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_tjS4G-eOMfsIWd65hVVN13V7YM92CmDJOoazmK1UcTmfk_jR0tH8N-0Hu2uKJaySMGjNkYWUny5kPeAA=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_tICATpfD9siqQJwrn34Qnf7mdoqVnmT7aa3Z0pbPZu0SoooJMcQUzRrpKITVBfHku96envechNL8K0vxhjN8aKsYUYm3EcKEeY4iSWDkW5AusE60pnp7-Ksd8=s0-d "D\hat{B}C=50^{\circ}")
e ![[;E\hat{C}B=60^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sl8tYLnFhX26YDSNtb_uUcMPn7FcbsSjpvFj0BSFq9TGDCGdrbgHIvsPKsRK3UDgFtMrEvpU9I3ar-VH9VEkCM-O2HmHkVi6w1TFCtHOfjvWZdwka6DqrvvZs=s0-d "E\hat{C}B=60^{\circ}")
calcule ![[;C\hat{E}D;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uXWqldpugSpHlARkw_8VDM4h3vRxIgWwokZuCUYNHxrD1qBX30DCQwZPcSMkBBLMy5G8IXxpCRQhNfgZMV7BAj_GLy9cD_EA=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_vI4vWGDp9Nu3OPaJvepWyRZXvwigPYtskWvKYEYRgCmVIlLOm-YZB1XNn5ZXLyhYKQ7xMoBLIMPvn_4lBQJQwcgvI7mgiKTtRu=s0-d "C\hat{E}D=x")
então ![[;C\hat{D}E=160^{\circ} -x;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tWUzPlR-gToqXwthLsJLDHY9XO8ePv9ab-Ul6wP64XM1GCD4C908u3LzCzwDU7p6C9NC0ZFXMuKzICYap_4-Q3hSozmTVvUr48PiUTc0Kjk9gLKswCAsPvqxJKcq_E3uY=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_sE9tlohPhDRtyRlIWjSC9AFUT6BDFB_9xrBAc_84dJ2S21lTHqdeIcPpiStU5Jex441YkUEYl_-mff0w=s0-d "CDE")
...
![[;\frac{CE}{CD}=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sLV0VKXKyycsbdoN8LVxOkcO17BH34GyufLFCiStiDr_Jel_gJA0rn9hUp6YLiL4HWgdKCYO23KOFsRZ1rEJbx49hmhtDeTpIaTuQlK9Ru3g=s0-d "\frac{CE}{CD}=")
![[;\frac{sen(160^{\circ}-x)}{sen (x)};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vEfl-QWojrnw6KbFJFpQLuRynpn68T2CVGvbDsl5Gok1LhizH8-2uhgLHR7xMZXsIsNGd1lDOkZpiMSnRd6C6t8mPCbOSRhUVE1ORhsRbW_xobffgSPw1yr0yZ5vZy8fKFgojoMs76_KqPoD1bpN7FGy6z9NBz=s0-d "\frac{sen(160^{\circ}-x)}{sen (x)}")
(I)
e no triângulo![[;BCE;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sWx5a_FxKG141AiRfxlJz0EUbg3GKjs7VNeDtgbSG7OcY-OQ03LwpQzBYQqzgI7hTupO-UZlj7vyov=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_tKJbbJMGzCMOf8bS0u4ZVkM7uO07nKDgf2OZHa2zuMRM-G7lQOrqZWyxu9kw16p8aSBv4_GPeWQOWNas1ZJy--LDfI4c4LK4calvm6z-14y0MBblxKW4F6NUFKKkC20Ag8RU7E03P7HwCqezA0gGk7nFDeRNDL6tFqWIQsHfEi3Hjgw6Ivl81HCbl75fncu7rIaj98U_jWWsWsMYRkSXCPmGYcwRe2yj47kpM_ruXiAz6hhBgzu7UStaJ_aAIeB2hu7BgvardoWLa_ZTWsEzc=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_sYrc2LjUKetF6k8DslVG5qP9j86Q7mQRZ6j6A64KO4WYtX1jlVXK55mxRZxX5WnH0Vv65R1HD22G1O0RVRdyL3QsfnoAfKLzU=s0-d "C\hat{B}D=")
![[;C\hat{D}B;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sYLGqPxSbAzsJj1WJl3GZfqY_wiafobyQRECCwBag7H888aDerOwifjrwq_esGN-33cC6hX0LnUwTdsuEpsZaD42Pe0TYBQw=s0-d "C\hat{D}B")
![[;=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uUnb5jtnZobQ3f-DXBB_i05f9P0ChIwL3b2GsSxUOggzqHzLxfL19Ub_qGEjGZWOjuJyGwMvUtMUM=s0-d "=")
![[;50^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_taFeaFpw0rA3fM7Tj1aVvztBIUpJk-s_Mbkjap7FDKYSICQihlMoBbzEEHkBfiNagMcfTFHfwES-RCTdfCC_Caqd2-7Zg5viAw=s0-d "50^{\circ}")
