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_sVMv7wskKQ9CBzKXsEYh8HiVcZYDRne_yHh0pQl9olQXsXmRR2AAD2at8DsdQDxPhorJjuzMvL3BiShw=s0-d "ABC")
ao lado é isósceles de vértice ![[;A=20^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tc6FSN7dxsK9RixwMonzv0CIabKpjLFiV7qg_xcGpiH4XY_ud82Mpsx-OYVMMsSibqvjfKkPPmZDa7JoWF1q8mChhU9Zo5ZOMWSt0=s0-d "A=20^{\circ}")
.
Sendo![[;D\hat{B}C=50^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s35_4x_ULcKebx4PpvjY6dLrmgRQ1bYr23VfzAgb8UL1mmld2d7ZATvdEK_zL3wHXKNCeGNwNik3gSnGmMvTOoN9ErV-9gtppz2ljQ-dgv2D6W0V5R5RfPoss=s0-d "D\hat{B}C=50^{\circ}")
e ![[;E\hat{C}B=60^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vsRnjU_56XuDilguFhZhC62vrHR5-WYDmxGVaCM2NhDRu2SBI13cPoeC1694LLHsHnotFCUISO9jJ2jbSouIunL8KvhaY4E4SqiW3T7yYirJn3RelCu-QPOCE=s0-d "E\hat{C}B=60^{\circ}")
calcule ![[;C\hat{E}D;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_stoY7jqdO--Qn0PDdLjjrHk74zY9TR4Ohfykd4zjBJ4Jn9kbP7R_caQhoVKn0lYZPkB1rV5FrqAUg5HCPZ46eOlRXofUL_Lw=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_u8xKW-SsRbKqOVAZG6TNmP_cEJtCLTmRgBDa4fuR18PkEhmD2JwA0n2ca2ZaJnned7VsOTPvroahSv-7wDGc1ywYV0VFOS0xfP=s0-d "C\hat{E}D=x")
então ![[;C\hat{D}E=160^{\circ} -x;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_umGm27dsWinLHZHgGX-AcLrr5vFeT2sfowjsUDdsYmVYlE4fd971D3_Ic1fna9YnHP6DU-4G9RTZIpv8_gk7SDNB9HHKgUkQPhr9CqfAXe93yipNzooNBUMKG9DyWi4i8=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_sJuwZabQ4NVHpWd0kgS4WY5U2bceXClGbR8welzd3rkrFOaG98lYBVPPnnTQbdDaMAbGrMVrJNSPdQjw=s0-d "CDE")
...
![[;\frac{CE}{CD}=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ud7f1OLErx3GPpdu8xCKboLbn2h6YMXziKslKspGmrCQwnRclOllNx5D3PDG7lNcOnrZBWXVkGFBeqeXvMvG6PYn3IMYl9TIJM0zHTrH8iOA=s0-d "\frac{CE}{CD}=")
![[;\frac{sen(160^{\circ}-x)}{sen (x)};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uGemU2Ckb2kF6F43fhH1qvG_LEAV3DSusJGMQhO0ZcZdy8THend85eOfiP-o0zOSSBhEzKo3dXtyRFMxDm7DTkAKzP7R-qZqMGpChefhfSg8FC4TNIbFgb2s2FTSHLLZMWxWC4mEOoJB6CPL9KYHablTG9KuO0=s0-d "\frac{sen(160^{\circ}-x)}{sen (x)}")
(I)
e no triângulo![[;BCE;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t9vLJIShBaQ5MyW-dTmE7YI-FNapPi-aCjn-wUTISsSNQ0kcXMqXNXQNQeTv6bqKTe0CVLQClBlEjh=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_voBBUSBa_4le0MmSv5AAWIHxKxt4LdzAbFpQhyj98Bc3XQaAxNlBtd5Q-LyLRFdt7EXO8Y0-qtKyVaujYdTVOc_UL5F92MwBnfHb48Px9bTdDO9XGtmzet6x_9t3_jKwdlD615sqJXCGjl9jB_gFwG7bmvaaSzDycXUe8XWE1fFkrFUF3xDDwz3oUlni-i_PIjdnMAVnjUlITkFfxz8Mh9YjAkITTG7nsm9YWkExyund6vg2O59cZUCsK8_OaGFIXNcvU6b3HXBCYoycPaNFk=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_sKQMYSrhkMjV7RwDXZ6XwznVktM4SqNgx_H4P6VNrOPxtYCuNZKx9Q37qEufz4-xE7T4w_EzKZtVbFb5UgCibNat_zYZFSEzE=s0-d "C\hat{B}D=")
