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_tp6b38DAE5TPTMI_iXXib7pdYh74bG61PORLSNGWcSZbFoUfmUpFPw92MO11shUPOJQaZ3QmYx8F2weQ=s0-d "ABC")
ao lado é isósceles de vértice ![[;A=20^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s1Wuo0lYcIfpiEU3M50SENf3SNOHf6guLhvzeuz0VxLYs9r_ut-yIgLQVVt9Bf9_IDbtwDxZQQOo9GLmwUnfWIU_ejzXnIqVmmpO8=s0-d "A=20^{\circ}")
.
Sendo![[;D\hat{B}C=50^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_suh7W_TrAquuvOkUG8KtspOl-hOMfYkgPtPR5zSYcXezE8knAPN0xPtICj-7wcxEIljgKtzM9WjyyWnKA9JO85uzwJa77b_GAcBatKWj8HQ2Z9WwXn_DbJcn0=s0-d "D\hat{B}C=50^{\circ}")
e ![[;E\hat{C}B=60^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s3h3bvvcbDMNM03UOneSNj_AYiCnxSSvQr3yPm91wglN4quVWSAb36uv5kZURuqiHG0rN_MltJ0h0DM0jg79jKI3I8pO5Qytn1ZLzT3O0jvzx54snFxC4fHks=s0-d "E\hat{C}B=60^{\circ}")
calcule ![[;C\hat{E}D;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t-G3vI0OaKX2YGjntWI-1N9Rdf18BXlWQ5Sn0CR7l9RBuXJsS_4HwJFWCybXVR6y5jVNYdzP6RISf4I0sNpNMFH634B9eLww=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_tiPabkb9GVci9W-mMVnGPIXVuCJikqL4aWA5dUFJOaaDfJzfh4r0cPXdjILSN37YSuJ5qVtakK0DguXG1BmWRY7M6PAp5hIZRj=s0-d "C\hat{E}D=x")
então ![[;C\hat{D}E=160^{\circ} -x;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u8j49Ll39NhBBuNTdxTjvd-4uxaeL1YVv7ngnrIH7E-3AvE6gm9YNZtpUCtVz0hCNt-CFYsr8kvLSRb3iEcp2wIutaOnjiPCvyHkwzCuuLPJqNLAJSlfqcKF0OWRw732Y=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_t46ySe4-LEP10vwDGwFMMvJSBEOWfTEEPKk3M8joVU2CyxcvT_mxPZwIf1ieo9Q5ZorFiMjJe6UwShYQ=s0-d "CDE")
...
![[;\frac{CE}{CD}=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sWM_ylREl8gERXgsh4VyBnK1h6Wb-GsFpHR7VRIJnC69GND9RTfPm7a712lL9xtObTiIWcv1tmXEDoG8evwMgofyMDOHfW_f1pm8bL_qM3PQ=s0-d "\frac{CE}{CD}=")
![[;\frac{sen(160^{\circ}-x)}{sen (x)};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v3gOvamCNkPJvrws9L-9-oeRfknLjUlCzkD8n5qwy-lQIta3yCs6lIzRHtQzjRdQ_z-xlopPsajYRz7E_LggZYwlkFUGR7nwTuKP2xjSCc2QtXlHdBOuCoj5v7Srh36ghkYcQfe-b5TD_p_bYYIO8KpOs1uRIf=s0-d "\frac{sen(160^{\circ}-x)}{sen (x)}")
(I)
e no triângulo![[;BCE;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_txY3yWhkGj9mnjG1gRQQjKBgL10bYvKHIyuBv-tYT6VVhXHid9qUwre1YGVXIbM9HLI8ljiPtMXBCV=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_ugyWD8fpa30ixXEqe1upiOUk7PJj3u2qLdYpzWsnf2nFKwijwfKJyJEqp-mZzN3lEwRBbrhZcwwuVdA8yoA-grpDhObNKyROrpvrAiABl8AZfHG9C_hv3hO2FRu537zNRxR8gdSX7Kq6p_SSpnLiYXGEmSmhJvGQ1Oo631swEKH2aiE_Fk7AmEZ4TFfFu7NrLzKMtB4M09oDafjXnet1avDnX-1QXT4ZeMoJQtriuCGtjLZn3J3im5yQxBdCjuWtaZ2rlGsXkBu92bj_4qScw=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_tKegIWZDIsaoLsRjEHSekKIK33NVmvS5E_TkGv9zN3aln0hjd6uW8b28jj2DYTs0NvTYEzXhUW5j9CjQqIRnqsDhO2rUkadyE=s0-d "C\hat{B}D=")