então ![[;\triangle{BCD};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_svNdKSdFy61Uzi19NTRKxXF0s0GAfGbu7VzElZLKbUoE9vSJoItAD7enOQ1udgqEdpziRClFQ658StBYTEWqtRMesGX082km8GCjM=s0-d "\triangle{BCD}")
é isosceles. Assim, ![[;CD=BC;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vDluRs2YmK9utiDoDEIYivni2McKpZ5TXjr0QJyJD6khxcjtu5t_2Bhkmxs8sOKd0EFHfd9UIybqRbSCc=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_sJtcNcLElQxuLeOdBR0NH4deeIZ3HHOSwbCAllmLcw5QT_q-JWpxxvFFng_qxSj9ZzDkPRHI91Hr3gWp5tdqVmpZApyLANyRzBnPy2hdLcxZvIWIwdkrcj2b2fnUfIQ5ytoKI7NK0Si-iGmrk=s0-d "sen(180^{\circ} - x) = sen(x)")
Temos
![[;sen(160^{\circ} - x)=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v7KwsIk4bMgx-tGm6RNEZCdiHXtmjRnN74q9mXr74lM6ViRk_lmGlC9KS0Xa01lLcPMS9SFuC2ZmD75hnjLy4ZerMMpxLRNZd-_zUsdKybNqzKhHn3Y7rGHydtQNk=s0-d "sen(160^{\circ} - x)=")
![[; sen(180^{\circ} - (160^{\circ} - x)) =;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_shGqWwm2heAFhir2ML8k4__t7BRBg_AsvRnUkltddCKIGmRmGzhFlLSOaLW_HiCLM5TnB6QNCXsMEhDEopFtW0v_leMau0qVSpwgImMoJw1Hiofg8HshiVOBdM10W_mr67Brzuj8EnnyJ2TSksLVhohRWgUsUuCNQSLwveYJsr4bdgzw=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_vBySLaUAjd-w1YhvdJ6rLhNZAQ5eUFPFNyBrHYhK_mxtMm9HK6MLddnzLntMDvpqr9bXI6VwPj6PbfzVBmhOc0zDiMZustm6umVdnWo_J2vBtYjuG0A7Ww5bn5jpXPKTFKbPaDPXD0ihOKe6nt9N5QHLw--97F7Bp9BNLIFD5zB36QUsmbEuZhkpDIDTFdndUK0hkqRGioxT_1ErdAnHyu8qDS6U4e01XvGQ7HyFyx4GwxrAr5CdYe1kHZvKI3X6bc6EHjeW9PpHKHQuPPlRLJ0vFyX4GPHPMm0Lc9mPG6m6eXhCAWLFWtlm_DB0h7vkVuSkgV=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_v4g0H8-4Qwsc-m_PjcwkaSJbaa02dBoCT7bX1E3d-Yd-6tTONq91BQlqQjMl1Sbr4RPuIvB__HjyPpCFTIFIBI6e7dwBI38RJ0Umg2xvprpn63HMnONe6Gro90v9Kemdvh6LQzb8tNrGHBu1zgASznog=s0-d "sen(20^{\circ}) \cdot cos (x) +")
![[;cos(20^{\circ});]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_urig02veB-U7rGmAle7Bz2T3S5_tyVDD-bu_6NYh_0TeBG73jbBwHAmqVDcSa15T8PzphhchCRnh09v92ugFYG2j8TlzQNMGdwOye2gf7dWrJCKg=s0-d "cos(20^{\circ})")
![[;\cdot;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tmHS2Ojvxsw3qChyTgN8IJbNmlFEggfhvjoyYlK4N66sioOpJEQ21uYcluDOR98mXHcsUbIYvsK1c-EU6RXzI=s0-d "\cdot")
![[; sen(x)=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t3j2wohsDwhYqCeCJA-Nhc8L4bSlUvc3RkKNyp0G4O55bEucZgx9ERJ_9pIw_T2wAbAXSS45C0mbN-iROWJFbI5Rcwj27l=s0-d "sen(x)=")
![[; \cdot sen(x);]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vdoig80dA-XO-hQgSDZzHEaqw-kDqefCStxB1yXPIHP_-8rYDrENk6kWKNJLBn5Id_sbxthWDcPxAMrcA3CwnCf341sLMVY4xYmECqrJA=s0-d "\cdot sen(x)")
![[; + \sqrt{3};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ttTwK13Q1UGjZv4Jolp2-V3WvuOby5XZUe15omopYvRtjpmJNnFn4g-sVYikDdRGlyKmIJDzFsnqyXDCrvIwgovIsotfbZqY32MzTf=s0-d "+ \sqrt{3}")