![[;C\hat{D}B;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s8YDnai2Gv83XsUg8QsyvV-VeG-nWYQ5a0KkEMn2i6zlgiirYOkjX2G0svhmHg7qNV49WwRUojH4DnhZld4mOM7c2jgKCfwA=s0-d "C\hat{D}B")
![[;=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vloOJHLQLm_Ip9iyGl9_vNHUrVSWjGJvCludHTNMD23Qg9H1QPve5Oyk75KPcT6Xy5ML1OFQhrCF0=s0-d "=")
![[;50^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uB9cL6X_6qUc3GAlKrmH51RjtqD7RWYgyAd8zLEnyvHGBxf7tZsLCPhHCBsKNhegcA5B7IMlHdb_qY1GNaccvCaOLeK7R9Sx5V=s0-d "50^{\circ}")
então ![[;\triangle{BCD};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v60t-RY6i5-8oMIsjFBeZ7THzgPswSZMcmz4U0J-MUoIISe0z06_zN40bnqW-vZPUY-pVMyYdt7Q_elIF_jNxDouIE1wsLdzwbaHQ=s0-d "\triangle{BCD}")
é isosceles. Assim, ![[;CD=BC;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tolLkdQwtPUfXJPql5jPgRDmT_nyF0VJ4MIPqNjhevPssOFfeAmvRtwJAxHvUiIGPQEUZPYgTl0VBkf5U=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_s5KKc6QOZghxjg22kDZ8alEia2ZS_NCjTTQr3V0kxCg9_t8uD8BP374QZYAp84rqaPmKcNsoAeaINDweaANPUIdRHeqH6zVgyuT_45wfOmkBRS-BppKkY6YqBB3xHDEZFuH_RAmC3cxSnq21GDg0oCr1WjZgPPJSiBcibofYptYuacZ_qA_okRKyc3kQCaYjL-klbwI7wZmu4FsZC4M88=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_vwPw61iSIBx73KoJ7L3idFEquCuktbqkN3pGfXTHA5LiDSyahjPoAYy_aGuEFaNG0cbN5PYVWKXJwZlKXRK8W4-bCi4dBEVcVSBlcDPgXMjtBhPKljI2U9FLZWSLJ--NRqMYsGR5vzXGY1Awg=s0-d "sen(180^{\circ} - x) = sen(x)")
Temos
![[;sen(160^{\circ} - x)=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sjBoIyQ29crw_D_EipupApiKljoE6oQVLdVgKIRHDToDukD06nm9ip1T8ingfhv6YkejrRDfUI_KHeUoDSWtqif73FKNEqC1KqLg3dcHb1X7SlfmiBfAm2V6fV6Bk=s0-d "sen(160^{\circ} - x)=")
![[; sen(180^{\circ} - (160^{\circ} - x)) =;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s-dnls3cu3pzhMbh_kYB98xu2T-dNtacuOYBiWkSEryleE-o4ydxi11JCgZuuJURB-bLHuK5nxlAXueuVZWuDX4psDhjBJRGEro0R7y46e1CNITdxEcU-LJ0AzWfr5bMlfI5OLdbr1W5W-Sn8TKdVwVTF4P-bA6lKqS0IjxmYJ12L0Cg=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_sm61tuzCWVd3J49powXkVWfsThgILzyG75_Dw0t_V33Q3bc6nMXSCKB5gpmGqJkx1Fs2lUyaLl3TFmjC5_yvbGeZMjN0zlY7prhJ0DITAwJKPmsJgLIBXA9cyAJvNkuEk5D6-7jK_mtgZ3T8-dh633JuyT1XplCkL79Acz9AF7MNyxxu7aEDzt3ajcNZSRTNBCyjcFR_qKFLu5pqOa68d1HBooL56MKFx8Wr8XAVz8wtSsrosKBLMIYGoLHsINjXptrnpmw1ABhS6GVUqnyoWzCFAekLCR04yUu6yp-fXxtwdPEZgR7rFC4dVxUgNwd2Bxy5cw=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_txPWSy7U1F3lmII6vLC25Vf1kcENzVexK4gG9XSohFh2qGZqs3OXiyMCmMPH-hevpM7G25BTBMaAGK9wMbnGTtMN3LgcT_UGnUyyv716F-HzW4HxDvnZH0DjcdYEn3U0_aQDvYvVtAQq_sfw7dFwdgLA=s0-d "sen(20^{\circ}) \cdot cos (x) +")
