tag:blogger.com,1999:blog-6933544261975483399.post7068474656690491612..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 891: Equiangular Octagon, Metric Relations, Sides, Auxiliary LinesAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-3755830711279337852015-05-21T03:27:03.047-07:002015-05-21T03:27:03.047-07:00that was really helpfulthat was really helpfulAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-63346248516626466282013-06-26T12:05:48.101-07:002013-06-26T12:05:48.101-07:00Like Jacob Ha, I extended DE, BC, AH and GF
DE lin...Like Jacob Ha, I extended DE, BC, AH and GF<br />DE line meet BC line at W<br />DE line meet GF line at T<br />AH and FG at U<br />and AH and BC at K<br />WTUK is a rectangle [easy to prove]<br />WK=TU<br />WK=d/sqrt(2)+5+a/sqrt(2)<br />TU=c/sqrt(2)+d/sqrt(2)+x<br />6+5=x+7<br />x=4Anonymoushttps://www.blogger.com/profile/11694885671410014993noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-37223995061580279022013-06-23T11:02:02.051-07:002013-06-23T11:02:02.051-07:00Distance between DE & AH = 6+5 = 7+x
x = FG = ...Distance between DE & AH = 6+5 = 7+x<br />x = FG = 4Jacob HA (EMK2000)https://www.blogger.com/profile/17238561555526381028noreply@blogger.com