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_sn-_0GYk3uCCKxLZoFbIhxYwzUtN-8xR8wofAyxI9WjD7Ncjt7Arm5KZwjAStJx46MXoXxionbl1m3tg=s0-d "ABC")
ao lado é isósceles de vértice ![[;A=20^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sjaSDsvGwf0Uwf2sFJTT21KTyrDGvoIXOarP9dO5RCSMHfwhRWTtji0wvvKg4iyV6a3zg-TmsryWG2AVcvwdZHpuJKETsf2oxL7n0=s0-d "A=20^{\circ}")
.
Sendo![[;D\hat{B}C=50^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uN_9LB8IzsOw0xpNP87l_5lwR9yATUCZuWMzebMHk6PKn-B7gs0OjeUJfOnTF2Sht8uNW3BEiqHQZgmRYoXikuidJ5dIM64m-jYyMDgxDto0sNDLDGhEZsggg=s0-d "D\hat{B}C=50^{\circ}")
e ![[;E\hat{C}B=60^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tXPy7bYxhX0qm6NAch_0tgr8EDf1U2ImL9n6OQMRqMdZWM9nP8rqyKWOLOtFApveKERT8XyznrSCwxpwBBNtPYR3i4mzNRQHVhBIJftZxfetUaOs7pVHstGMs=s0-d "E\hat{C}B=60^{\circ}")
calcule ![[;C\hat{E}D;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uLWlpiZTYrKsKQqGJKoFNnIfVLbw8nsep3hS7TDDj8ukpYeFW7ume8VEWaQbYi3xSREJi2ZRVyx-dZDaE-F7SVv0Dfchq7rA=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_sO51cu_mpvHqGRFTk8uDfGY_zqVyPNDpj5IDTWxwkLR0LDL4QX3ThoOBf_Dl8Sw65FjF6YpKgxM2N_-jCtA2BgLB73YwXizlgy=s0-d "C\hat{E}D=x")
então ![[;C\hat{D}E=160^{\circ} -x;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sv97VccnUNwArkM30aZVdZe-JACDeiCh3tUbjnM7EV7aOLNyd4y7TJQ9aCha4AqUhzz66U4eymekZpPNRCcwKpH0MWLlYIsZPtyNn8AMhKO5DcsT79LrRR7PF6uLX-tDE=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_v3x1DA7pCORykxPDPfuUoSsxcMg9Yg6RKKkowjCPjz2d2_Zt0z4tO9xEMnrouDsGnnJChnnpxJVJSusw=s0-d "CDE")
...
![[;\frac{CE}{CD}=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_usSslNNG7iRJBISVlbEpKjRtcx8Qs2XwhQZnLTkv07aT2iW70_aFGMzWTjDkZsFH089S-yDSUETehkXKc-QwlDZXFo2e4F4C8cXmgwJplQjQ=s0-d "\frac{CE}{CD}=")
![[;\frac{sen(160^{\circ}-x)}{sen (x)};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s1vEWMClejNH1NsUn1NUw58DfGfo5M93dpH5s1noMfIqQTjkQZ89IuJFXIc-ViqL3h1iPei2XYHTC_zmy-vxCc0yakmif168MtXub2DCJtOCEfu6T1rROlkwh21wVyJ1fMvFpBt59rVL9W0P-hP-kAAD5MbvfA=s0-d "\frac{sen(160^{\circ}-x)}{sen (x)}")
(I)
e no triângulo![[;BCE;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sFvpHNgVQvB7wjHbouM00cy9Oe8owvK8Yzi4Zup4GqnT9SztL4haj0OC7NOB5EiCj6MnapHsy8wHhc=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_ucY8lb1dfp_cEaR3aU3LVg0HxhbKgbGJYP9D8DwK2CqB8T7ruzkyQXeWRB2qdmxzo4ITSDeSRf3K3-pWQem-zS1JsFvs4kKu8imAlbPuY8aLBtaMDRzEMqJlX1RV8gpkVZexKPEWIIcgn1IcEjeeRCSzg6N1uiFtP7-6DA9LiLr7fyDG3rhK1bMlihlY17DszEB85J2CulLgBjLNkrqcIh6WMJ4YrOJlduAs0CkxFU4vF1DGQenaB1KrmtKWwuQU2ZLS0TF7zNuxZS5bOlkkE=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_ssn5MSh8FNQKdf7HwerKst1--_thDw8M_NP_DZmJqVKC6TcZp4sjMBpDMfMs-0yJMlsOdw6psXqwVQpIdqOf5_IuXjvy2SPRM=s0-d "C\hat{B}D=")
