tag:blogger.com,1999:blog-6933544261975483399.post6098751810027816237..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 933: Square, Center, Transversal, Perpendicular, Metric Relations, CongruenceAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-6933544261975483399.post-21050664384013634832013-11-28T08:35:35.005-08:002013-11-28T08:35:35.005-08:00<HCO=<BOG=a because <BOC=90. Both triangl...<HCO=<BOG=a because <BOC=90. Both triangles HCO and GOB share hypotenuse and the angle a so they are congruent. Similarly OCM and DON are congruent. From those 2 pairs of congruent triangles we have that BG=HO and DN=OM, so the answer is 2(4/sqrt2)^2=16<br /><br />Ivan Bazarov<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-13168795545391236952013-10-28T12:29:40.385-07:002013-10-28T12:29:40.385-07:00Αν Α(0,0), Β(Ο,4),C(4,4),D(4,0) και (ε1)=EO: y-2=λ...Αν Α(0,0), Β(Ο,4),C(4,4),D(4,0) και (ε1)=EO: y-2=λ1(x-2) , (ε2)=FO:y-2=λ2(x-2) τότε MANOLIShttps://www.blogger.com/profile/00151249027693207153noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-23583863346340915802013-10-22T17:05:06.262-07:002013-10-22T17:05:06.262-07:00thanks for this kind remark...but...
my post does ...thanks for this kind remark...but...<br />my post does not meet the standards of what can be called a 'proof'. I guess you filled in the gaps yourself. So 50% of the proof is yours ;-)<br /><br />bleaugAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-65529236196092068892013-10-21T22:27:28.127-07:002013-10-21T22:27:28.127-07:00What a wonderful proof!
Thank you, Bleaug
What a wonderful proof!<br />Thank you, Bleaug<br />Ajithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-8478834425228825382013-10-21T14:07:06.618-07:002013-10-21T14:07:06.618-07:00http://www.mathematica.gr/forum/viewtopic.php?f=22...http://www.mathematica.gr/forum/viewtopic.php?f=22&t=37369&start=700#p188400Μιχάλης Νάννοςhttps://www.blogger.com/profile/02379101429577964881noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-34340161742460433842013-10-20T05:25:36.427-07:002013-10-20T05:25:36.427-07:00http://bleaug.free.fr/gogeometry/933.png
bleaughttp://bleaug.free.fr/gogeometry/933.png<br /><br />bleaugAnonymousnoreply@blogger.com