![[;cos(20^{\circ});]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_srCKMhHAVhy9kl_J2Ze9ZBsS1CuDaLH_IZZ_mJQhCbXyucmOEbTASACnPfPLWl0-6FWsCWoKFwlhA9wJRBc7EJE4t6jw1yAhN7KKmeINkzGP7h2w=s0-d "cos(20^{\circ})")
![[;\cdot;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tbs1Sbwxx9I1GfUE3eypbJjNbIPj2q0Tl9wkkHQ5P0bdz1QBfjuM1OUX1KcdzQs6oFqHpsSl-jc78AxrZG4oA=s0-d "\cdot")
![[; sen(x)=;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uY2Q75wk7UPNyKstcvxmBFAmpZw3gJVEN9-EWJeUcI2UaELRURQHf3VIG-YHDOidpOdzaPcP82reHuHjcTP1cvMiiqm1TH=s0-d "sen(x)=")
![[; \cdot sen(x);]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_srEqvcsMCy9RgFIEmDzPDHLezlT69Fk_ddunab212Ubs1VThtJNHrrMFet7V3LnpJjv6zX9PQ5XRnXPL78-Nv8JO-dWAZFXMV1uxD1hjc=s0-d "\cdot sen(x)")
![[; + \sqrt{3};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vudXI56cmDgKj4hBoJUYttN-0MPThSBf1NY4g5DEnlVbTPzknCqzzzkR11S4aXhy8k0T-J2Jci2ChoObFUpexHu-il1rI_AmJKKyJr=s0-d "+ \sqrt{3}")
![[;sen(20^{\circ});]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uW8YixzwjW3gr9Fg1nZr_ELYTA4n0kFQ6qKClIAh4e-NYg7M_11HJeCsfa2eR4TYwM7--XFLXXHlOss71169T6c3F_zjbnbg-0J6Cla9IOPK3JHw=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_vs8w9O-098kWj1w1cDUkq5jlPkmL5WHlJ3NqAYBREpHgyqASzky-ijOcxoxRWVL7kGXngVMSgAJaetnVXidUnecSmBGD7ILzUiC3AYFQNQxo11gU0aO_7FIf2Sq_VnrNTI0PoPY02EBl5OmCKMM5u5pqbP2ADjfqOK3_bCidgzUEOGnmd13NXG7hxVdmb9RPwNl99mDya8biHuDx8XuBLOv4gr6rScNvaqkTamS97v2Bef=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_u7YV36JSoa__Y6AoXZR1gtV07L0hQWJr5FYNp96jCJc1YWIq4mv8LsbcCAKQYsla-aLasfp2LrIgWMBvGg-1jEfe2oPW-sZD_vl_ecuTaDKBE=s0-d "ctg(x)=\sqrt{3}")
daonde![[; x=30^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uX1HTJV29O-AvDGbNewe1r57qdvj6MUa2CPtxLLtXn-2kMg5IgupJCKSFOHmt7rchDm6TL1H1Isp86Z5PZ_K2L-MR9lR4fuMPQ7a08ShBv=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_tKz9s6GwvCT_ULkErIzIchJUksLkXRO9D7RvPKhuMmAXL-cTt87sd4Jn7R6kOpKbBnjBvUwls2rYOF6n1E4xIwpqJ4gPOtJ9aTq8mM=s0-d "ED'\parallel BC")
. Seja ![[;P;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tzjeuroGWxAfMELODBkiAJ_s6sXhf52H_mTYUD1POYGfwgY3s993R3SjIvupzniy-XReCna68BYjc=s0-d "P")