![[;C\hat{D}B;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ur2oe8fANIygrnmidmU7S6wKuDWXKtmHMasa_UOwZ8msaeY1sTQGrX-xUEa8zTi0N502RB8WjWL7_1SJHAiNwJqXqHCFwMiQ=s0-d "C\hat{D}B")
![[;=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vphjM0O5SmsiG9Zz4iv1QkxNMbO5dNamqhwo9Funti9xUabDiQrS6aJ86dqZ9F797SxYaIRsVi89E=s0-d "=")
![[;50^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sH4B8vINGoEyFdYzkRVfcad0O2eokzGqIz6BMT0ilRYZRlUWNI81pJmNMfNUnrnJLc3QBiacFxxm7gRrOiffQK924h174IkwJr=s0-d "50^{\circ}")
então ![[;\triangle{BCD};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_teGuJPtVosCbLytdGcQ3P-cEhVLWFmE8wBKxi2lbI5hKvXDNiaUf_38quvhVRacR6_GK3F82ozghZfHYt7N1VtOTwUupg1nKYUyqY=s0-d "\triangle{BCD}")
é isosceles. Assim, ![[;CD=BC;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vS8_LNK5xkGQXJDiOevsvcNbxbYajtNcfxGTLCqA9sBJSYBIbQl0wKZp5nMH6fRN8F96K5uUzZWLqCScM=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_t6X7D5UpeTli_2jx9WdS3LRFi05gSzIqA-E0BDojruUgW4jA5VZB8N44i6DGWehHj-FbmBVY8omDtxE5FUh16WF9jUnAcasJZLTAPHm3y-k4hTC-aggtV12Iefi1DYIxbUae-pDIItSE_idsMI1Yp4jo4cnBAF3pIA7AvlD_IoZVu3wU_cY83Ho5lZNALW9fgrgKBOEcblvYveWL9vFnM=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_u5NcrmmH152BNOBgTKZ-pyWCiCrkEP509OBXGIcusMCcD3etZX1kJiDyJ94QdjwecAMyW2TBU38EyMiPY9ehY3pj-IZOicGNf0DFWl1y3TPVJ4C0rGakiRi0B-uNegSwMYw7_f1bonhUt2XKY=s0-d "sen(180^{\circ} - x) = sen(x)")
Temos
![[;sen(160^{\circ} - x)=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sXfwNy1651-g550TO4xHzQA_C5aEQtzUabWNbnAD40H4OxsEnA_2fPIPPp-aosMuRSC5MOW9d4OF05P8vSzjjTcM1vtQ89h73TJCZQzFr28JmQsGDfQa0V1xgkZYo=s0-d "sen(160^{\circ} - x)=")
![[; sen(180^{\circ} - (160^{\circ} - x)) =;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vt-WNdlHGT2nfG3CvvHaAxVr_69t5vaz902stlAW97u-7yJn5BPGv-ijaKfFfIehy-RC1nhrr9DMuTRDhSuFMoGpnwnBbAilvGnugbuSy-Kjwe4TVk9ByrghW9I8yil-hypoEj5cBbkiNay8uzCDnL0y60DaWyJ2liFycbMXp1tpwzEQ=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_vdcUvddLAJAoGWgYYYCYy9Os3YLvEoP7W_7Qd1OqnX9Y5aEyMURoPJyFZC3VpltKOvFZz4mKhdaVCN7UBBEY0HgN24tBVyDAIBo8AwzOhUdFvMzyx-kM6fS0uLakzEZrebERZyRydLAnJRgaG8qbOKimZ0dANsWDgJZmWYjnpI4umc5nzXry4LE2qZCOMyzR2VYDpDMzoiFqy6LOeCyfZq2jQnq2LU7xVolej72StffxRcDldKufsyqdu4xmRBk-jDDW_akm2jPJ2Z7NW25i172TBfBtrJLD9SNIk_hm0Izr4c49MijGijThAWsj3TLceQ_lOO=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_tZN4n3piKLe7eaeANydhi-i28qd1LMiK48D4qqPFiVk9cApbwMZf8CgnKeHMTLz7TMVsTyXjwMFK36pHiT1DVnHzKjB0esqt2VCWu1f2whN4iXEV0vIzmLZjkjmzam9BR87RCaAScs-v1XIBR9HQ-2qg=s0-d "sen(20^{\circ}) \cdot cos (x) +")
