tag:blogger.com,1999:blog-6933544261975483399.post5600540486254752611..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 369. Intersecting circles, Chord, Center, Angle, Congruence.Antonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-71896809594354923082023-09-03T02:12:17.103-07:002023-09-03T02:12:17.103-07:00BravoBravoMarcohttps://www.blogger.com/profile/04632526355171968456noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-86078013371004132332016-10-13T11:38:23.169-07:002016-10-13T11:38:23.169-07:00https://goo.gl/photos/V1K1LviHSrgYLvjy8
Connect O...https://goo.gl/photos/V1K1LviHSrgYLvjy8<br /><br />Connect OA, OB, OC<br />Observe that u=∠ODB=∠OAD=∠OCA<br />And v=∠OBC=∠OCB<br />And ∠ DBC=u+v = ∠DCB=> triangle CDB is isosceles => DB=DC<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-3350626345307616582015-07-08T04:44:52.057-07:002015-07-08T04:44:52.057-07:00< ADB = < AOB = 2 < ACB implying that CD=...< ADB = < AOB = 2 < ACB implying that CD= DB<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-27618604344953436222009-10-20T14:13:43.304-07:002009-10-20T14:13:43.304-07:00Como OC = AO, então o triângulo ACO é isósceles. A...Como OC = AO, então o triângulo ACO é isósceles. Assim, Ang(OCA) = Ang(OAC). Os ângulo OBD e OAD são congruentes, pois seus vértices pertencem ao mesmo arco capaz. Como OC = OB e podemos ter que Ang(OCB) = Ang(OBC). Assim, Ang(BCO) + Ang(OCD) = Ang(CBO) + Ang(OBD), o que podemos concluir que o triângulo DCB é isósceles de base BC. Assim BD = CD (demonstrado).Valdir Marques Faria, Colégio Visãohttp://valdirmfaria.gmail.com.brnoreply@blogger.com