o ponto de encontro de![[;BD';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tnSY9wAzIoQzfc0RwEIBWlhjTs4iNhWqHcVSHCDq4B8lfleNjP0wQ6FMIRGnGDXntrOHZnd0JL2mWkUnHT=s0-d "BD'")
com![[;CE;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tUbrTmkb7A2wCOP51ER2WW1qoHelHRd5fUtWT7mjtfCZJERv0cKryuQqR8Il9-dl6sMi6iB0-B2EEw=s0-d "CE")
. Assim, ![[; \triangle{BCP};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v_7xD4ojIkdMxjG-sWNXsvo_WGzu9QVmqKwHVpeY2yueB8UXO_c8vDWV79-oFsoIFHwOr7McTcRlx3g8jJw97fhRqXknFp_13WmCW_iS3R=s0-d "\triangle{BCP}")
e ![[;\triangle{ED'P};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ufOYMVPSRE2ggTmEeHmMEC2HLkoO1aa0ImeP894LNKPodKYmTN5vyEbbHSw7VSDt_adu0RILe0861ENmm3vQQcX5n0fBPKREamdOyaPGOZ=s0-d "\triangle{ED'P}")
são triângulos equiláteros. Ou seja, ![[;CP=BC;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vX8Ko5I8NCtbWeMxNDHfWu8CtSnKVLB6kOu3bFAiKlzl5C8zuNSEpx3ryIkIKjjkz1xZp6HGhwGGa7HaZP=s0-d "CP=BC")
como é isosceles, então ![[;\triangle{PCD};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sRNXmcxMtzkXmX8cj-x5LDo98W7O_moBLWMFWn7iDaMZ68SKsl49f9V15jclO1iJsLZRR4KNchy1vDlAXXWpnfs_9soRz24agWwJ4=s0-d "\triangle{PCD}")
é isosceles com ângulo do vértice ![[;\hat{C}=20^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ulaMvZiBcnujbAld6lycU6FKBDDxBT733J9gqpCXmKyeYrwDbCDXQDtP63Bn90049bnNjNF0sl76yjuZsjOz9EGj-qn6wbl_7jYntNWWHBRZ8gXz9P-8Rl=s0-d "\hat{C}=20^{\circ}")
, assim![[;C\hat{P}D=80^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uP_BXDkpuuYmBmEXiLpall6F0XIOF82gkq_EiCJfeiPHwZmnOloLNnmqKWBfhfVMoStZLMytMp1mIERfLbO7y1jyDU28rS4s6RGXbsCohe37H6vXvQJlleuhk=s0-d "C\hat{P}D=80^{\circ}")
como ![[;E\hat{P}D'=60^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u1xcpy6yQ4rUbVxQTzSENxSqNOCimxZj2BTD7ExgzLP9aqNGdvi7oC9CulnrL6VXfg_oYFMhhLvzwwOW944wtXBT2lMA5FJRjRPZ9xGonYDz3cEXwWXcFdo_Grs7c=s0-d "E\hat{P}D'=60^{\circ}")
e ![[;E,P;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uOmPfeDI1fPKLsKAt7EbX3BS0wSAo4LjyJFqVsQhaJKCN5brLfg8Ng8XWQYFzmiUK9kJbbWBGZj70azw=s0-d "E,P")
e ![[;C;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u0keL1Jc2chUBIdxk167T2CVaRWXJhlbyfD2zjItYWW9eKtAdrr5Uo2jZXqyMaLQpgtbi0bCHEhA=s0-d "C")
são colineares, ![[;D'\hat{P}C=40^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vSeOlZL3A_tn9Yu8DoGZV0jQTSMjAQxkCTBNvgAZ9hQcZr1MyEIPCrYWT6XsDu31QLfNoY4_FT4WIGJrYhQdmnLtKj33cg3Wmzw_h3Gn4cpg2B7qaliCvFYF1NwTM=s0-d "D'\hat{P}C=40^{\circ}")
.
