tag:blogger.com,1999:blog-6933544261975483399.post7037450127439751587..comments2024-03-19T00:02:30.728-07:00Comments on Go Geometry (Problem Solutions): Elearn Geometry Problem 233: Parallelogram, Exterior line, Perpendicular linesAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-6933544261975483399.post-41733259196669255882023-03-09T01:40:03.493-08:002023-03-09T01:40:03.493-08:00Draw a line at A"B"C" pass through ...Draw a line at A"B"C" pass through D // A'B'D'C'<br />From problem 232, AA"+CC"=BB" & d=A"A=B"B=C"C<br />a+c=AA"+A"A'+CC"+C"C<br />=AA"+CC"+A"A"+C"C'<br />=BB"+d+d<br />=BB"+B"B'+d<br />=b+dMarcohttps://www.blogger.com/profile/04632526355171968456noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-82643631763417219232021-02-02T14:14:27.327-08:002021-02-02T14:14:27.327-08:00Extend A`A to form a rectangle with C (A'ECC&#...Extend A`A to form a rectangle with C (A'ECC'). Let BB' meet EC at F => BF=b-c<br />Join AC to form another rectangle A'ACC'. Let DD' meet AC at G => DG=a-d<br />BFC and DGA are congruent => BF=DG and hence a+c=b+d<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-88002761173336962532021-02-02T03:27:11.776-08:002021-02-02T03:27:11.776-08:00Proof follows easily from my proof of Problem 232Proof follows easily from my proof of Problem 232Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-46937862970694159672011-02-05T09:31:43.880-08:002011-02-05T09:31:43.880-08:00Draw D'D''DD''' parallel t...Draw D'D''DD''' parallel to A'B'D'C' thro' D.<br />where D',D'',D''' lie on AA', BB', CC' resp.<br />now, D'A' = D''B' = DD' = D'''C'.<br />so, it requires us to prove that AD' + CD''' = BD'' which comes from problem 232.Ramprakash.Knoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-39284252958399182012010-01-05T08:44:25.219-08:002010-01-05T08:44:25.219-08:00like P234, 235
draw AB", DC" // A'C...like P234, 235<br /><br />draw AB", DC" // A'C'<br /><br />b-a /AB = c-d/CD<br />b-a = c-d => b+d = c+ac .t . e. onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-4672776572611617162009-02-02T05:30:00.000-08:002009-02-02T05:30:00.000-08:00Let be O the intersection of AC and BD, and O' the...Let be O the intersection of AC and BD, and O' the projection of O on the exterior line. Let denote by o the lenght of OO'.<BR/>In the trapezoid BB'D'D :<BR/>o=(b+d)/2<BR/>In the trapezoid AA'C'C :<BR/>o=(a+c)/2.<BR/>From this relations we have that :<BR/>a+c=b+dAnonymousnoreply@blogger.com