![[;cos(20^{\circ});]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sTRPZxTrE8FhDHbIS8GK9kEHMEn5BxrehEd2V2taYWYlKFsXcsPVQdxftyggHJps-rj_1Ojbv-8mv3720DMA__lnPmFYhaPmkv84yUd6SkpALIcQ=s0-d "cos(20^{\circ})")
![[;\cdot;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sYOAAAKAaNsrNZFNoh7lhpNimcVZHyW5VYKuQ1SOI-R7zaQO38Mk0OhyUch_futEnwynkiMliuzmrNy0VoIcM=s0-d "\cdot")
![[; sen(x)=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_syacPPRUHHewylFjTtUehUY7ga64QJkl7HyDtpE1_7dBZszb_Ly-KRb1jRQ8hxuoiPw62Waf_eq-cns7UoJBcHEIpWm5kN=s0-d "sen(x)=")
![[; \cdot sen(x);]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_toWBaqI-hn3F9Jp6L78AoSNDeGW9bkdMqFyEV04qRkYgc48amGmZOyLcPQZ_75YYp0c0gp1qjsVEF8cWWqkEObmmepO08dhOztqpp_L7Q=s0-d "\cdot sen(x)")
![[; + \sqrt{3};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tJ3v8NnAXMW6ewUGUI3LXEWOO9j_k3Vj8OqtlqqHguaKTtPsKwz9nR2SR7vCra5WBOau8F5Zr18al4c0N5qWyMm1sBPg3P6HQywyd-=s0-d "+ \sqrt{3}")
![[;sen(20^{\circ});]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v2I0S-BPwGhWJBXIVDxDv8W-Z9x-Gg801sGq4KfKPX3v3O9zotnUFnYU4nByHxmtzs5fHyjkdIBv8_3DjirQxs_5xaXNk5FTQiaEoMCWGgeu5xAQ=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_s0YHaahpXdPnpQErF0zFkZXeBMU8aYx5CQ7g8qkq5FbATGz6NhX7YQyhlTAvOrno4gt96OUekahYlDEv53ovDhQVTLE6FmQVBzPYhpfQyg-tGz6kG0HlBgfysQ1bsDjdhEylUqoH38wWf09TPAx8pL1--kniXPJUkHVswwGuQDDBRQicQLsafeIVfqGGghm1hDUm6mJe7v6iVslA9Z8nSPm91jrFqYMmsz_4gTelSMWxDK=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_uWbeXE8cmtgd8eePMNAwQ2o2kWLT1lvsOanSH4EjQ99xUNFP1ldXSIAWObZCOVbVpxt8GLKwKYexDNyN0O8L7tr0c4Pl3jgs-GIJy6GSYhAyQ=s0-d "ctg(x)=\sqrt{3}")
daonde![[; x=30^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tOCRX82NnohaqkRbRvkuWxL1y1GchcBQbSC8aPP1HFn-OPi-HvtBw-YzezICtKxUDsTmiRUIPLXw3I6MqvoA_xcMAuNC1YsBoug9Rnqyyv=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_thteLrJM-EJhjCquNxqhE-xlwp4B75aTpVDpgz1CnKbFYilvi7na3LD7sQABZ4rGYGiJbimUzo3Q-acojQm5e4qGuONLxFen-FWH_o=s0-d "ED'\parallel BC")
. Seja ![[;P;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ux7iagvWhUk4BQjdLMFihTHBqpp8VBdcFL7E3keT2ULvMAcCBfXNS-v4QLQ-B9hd_0NFnMsV6ndlM=s0-d "P")