![[;C\hat{D}B;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vL6wtU7oUhkxM66dhTg6yvxz6Uz-nxvp7i8_7yy-_MwOmzQ02RONA458Co97PkoZBYTSQe9yv0pM72vOVJw3-mn-LSet2BtQ=s0-d "C\hat{D}B")
![[;=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uN4FUoo6-ToLdoPbb7QQiky50zDP26kxRNjvyHAcnULlW6frhRdNl9_c5Mqxl8y2ZW-nOT_Ls1hN4=s0-d "=")
![[;50^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uyEi5SVtyVxsGBv2TlTPbuBE8IZm-Rr-X1okp4CuDrW_-NHeT4kKbe6F40tC3caSaUDQ1cAgXU-RDM8D1lz_ONDWU73kuRBbEi=s0-d "50^{\circ}")
então ![[;\triangle{BCD};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vEMwxVrLk1W5bR1oEpvHrnLC0TSlT_rucPLGQ8fVvIIzgKEKSNBRyxsPfzwE6-iJcnBI4BMUOPkcrNeWc7XsMQkK_to22NCR8qjx0=s0-d "\triangle{BCD}")
é isosceles. Assim, ![[;CD=BC;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uvBBi1hyIahW7pUM5YMdUpHBWorhLqwyngIiknrfIHcNhGoxmpITwaBddgTv7HxIDh7osqT4m-OpLMZ_I=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_sp0-ZFzsmI_wDbdqxZcwzJzOF2ILliHzq8DmPVfKpfhiW41IipAL5FEuh1rQ_xMT5rd1N5ADzNiMJOsC0dqb_sBKUYRAtDwjCTjuu1z4RG868_3dXzeLmP6EFDwPhKGOUbZJOQiIhHgITU61jZGsclPJsjCuAzsI_eOjhxCsKh20-gU6fJ_G7IE6gNFMwCPIVfwwOWRgku1MM2X2SqUIY=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_s8m7FoiBk0kIxVY8rCeTKE-k_MZ4P6C0I0jVKRuakiQRFNO55sdl6i4MKzkPiRqDYjyCDGP1aWZA09N84VuVSkZ_crUV_6uuYURoxr3RF65O3gSAonvEyWtylqRZpvjKGM4YBZSTKDFi8HGtg=s0-d "sen(180^{\circ} - x) = sen(x)")
Temos
![[;sen(160^{\circ} - x)=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_urtGZEesSfCKQukeSTpg7c7CoPR-_X0-ivUiY05KwSoX96ZQj1qAQD3ezyTOKNyqLQGQMtZaf3c8Mcaz4BFUjJh1YaQI5QMTGkVbgbXh6n2Evz0fUHNjSGVf0WXEs=s0-d "sen(160^{\circ} - x)=")
![[; sen(180^{\circ} - (160^{\circ} - x)) =;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tnhWf7iuZQZ04D-kivAHozCl6w4LYHP_YP8-jeeweP7QtozIJnXyAPfdLEU2uII-GDWSa6awccN7vxg8mDMve34jaB4NwStSGV-n1tE5P3hdrgx-_9SU95L_YlVZ6mv0x9ZQZjGjPYhCWJtblnkoO0vklhUNoAqUt6JNUI2yIZdJzLbw=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_tZbezbnUg2DVCVtyGMiFI3MkKcHhrwToRaLFZsYD0rLRLKZNiu3e5xRw3wB3hBN5zCaTWA-RFAXs2gdFFnV_6dRz16hvHUeLLxKBzhgY7OCfIIipKTLBJhI70_7El9rrL0El9AXtGIsijTI7No7KCsd6ezJ1VohNiUuDep8s9eQTTanxw_3LDOhibtSlRtGINgmZpy0SaUlc-HPFZosdi-kF0K-zK_hmvPjEsKS_HC_0Q8eeBvpTrdfKwP-7yPt6Fg9Oz1nqQTcQUSHNu59YEGviOsE469Emx5N62q5icEEDCtCF2Cu6JMmADADz3PvuoRIjUb=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_tV6PPnZTDy1BQWGCOi5JwI8TUCmL31X8y1il9PshsH-MyGkoiCjxqHkKex1eHtt6S_ajueSStmoJqJBiohXDFfrHUnq4apgxP6HWIX50ay3X8Xie7BLSOdf4Zi2uTL1DpKKWpVX0a1hVwcfHizylj1aw=s0-d "sen(20^{\circ}) \cdot cos (x) +")