![[;sen(20^{\circ});]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_un4Nji5YwSdplzq1N6V3BE6qvbVse8OPBo2D2iTw3yIKX20JZvl9jQl06eyznTxrKmKJLgSSLBmRCeugtW5tgwApqcrZM8ou0Hh_JqS5wDLsHf1w=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_t1kTYbPeqHFOF1ptM4jxlVE7kr3I_8llhDrDYC_fJI2hGyNFDpnHVs0NsKqGtOpDtlBMLdbNqBPX7v0zJZPd2oCIIP4iPZBKdbXcGsublKw1QTJ70ykigF4ETlp4nhBexVdpQfzSXspMET6QwznO7nWMF0mM-9QbcC3p365JzU436c_1mX-_7RvXNA3jaDz_hwsYU-HVcGKUD__ZXeWeOhMCfW8i4c0XlEoVcF23ZOhmsP=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_seVHWoVLxx481-Ti-Uhh446SA5RcJWcFkWJZQZ4GuluKkM3dD6EkabCI9nxl7Eqs0NXtWRcECHdKie4Z8Jw3LsC6jsqJTFpdoGIHHrCDl4=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_tppmPNvK8Z2isXsZMxq4fOdVkIRQWyVyp4mDN0cBgcArqPQr60jgdjWtq6PptkXT5X03SCFphfXFHcwXIg72p7FgOcXglzzurzvAcm=s0-d "ED'\parallel BC")
. Seja ![[;P;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sd0q_c8ODqM_uqQ9auqx5GG7BweEJj-sdxZyaCNkTQaqzmfQ2K3ATatdr_B-NbLRPIwhdctbZoYx0=s0-d "P")
o ponto de encontro de![[;BD';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u3E1mlXtSWzpxhfG-cbMkgx3855GOf6TfGLyKhdAN5iPiF6txjn4n-Unrz0uWSDvLXoJAFVlw1gKngIfOj=s0-d "BD'")
com![[;CE;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uLlN9JUHTp0E6UKwOctDJGmVo9bS32pJx0kgEB1PVSSsg96fn95LMQL6NBFqvmAdpQjSXYnd3hqVXu=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_udQo5eQ0tXsOdnDZFyGHFII8xN12VwhbOG4cBSQywPynOZdWyJ90ssO3BvNxfB1F2CY_Wn5HXjZgm5Opo_Q02bMiXrIZ5cwiFXcCSDOwQM=s0-d "\triangle{ED'P}")
são triângulos equiláteros. Ou seja, ![[;CP=BC;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_txDMl26ouUywHYrPAKyuRwH-dBMljwV5DV51KymrJSKiAgRJX1PvpKUdrZ5dms_l3xo6-BlncMxhfxph4-=s0-d "CP=BC")
como é isosceles, então ![[;\triangle{PCD};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u4EZdrbr2UVtYr6kOiWDEBD2dFUUssgFhl9q_UJUxVJgD_J-W9nZq-AumtTYVVD9Au0GqJuQzWE42dPZe179x9fz5OlJW9t01KeZY=s0-d "\triangle{PCD}")
é isosceles com ângulo do vértice ![[;\hat{C}=20^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sddGCM0ImOdB3U13H5ifxb4PWkdMX61srAp8mVl8tf1qR1hyTU0-pLcTQ08SFl_zBzE90oXmOsZvOcJw-f2uK0flXx05CcKDpbI4QxwYJUrr3hFNXio3Yf=s0-d "\hat{C}=20^{\circ}")
, assim![[;C\hat{P}D=80^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s7ZrbCLhjTCB3wK3hPWGEHB706-s5zseisGN1guphFoTRGng_ywnbbnscjNXDRT4qYvZYjBSdBIE5cGvmcaQJdaFZkuzq06c23AIGRUF_6DMLegno6srO8m70=s0-d "C\hat{P}D=80^{\circ}")
como ![[;E\hat{P}D'=60^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_udrt06F9EPmBtcaNvIn0Kei0Ht_ws1ZFCwhzuCPFKESNXYnh_3qhdRAWvsqKsiwfF9g5gAoCjebkUXTna-Fu_WpuaQ9uXQi_tDmY5k9AmP9VQTp6d1IYnnTIJP4AE=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_u6akq6RstXSy8G5DBAyluTBNnOE-vE1bO6Q93nmJvRZzWBwJ7qtX7NZ5geBeAW1UFUFXYaxVE8bA=s0-d "C")
são colineares, ![[;D'\hat{P}C=40^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tuDEQgOrhb8D6qDkdjcvm0HOEzU5-67Aypo57uLfrkpu6YyICH3Rjv1LMIDcZIuAuYTTM39xT4QZT1smXoQ2qb42SnQoEsP-ioGo9KJveIt7En47shemqSAxffkP0=s0-d "D'\hat{P}C=40^{\circ}")
.