o ponto de encontro de![[;BD';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sczg6QHQ5RkOlst0kWLmfyq5um-IiZblWRyKzN6GkKtjY-kp5imh6inEsMQDX2iybzwHXuTvw06TdIr8Qv=s0-d "BD'")
com![[;CE;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tBQBhSja_R_XkRwLyA-cpY5tvPWc7Ticmu8aTz8pu5vvNIcCGqaOjBy9Rr3TBPZ0ZY6c4LrslRp77h=s0-d "CE")
. Assim, ![[; \triangle{BCP};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_thxcG79P340vdWiBdfnb2DaxoTJOB6oxPWKKttlEsAvgMvoGMwyHQgkzx2sE-gL8kYNiZ9N_DLbDEKld7EVGmdgM43GoSC57pdRo-4Z-Tz=s0-d "\triangle{BCP}")
e ![[;\triangle{ED'P};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tIthtJJADLkTi4vNJ91-Fgfgp40u_SdTMlt5BHE6KMNLHdiuiSDDwjsQgHihNhwT1HYiv7lpBLxhpqdTfEwbvGAydFXBBtOQPK-lVBUVUm=s0-d "\triangle{ED'P}")
são triângulos equiláteros. Ou seja, ![[;CP=BC;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tpdCD3W75KZaPYBO7XcfIK-NE_hE3cMC7r84yOU1BKr3KK8Wku2HkC-CXwk53oq8mP_TRccbY76w50_lY_=s0-d "CP=BC")
como é isosceles, então ![[;\triangle{PCD};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uKREVHbPrJKRxZe4Ft1PFaDgjqoRAMOqY3w4-MXwYUuqkMaEZr7feZRlvcmCWvQ2O0_vs6iN8cqho32SU7-oWNjkGKjfkX_ra9cbU=s0-d "\triangle{PCD}")
é isosceles com ângulo do vértice ![[;\hat{C}=20^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vjEg3aYmFCDU1QWV6kEgM-rZB7ZCA15g1Syvvwhoe7AMncJJ4--PrwAIfC6ApuAKrnqHMJcKfbTqDRoBNB27Xjq9_B3PKpfYhqETB48L-aP3zdaNstght2=s0-d "\hat{C}=20^{\circ}")
, assim![[;C\hat{P}D=80^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uSMVdqHPb8ZiL91mgvpn0O1Gk4YDCUO5JweW8mW1AxvtS4yQp_JwBHCbtBH4IMjtdz_KVgK2BIPw-uln5QnEZ7ktv9zp9MZLGx08MwD8JfNqlF-GwB7AYaoGo=s0-d "C\hat{P}D=80^{\circ}")
como ![[;E\hat{P}D'=60^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uKLvAyD7uxR91TRhq6vUoxURF9SdYs0QLwvp8mpMYqsthMQC0dlpwf-E5JqiiZS9sCZLfTG47EgBbavvDpvmQhgokUnlWlPbq-BDas86U-YSd6OyX-b3Z66Ql9JB8=s0-d "E\hat{P}D'=60^{\circ}")
e ![[;E,P;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ttUwzFMetVtLTCnzTUpcfDfNExpd7DoXk9q8bLh9YfQQdU_T60DaLAJy4Q6PwcDtLxsI7cktpBrTewVw=s0-d "E,P")
e ![[;C;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vwN93x1Phghc29TgxL6LMTQmC3imlLvw-QlzndtW3xK5iImTmQMDiLK94X99uDBXvRrzZ4Mfpv9w=s0-d "C")
são colineares, ![[;D'\hat{P}C=40^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sVadCqquGncbKT3Oonn6ETGctRmLvfj94YZgViaVYO4x-dH5aDW3qOPYwVGOMXHguOiBPANHQ9gAAhpfywKxE5-UftK74ZNYZfIAiND3O1p3Te4DgIlWKhtzJ682w=s0-d "D'\hat{P}C=40^{\circ}")
.