Como e ![[;B\hat{D'}C=40^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vFK_xoMCzQ1SwBCRwJAG4QArTQz0pJIkzRACTA3LlYIhE5ZiUtPETgjpuGWVVxW_t6ngyx9R_JchwS1cRNPKZC6ADgHQVmTMKc9zn7pVDyIJSo4gG_h0m8BT9sN10=s0-d "B\hat{D'}C=40^{\circ}")
![[;\triangle{PDD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sNhriRNJVyzh9QXY0Oj05WIM3_iDV7aV1xTpJyf25SsELQS7h248z5VqRZmSS2z-uHzvglmOM5tB3o4e2_LD92XYYyIxQlsDuklugdcJk4=s0-d "\triangle{PDD'}")
é isosceles com angulo do vétice ![[; \hat{D}=100^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_smi0Vqf-4XZKiSV4riG-QqIDzSyXI8sT5kK7ji_b8ltFgH6LkZ8b-LYHt5UeugcGtkUpGBSbYCCONA9-mTGF2zFS4exW7PWEQ1NhjFEIpaI412NdlQM6FtvJJdww=s0-d "\hat{D}=100^{\circ}")
.
Lembrando que![[;\triangle{EPD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tZc8LA39l0J3x9sj9fq8qbyI3K_RTH38H3LczU5azUaX7rzTYY_6fGz0JRLlBNNZItZ9rdyEiuc7ksj7UmmXk9AGW_TaJUl_BlhYewNZ_X=s0-d "\triangle{EPD'}")
é equilatero. Temos ![[;ED'=EP;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vBpUxIHaw1EDWYsHjdHJD3RqEmVNpNVkSKk4amnxJXw9nsP45r2l0ptQ2wbscHgCEDVqTAzjTlI1zNR-vvhxCG=s0-d "ED'=EP")
e ![[;DP=DD';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tDYIjbJW41OFe6a5aqY2u3QhR7ycJP_AxYkcSlDkE3d_h_xpR7y-RUo5yF-NtTy4BYPUg6h3O7sxuOPwQZzy2D=s0-d "DP=DD'")
assim, pelo caso LLL, ![[;\triangle{EPD}=\triangle{EDD'};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tjVk7mf2z23KjraPmAeSsNkxB0O-SyiiSNYCDm5nI5hLWTevD0hMrsCKbzTfIegansD1eYKHFDnytEt9lzkuTsqfN8YLrd2qPF1va7EfgNz4g17G0YKYWkU2-A8DvC48CU5Muk=s0-d "\triangle{EPD}=\triangle{EDD'}")
. Por isso, ![[;ED;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vqMLR6lYftHDp3aCIXC4zd9lkdoABbnn9J47fVKJpjVXuJBHOABTvyCxYTeaoFMHRzVePEYFQj1GDS=s0-d "ED")
é bissetriz de ![[;P\hat{E}D';]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sfu1hh7-b0Ydsx2NTYJ3csyuDxSF5WWhoFz50FDpexc6CxcwQln6Ygfdyq_BpEUIRkZYGxTW-kcUGbPK0h_lzkWgD726X_KZFIWQ=s0-d "P\hat{E}D'")
, logo, ![[;C\hat{E}D=30^{\circ};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tN_iuvovLoPV794ku67n40kYdA1oGJr3J3bb8fmMHoi79iZbAIHn10YE5CWFR-Eb3D8dD6SwJ6lpD7i8IhsUpU8GlMmOlzfG31PDam1T_SFTq4XPLPi7e8kw=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