Como e ![[;B\hat{D'}C=40^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uszRdG6iepJ6F5s4-ic2Ehb-52sAaTIEJzIkWU0a1Y6KsUHyJPJwzAM2j5SGrTiOXU8khJXX-gGxJv7VPursyZmCJutXKXztLDru0gnlehqUlXKkw-yxVFleluh6c=s0-d "B\hat{D'}C=40^{\circ}")
![[;\triangle{PDD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_svmfXyaDbZxhkAnFiSUOllNRcpZvL6vCNEh9LdyvVQdkWZaOZj9gA9hhpkuUYEbDTAPCPAZ-DzMKAfwX43zTayvHcGMDiU316AqhTerw8V=s0-d "\triangle{PDD'}")
é isosceles com angulo do vétice ![[; \hat{D}=100^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vKIJqXEou8t-ZvfuNRt0xWOjPdRBcAC7sV3UNHN2oBkbK6zDfmpZBBp_M4GLmHIBZJM0I9NbHUdjImIx8nGs_TtXB4cV9-pAzeIh8KpZczjW8QIImbt2TWsQN1nw=s0-d "\hat{D}=100^{\circ}")
.
Lembrando que![[;\triangle{EPD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uKorx5vsZvoVaIMmxB3UZYk23o4PmH5puyXOvmzwpeRm7qGFu089aR10xilcTFFqcfgv0mgn2apSl3tmjw7ePg1NP4m0x89MqonBKjSprn=s0-d "\triangle{EPD'}")
é equilatero. Temos ![[;ED'=EP;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sfU4DeWs1UJmSTDNI8c87VevCD-ZWJ10xUzi9a95atC311oev08RrWA5XQQU174Mm8f-KNWNwwXuM62bgfneGd=s0-d "ED'=EP")
e ![[;DP=DD';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sumAcxBrrwjYR0euXZ-wU4VSJktpwD_KQj6SnGnOynCGVCUwoQKv9WMbq5nEHFSedIOFLOSTCCGH7jTk1pq5Uz=s0-d "DP=DD'")
assim, pelo caso LLL, ![[;\triangle{EPD}=\triangle{EDD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uoBJI3-AcJ3Cpf2K2gV54AQnwzK6qzI-0O-YCdARc8yGJATCEyqWI9a_iUpwrLWblhtX1UE0VqRhaLrSNFX4dqVGm6Cb7MNM9OfmAs9lxKvwtOOat2wuO4YOpdATsIkXUl7HLl=s0-d "\triangle{EPD}=\triangle{EDD'}")
. Por isso, ![[;ED;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_svFHOsK9WZ7qvqc6PstTkQkxgZ5r2HVL-1RNtjecEW4NLWCncd5Fny4VG6UrIwy1ksM6jmWZJItBF5=s0-d "ED")
é bissetriz de ![[;P\hat{E}D';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ufGCCNJdXazsP2GxiM1Dx9Q6fGi5Kk1RMQhl_jeriFhkEv40LhKVBe1rK2IszR_NQecynH-0zmXsoFav9j59BCU2byfkAnwnG1nA=s0-d "P\hat{E}D'")
, logo, ![[;C\hat{E}D=30^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uoEA4SxzNmqnHLr7kmLzHNP6h-IaOASguWAQgr3W5vfjKkYA84kbXDi-WG9Hc7QEOuw4BfOy6B6f43cQdi7zcGQ-SFXFe4O28C2E3cBfWlSUJZ3Md28FTa4Q=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