tag:blogger.com,1999:blog-6933544261975483399.post8279717235775767932..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1012: Equilateral Triangle, Incenters, Inscribed Circles, Perpendicular, Concurrent LinesAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-6933544261975483399.post-27268192503784174692014-06-09T20:30:27.103-07:002014-06-09T20:30:27.103-07:00To prove that A1A2, B1B2, C1C2 concurrent at a poi...To prove that A1A2, B1B2, C1C2 concurrent at a point E we need:<br /> (C2B)^2 - (C2A)^2 + (B2A)^2 -(B2C)^2 + (A2C)^2 - (A2B)^2 = 0<br />Which is equivalent to C2B + B2A + A2C = C2A + B2C + A2B <=> C2B + B2A + A2C = 1.5a<br />We set AB = AC = BC = a <br /> AD = x, BD = y; CD = z;<br />Since C2B = 0.5(y+a-x); B2A = 0.5(a+x-z); A2C = 0.5(a+z-y)<br />=> C2B + B2A + A2C = 1.5a<br />So: A1A2, B1B2, C1C2 concurrent at a point ECách ta nghĩhttps://www.blogger.com/profile/03723897929818557669noreply@blogger.com