![[;cos(20^{\circ});]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_smQ0uCokZXFiRaD3Ktfkki1gWeQS9oyJkxIJZoPz_XitYwzA9GCTGeijapKmfAlUAIBZD4d3LHoe_csLjfHbjtJO8mEvn5naqO3_FulFWbBSmkog=s0-d "cos(20^{\circ})")
![[;\cdot;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sFtbrQ2VJP91-p4n4uJq9EQ9RqTthGyO083lKB6zTGfwc2Of8C6EmRVk4xAOp1_JJV0ZXhBQ8pEOjUZyHgkZc=s0-d "\cdot")
![[; sen(x)=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uWwi7mXmQiCzYLoVejIQiI-MU3JAqdxd8FLRlGsfSnZLHwmbjrQOjIkVfONUE0to-Z6AKghJx_vG6vAfisW1_21g-FiA3_=s0-d "sen(x)=")
![[; \cdot sen(x);]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v62wSYGPInZPV4HVvJMT5u_YK75js2OPEJ6mZkomLHGC6qMXaGrWrIgqWDkk9Xy9akHeyohe_zXBuSg1bc5uWGg1HB4i-pUTyHRLcx1k8=s0-d "\cdot sen(x)")
![[; + \sqrt{3};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u3ATh3U00XgvqLRcQgbbf9f_OmefsCoHyGbD93WChHCA_EdTaqyO4OVG-PA_AjYMVXHwdVK54ViCx3mXO53mcR527U6WtxThS_xiTV=s0-d "+ \sqrt{3}")
![[;sen(20^{\circ});]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vTN0ye9jkGNOl3-Ey4UPBizignh5pwnSDJxmek2tbwFv3EkVTT80eGGzsrSF7k-673wM49ojZ6ImfBdk8KXnrrAK7Q-EqQJ4RWRJ32GMe1GFgXXg=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_thPYCCDxa744FNM_RXjDRWQsdNwK0eyVGDovTzgxqc2zEHazKUT2zSUTiUzUAwsNpnrN0eOCrsMYVQAN8DEvzroFqmmSBgC2rGgqrc-gkiaiScWbFhHzknaYOC7nU_RFW0UR2sUH_0BElpY-L4uROe8OoZKdyM8mHSBnab7Ti83OAmMPwFB7SotIaHwhPgI18jsL0PE93H8TouXL6IDT3UItz5c3kmHxiTz9XQ0hISrha5=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_vou8O1Dv8gb3UhdqOTRrG9Il-nSQCeYXP5p7mWHyDXRA-hSPb6bM9-Th92EUcJnvdtD2G_NWV8kNs_RVz0M0pdkOIIn0pgZjs1XudD-MwdRLo=s0-d "ctg(x)=\sqrt{3}")
daonde![[; x=30^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sAqKM37MIXyVUSckNicwBu8Z6aCLgtCQUmdUTLQ9YjDBWmtEQjR3skjVPbAL-FWdoo3OViIa3lkY0GbuIvtsPmPnO-8mYTD2T5QVL7ljqi=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_tGfDZuF4LRjRDzQCMLDISlYbS9cAmuQEe4ctr7xV2--1VE_7RCp0pYNIzH11BCgHDKakNGjPM0PNnsx2zqxcySMi4-t9_C2IpmS6F1=s0-d "ED'\parallel BC")
. Seja ![[;P;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vcaDmPrBEBe089gWZaJSbz3uOdL0akOI3OE2BWdgTfm1SBs1f5gzfgG9ep489gyFMoFQ9huSESyHw=s0-d "P")
