tag:blogger.com,1999:blog-6933544261975483399.post1304236967332301651..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 592: Triangle, Incenter, Incircle, Tangency Point, Midpoints, Concurrent Lines, CongruenceAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-30008080011754639622011-03-10T01:17:29.833-08:002011-03-10T01:17:29.833-08:00Thank you Peter Tran & c.t.e.o !
Typo error t...Thank you Peter Tran & c.t.e.o !<br /><br />Typo error to be corrected thus:<br />"Let MO extended meet BE<br />(the altitude from A) at F"<br /><br />Now BF = r = OD and BF // OD and <br />BFDO is a parallegram etc ...Pravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-5717191049953285442011-03-09T07:42:37.436-08:002011-03-09T07:42:37.436-08:00To Pravin
Why altitude from B passes at F (accordi...To Pravin<br />Why altitude from B passes at F (according to P591)c .t . e. onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-44763624232930357732011-03-08T10:35:26.900-08:002011-03-08T10:35:26.900-08:00Refer to lines 1 and 2 of your solution “Let MO e...Refer to lines 1 and 2 of your solution “Let MO extended meet the incircle at F.<br />By Problem 591, BF = r =OD ”. Note that per problem 591, MO extended to meet the altitude from B, not the incircle . It is not clear to me. Please show more detail and explanation. .<br />Refer to line 3 of your solution “ So BFDO is a parallelogram “ . I think that it is incorrect. BF do not parallel to DO . Please show more detail.<br /><br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-18424577730366031402011-03-07T18:49:42.890-08:002011-03-07T18:49:42.890-08:00Let MO extended meet the incircle at F.
By Problem...Let MO extended meet the incircle at F.<br />By Problem 591, BF = r =OD<br />So BFDO is a parallelogram and its diagonals <br />BD, FO bisect each other at E, say<br />But the middle line GH bisects BD.<br />Hence BD, GH, MO (extended) concur at E.Pravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.com