ![[;\hat{C}=20^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uJ1ATPttu6GxJQgUeCVlG3jdNdSkB81t5xsrlde3AAr5kbOawDsP2xf5RB4EdAAEtUxUQsfPZfMLQ92DNLJR0eT--W838TcsqKFSNGl9nOW7GDjL_OP0E-=s0-d "\hat{C}=20^{\circ}")
, assim![[;C\hat{P}D=80^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uHjSUU8-501z1oQnhOg_DzXrvXUZf1vmkEiC8oXuZO3eurwutxP80TnSUN627vxxstDKVkJs1OsQg6mqwsvTGkB_lK_DXIW74N2GI2sXlhCIEsmXHUCSzbgyE=s0-d "C\hat{P}D=80^{\circ}")
como ![[;E\hat{P}D'=60^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vJ3KBdJKZuNR6ol_quz1Wqyb3amnC3CY7Us110NGz7xszC3SzcU9_JE0iMiK_XIpuyrBAkeEc74pS8Be4DsQtPsEtPA1kDaCnZh-WsFqYUwi06njjtvyrCGIK5oHk=s0-d "E\hat{P}D'=60^{\circ}")
e ![[;E,P;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vTN2S4mr7iYdhNTTyBeP1UVd2nsPVxeKaDlkkjHjMKVKxdNhVqHFNMH8tDSWDuroXEe25d5Z2hET2t8Q=s0-d "E,P")
e ![[;C;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sX11SkiR2BPtxQbelbOls7Jeshu-71MXZtOYFD7N_LGfk0u8wlaNMzbnah8o2DKRSLf1vbX7DmnA=s0-d "C")
são colineares, ![[;D'\hat{P}C=40^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s7MJoX2mrJ9TtLGEutIvinIj_yoCJUb2KU6doQ_mOce43T41hBjM4Qgsi8i7I6VOongijHenalvcbNjwgnzSbAo74BkqUWfAhrrUqBVZHWYWGqeBIXsexvaW8aBJI=s0-d "D'\hat{P}C=40^{\circ}")
.
Como e ![[;B\hat{D'}C=40^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vBUs9vrzJYSvPC3VmiQvAuZ2EvLw1nJsEZOzrNmuhpFyBzPNxkbMjSSwALf6QqXlfp0rJNcPYE5bk2mM3_lIdqFEY95Rm-ur1042KBS0KGzjnbDYc0XVtfKQ5h84o=s0-d "B\hat{D'}C=40^{\circ}")
![[;\triangle{PDD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sRfVRVFDvFI4piPj_AaztxWpw5FuCAY6XupRrAn6ssaLZEYrLNxoM24qPFn8R7RiRN4pncCKEmkegA8LHnZAuKFXVuaHpTLUqRkJiu4gqS=s0-d "\triangle{PDD'}")
é isosceles com angulo do vétice ![[; \hat{D}=100^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t8JYG48L_g6HXDg8u_THe9fHjF0BKM56B6TrdA12aCkVZ2PrcYDT6A1S-zQfo7FA_QG45aQv4Q1Qz3RYoFf4L928-mBJxN9mY6UTXcv1C-eOEgi-nZSApSGtm-LQ=s0-d "\hat{D}=100^{\circ}")
.
Lembrando que![[;\triangle{EPD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u2fkpzgxVtCravo9dvCviEMIz75J_WJh-DvvImXnQCPqYPdoEy9XfMQX6cCCmdqi8VL-ua0RbN27i43Vgd0iZy46K7PKrVUQ3bj-BvRrSD=s0-d "\triangle{EPD'}")
é equilatero. Temos ![[;ED'=EP;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sr2cALCV55QJnltUnVsJDkh35TX5d-RS4sX4uZec44P6E2RM0SQ7P5gc4XLoLXNWZepCR7cYnfIDh3823jQu-o=s0-d "ED'=EP")
e ![[;DP=DD';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uyin-SouxDRK7lX3FTAeT-fRsGfFl1UsHS7PQZp3RB9GWLTqUTVFCyVHk10c6Bu2P3bCi2Vx893KXzVZG9a0Kw=s0-d "DP=DD'")
assim, pelo caso LLL, ![[;\triangle{EPD}=\triangle{EDD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_so7_JUKaQqFvHZz0AEC-8NNo4egEQevWZScjCiOJk_OAjn8Eigi75c7iZHKXPRSq8LgGkXCCpD2aetiJLfEMWuN9ZF4A0lLC5bB-d2ECFFz2n1jkg0t_z4mN0AN8PKKpRU2Rsl=s0-d "\triangle{EPD}=\triangle{EDD'}")
. Por isso, ![[;ED;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v96Izk5guflF-RRxtk67i4HysrsQz9CZ80bWymmYLfRS2qg-mBjgJ4wj_ttz7gU_0JSPiNlKdobHm9=s0-d "ED")
é bissetriz de ![[;P\hat{E}D';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vHsbqblfdTZGVY-xWyl1ph7e-FheFB9kDVtN1KlsnPhQuHOgIA3wWMXFMc5YeBc9XzZkCs4TV8AYV19A_LKJGQ4k2PjoiqrAkd0Q=s0-d "P\hat{E}D'")
, logo, ![[;C\hat{E}D=30^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vAog232e9BCe8OACNfXaJSB0RNRJ2V6awY7K6oTjMfgec1W1-aR0TdsV6_aRBSLXPjSq46oLEyxqE0Om6vDhkefUus6jz7QhEf02AfrbnA_nXeMrJSgyhdMA=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