tag:blogger.com,1999:blog-6933544261975483399.post8495400226486959058..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1117: Right Triangle, Angle Trisection, 90 Degrees, Parallel LinesAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-50270740980440558372015-08-07T01:59:28.944-07:002015-08-07T01:59:28.944-07:00Kindly refer my proofs for Problems 1114 - 6.
A4...Kindly refer my proofs for Problems 1114 - 6. <br /><br />A4 is the in centre of Tr. AB1B3 so B1A4 bisects the 60 deg angle AB1B3. So A4B1 // A1B2<br /><br />Similarly B2C1 // B1C4 considering that C4 is the in centre of Tr. B1B3C<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-8798196282409731232015-05-02T13:36:55.587-07:002015-05-02T13:36:55.587-07:00http://s27.postimg.org/m6mecvwgz/pro_1117.png
Per ...http://s27.postimg.org/m6mecvwgz/pro_1117.png<br />Per the result of problem 368 we have<br />B3C3 is an angle bisector of ∠ (AB3B1) => A4 is incenter of triangle AB1B3<br />And B1A4 is an angle bisector of ∠( AB1B3)<br />Per the result of problems 1114 and 1116 we have 60 degrees angles and 30 degrees angles as shown on the attached sketch ( see sketch )<br />So ∠ (A4B1B2)= ∠ (B1B2A1) = 30 degrees => B1A4//A1B2<br />Similarly we will have C1B2// B1C4<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com