tag:blogger.com,1999:blog-6933544261975483399.post6780762345322421815..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1029: Right Triangle, 90 Degrees, Angle, Midpoint, CongruenceAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger11125tag:blogger.com,1999:blog-6933544261975483399.post-75457854493117513312023-07-09T10:41:26.250-07:002023-07-09T10:41:26.250-07:00circumcenter of DBF will be on BC, and circumcente...circumcenter of DBF will be on BC, and circumcenter will be on perpendicular bisector of DF . the only satisfy these is point EPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-78313707678817320292023-07-04T13:17:53.191-07:002023-07-04T13:17:53.191-07:00circumcenter of DBF will be on BC, (correct), E is...circumcenter of DBF will be on BC, (correct), E is center is not direct clear.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-62040064598687799982015-09-02T04:06:14.560-07:002015-09-02T04:06:14.560-07:00Another method - let GE be the perpendicular bisec...Another method - let GE be the perpendicular bisector G on AC. Show that < CAB = ABG = < BGE = < CGE. Hence < DEG = < DBG , hence BDGE is cyclic and the result follows<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-59146569627880118812015-09-02T03:55:56.996-07:002015-09-02T03:55:56.996-07:00Draw FE perpendicular to BC to meet BD extended at...Draw FE perpendicular to BC to meet BD extended at F. <br /><br />Since AB // FE < EDC = < ABF = < BFE = < CFE since FE is the perpendicular bisector of BC. So CEDF is cyclic and BD is thus perpendicular to AC<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-21836466138904945352014-07-20T14:46:18.670-07:002014-07-20T14:46:18.670-07:00To Bleaug
Now it has been included in the "G...To Bleaug <br />Now it has been included in the "Given" section. ThanksAntonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-60311942457124687662014-07-20T14:17:36.699-07:002014-07-20T14:17:36.699-07:00naahhh, seriously?
then you should add this in the...naahhh, seriously?<br />then you should add this in the "Given:" section<br />bleaug<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-68931890301968813282014-07-20T09:41:38.879-07:002014-07-20T09:41:38.879-07:00To Bleaug
Problem 1029 There is a condition
CD g...To Bleaug <br />Problem 1029 There is a condition <br />CD greater than AD,<br />ThanksAntonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-68810272229692371102014-07-20T06:15:57.435-07:002014-07-20T06:15:57.435-07:00as stated, problem 1029 is wrong.
Given right ABC...as stated, problem 1029 is wrong.<br /><br />Given right ABC and BE=CE: ∠(DBA)=∠(EDC) does not imply BD⊥AC<br /><br />Take D' on AC such as ED'⊥BC, then ∠(D'BA)=∠(ED'C) but BD' is not ⊥AC<br /><br />bleaugAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-63091141689819671952014-07-15T20:42:12.999-07:002014-07-15T20:42:12.999-07:00Hi Antonio
Please publish full of my problema.
P...Hi Antonio<br /><br />Please publish full of my problema.<br /><br />Prob 1027 <br />We build AH perpendicular BC and DF perpendicular AB <br />CH=HD=1/2DC=1/2ED <br />DF=DH=1/2ED<br /> α =10 <br />X=90-4α <br />X=50 <br /><br />Prob 1028 <br /><br />BE=ED=EC EM=EN EN =1/2 EC <br />Triangle ECN < ECN =30 ECN = 90-4X -> X =15 <br /><br />Prob 1029 <br />We note <BDF = 90 <br />BD perpendicular AC <br /><br />Erina New Jersey Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-65961303222087723282014-07-14T03:15:52.867-07:002014-07-14T03:15:52.867-07:00Is this solution is correct? :
1)Continue DE till ...Is this solution is correct? :<br />1)Continue DE till point F such that BE = CE = EF --> BFC right triangle and E is the circumscribed circle <br />of BFC and since BC is the diameter And angle ABC is right so AB is tanget and so on ....Anonymoushttps://www.blogger.com/profile/02970933367484538195noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-30153413872864963002014-07-13T16:27:52.380-07:002014-07-13T16:27:52.380-07:00Extend DE to F so that ED=EF => BDCF is a paral...Extend DE to F so that ED=EF => BDCF is a parallelogram<br />Note that ∠(FDC)= ∠(DFB)= ∠(DBA) => BA is a tangent to circumcirle of triangle DBF<br />Circumcenter of DBF will be on BC => E is the circumcenter of DBF<br />So EB=ED=EC => BD perpendicular to AC<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com