Como e ![[;B\hat{D'}C=40^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vIV4SA8IGKM-dPGLpbKAXh6OOu9gOV22UaiYuZUn7FMVUXk3q2jHlEVzH8_c9VkEay-csJwmgXsIFogYKYZeLV3D2_Wk_h72XrBwne7njlMTmASS_fDy_V_Ns3aiQ=s0-d "B\hat{D'}C=40^{\circ}")
![[;\triangle{PDD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sGDBRWD72BOlx1kX01IO-SVrQEiUpPlxmqL8rXQTHzWg1wuANLE0JPbVyPKAF1k_z_DEroQ0Dd90axe8GEs1lMV2HSwFl-Q22ywGz_8PY1=s0-d "\triangle{PDD'}")
é isosceles com angulo do vétice ![[; \hat{D}=100^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vJC5hRlNTJ-T_aDsaPgaKRg4uC5PXyBjiOwuwsuQ-1vI_AQ0lqKoJfwzLQFa6Wp787vRg6poRe8e0YQO37wCaCu9mYqtTj0sOlAVvSo9TD7-Q5e5xtNFN5nb5M4A=s0-d "\hat{D}=100^{\circ}")
.
Lembrando que![[;\triangle{EPD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tWlhR1ajgpM7NcP9cdCAxkuOIlLY4gTtTmdo-QlAhT-R_22ZrTXRd-hNWH-LjhPTiPI1quXSkpuqXYXQDaDJtyRyb2T3-xXsXLvJy_BkIu=s0-d "\triangle{EPD'}")
é equilatero. Temos ![[;ED'=EP;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vcrGi6nHFFdsYns7PXFlgBilpp0tiLYedsIEb5tA8arZDvMUwhpuVb7IiRqNRzOm7vxzk0tpPay8kx-s3sSxaP=s0-d "ED'=EP")
e ![[;DP=DD';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vphFi5-oZ-zMpsFaMbp8fWyTMr7MHBhKJJT-d1_GqpMpBtlHP64oPDZgoAumoMlh6QCGRaqQTfaPMsdA4BNyga=s0-d "DP=DD'")
assim, pelo caso LLL, ![[;\triangle{EPD}=\triangle{EDD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uoRE3qHnJem8vzVxRdz1UBR1QoghOD5Zv612Y-YT-Y_3hrS4oLwtaKpZFkjXChFd0pl4ypaSPAK1pXA8CNwD188JHkLMhlQUsOHJXHWd9x_yIt8VR8yVwpsr89QpTPmDYd7NaH=s0-d "\triangle{EPD}=\triangle{EDD'}")
. Por isso, ![[;ED;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tDQ1vBmmoTUW1aWgAWipcHDJyo12Z845hiR7UcBhoYZLHzex4MFAFSezgnoR0SeOQ4Ut60Ldgu_ZHD=s0-d "ED")
é bissetriz de ![[;P\hat{E}D';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uuhhtcFZio0keTbiG7_0FVAqroN9yHRPDtEGYdRYgYyD18s5KijB_HocQibwx-l46zxE8sQpx4V7yH5RZ17BBr1dRYx0e__HzsrQ=s0-d "P\hat{E}D'")
, logo, ![[;C\hat{E}D=30^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s5v4BpPsgoHeILqcYLl4PLcgEcpNOLqJHkTBm-nwvguXVpW4TWGEm9Yb9UUnSc7QMiX9-ea_WTaEiJrBnctsqRLNs6FD2uwetd8jIsYABXJx7RXF7OIiEMJA=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