o ponto de encontro de![[;BD';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tyTQ8bsVuKw-viRExNJiGDU1QqyI-03HbVS3GSKxzMKZl7bhkjCxdx6l_w3A3t-iLApGEZjfZRS-wUq297=s0-d "BD'")
com![[;CE;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uyYLdArif0VTP2ZBf_Y3p0-ldidE6Swii8b5BZ_PjNo1wf9-fkL-gM_NM1EZdo15ycpgNP9nlvExK-=s0-d "CE")
. Assim, ![[; \triangle{BCP};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vdzxfulOJ_Va0QkfP8ba_qwvKrDYIlGW7xIGErcJ20AC17v68l65EQSGfpfktfQBAeq94aVVyMl1Y0itjZaZ4zyn2QvS6oqM-553X24azN=s0-d "\triangle{BCP}")
e ![[;\triangle{ED'P};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vvlJOZZbCs0OPybDFR9JphtuJtYgPlCJtFTeHIbIfFgD3C0dLJzDfxOjHkX46Xls4rKfM-6AKxgT7UZkssgzmVKkUwQ8-lyAbPCuLX-Oo-=s0-d "\triangle{ED'P}")
são triângulos equiláteros. Ou seja, ![[;CP=BC;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sOP3RSe-1YWxpGA11z8tmMXfxReppdOGMnYx_SBFs_fL4zKXL8Ju1CFU3Mj72XIoVZ-F4ozRbgHCVYiTp_=s0-d "CP=BC")
como é isosceles, então ![[;\triangle{PCD};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_snF_VJHE29gbbeSyBBUAP4frbyiu9tqM5imQmPbgHRYa8rT-XEPglqwXgp4LbNziwOqEePpcUJ1wtMMGtzvMwUDdwMtlHk2rkq55g=s0-d "\triangle{PCD}")
é isosceles com ângulo do vértice ![[;\hat{C}=20^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_usBQUtzZTG5xNT5czEAPAsZz5qgxDNspwsgKIbRG22TrFRffBjZwcAjkMG5Sgqnpz98dkFE1CW2IbYCIxRYJldb_aamdi1vqjqjJ7bCi5_l765lPTVZ_FD=s0-d "\hat{C}=20^{\circ}")
, assim![[;C\hat{P}D=80^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uqBbGuvhsayufv7BZGIxXFNeYHbVv12gJNHWR2siDbGNlPNmFu6IQGczDeWaSHMCJhMBrLX9Gt453wFumO16m0dGXQikPPm-I1660Scn1eE292R2vjyaA56sk=s0-d "C\hat{P}D=80^{\circ}")
como ![[;E\hat{P}D'=60^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sZktoPvNyx5in31J9ZwnJD2KTSmCB74C_7V0HcxwY-fYx3pD41GUPt1fhtAB718UB9VmXpiFsA36Gfi7BHiR-xW6Lfz_W_cKaaKULlyrUBShBErHIGKwZcLQEfduw=s0-d "E\hat{P}D'=60^{\circ}")
e ![[;E,P;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sCwejx8OAsDkIVhYDKMp8ACpVUrZXllHUbEy-JV0hnoHGTl1MBNbu34cBNmUyGNr0pdSy5m455akamEw=s0-d "E,P")
e ![[;C;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vP6aIjHhn2TKb9321BMnKr-GSRVCJUnHsTQ-SpaI9LtyJnDyUEKGq-KSjyv9UxC95t4Pv2gUijQg=s0-d "C")
são colineares, ![[;D'\hat{P}C=40^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u8jvVivpuF8o4UDl1iZQ5agqrioAjlQdvCxEnuYf6eIp-15NwnnzaFHSKsP24LOx5FUmJQeY9Y2aLBc5V6Z74CKHhFIFXYq2kJlzdef2wRJKbf3DYg3XT6CwgYMwQ=s0-d "D'\hat{P}C=40^{\circ}")
.
Como e ![[;B\hat{D'}C=40^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s7r3j8tYCe1jFlWrlHPGDQAUuZ4BtqyjarMAVkXJ0KXq2Vsb7NmTgZmVaJWMi2jO8CiRVgtaNpO43B3_iy5vU_KNf-EqoEdjjj7POXNKmUOICVqiJSrHAZVEUnGHk=s0-d "B\hat{D'}C=40^{\circ}")
![[;\triangle{PDD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sBa839E05bvC3ghTsr9hZZrnWxOly9ZwA9NK9dYcgDvCTK67iXJ-WqUnTXpvTlf2HtFXX7co7SNwJTSIoWAxdlAwOuk6_XHN0hiq6vjyDf=s0-d "\triangle{PDD'}")
é isosceles com angulo do vétice ![[; \hat{D}=100^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s4rgyKerRYsAqk5To38RCRmLzLxJpUZizdPAIbJ7BXqiUIOYZGZW9vaxD_OBKypVKs_ucTROxpNLpY2oCJFJEtF_-4rcPK6HWilxaoNnfjiefDZ1BZ-5PMADXD0g=s0-d "\hat{D}=100^{\circ}")
.
Lembrando que![[;\triangle{EPD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u0VVK4uQYDj7YrL5acOFNWqalNBylkhyLB5JmnJFWwDdqDzWh70ahrAIF48xgz9e8lBVLn-BsnOObKtGhjlu3NsV_x1tUfl4kEIfsSqp-M=s0-d "\triangle{EPD'}")
é equilatero. Temos ![[;ED'=EP;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tdwiOH-0WHAxrHzCFvqA_EfgKPPD20m5L5Z3XLf5hNlo0GIPanA0vatCLJ1nM4zIf047243o97i2ZM0KIVH717=s0-d "ED'=EP")
e ![[;DP=DD';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uxvcPhIOD2_TiLQQvzB91kjSzRnUJv51Uuc_LeTAkW31rzeHwn74ppFNb6NagUAycsmJjxE8yyxO80yZd9QaIu=s0-d "DP=DD'")
assim, pelo caso LLL, ![[;\triangle{EPD}=\triangle{EDD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sbsBl3ln_SG06A_Fisy3MyoWmXpljG4tFCp1hdN5HESEbDutG44TremruJSkyNxLvlpU3IqzbQzfuawZ2wd718lvcC_wYBm_cBFeZljT2wukXtux6Lm92MSRKUfFKnWKjMLfYz=s0-d "\triangle{EPD}=\triangle{EDD'}")
. Por isso, ![[;ED;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vdwWwl4HXhyou8FyHCErFCypScgCVx82MlEG_y2aj5aeuc080f0xDTcQ8MM05l_gxTh2BT2GMjkO6g=s0-d "ED")
é bissetriz de ![[;P\hat{E}D';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uZ2sqRkfJpYjZdyhNKpWH25DB8I2YeBldGA_Mk0aGU9pvdh_pr7ec81jGK3PTT6B0LeBuq0uu8BUh8tav29AfMrQdaQ8QLBh8H7A=s0-d "P\hat{E}D'")
, logo, ![[;C\hat{E}D=30^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uq_gtSOJmnz3iRHsunFoB4F3QUpqvPkmNHcvZp_MADwUshHrQW-lbIMn52alC-OoadePdXHuGEGmPInNaJTjJqwSH3pJZlEjrwYMhA0pVPcBKOY2nQJYjs8g=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