tag:blogger.com,1999:blog-6933544261975483399.post3408069401515762233..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1128: Right Triangles, Medians, Parallel, Angles, Congruence, Half the measureAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-6933544261975483399.post-39877958972473717432021-04-22T12:59:59.737-07:002021-04-22T12:59:59.737-07:00For ease of understanding, read @ as Alpha
Extend ...For ease of understanding, read @ as Alpha<br />Extend AB to meet C1F at H<br />Observe that HB1D and HBE are similar => m(BEH)=m(BDB1)=@ -------------(1)<br />Let m(BAC)=A and extend BE to meet M1B1 at P<br />m(M1B1E)=180-m(MBE)=90+A=>m(M1C1B1)=m(M1B1C1)=m(EB1P)=90-A-@<br />Extend C1G to meet AB at Q => m(AQG)=m(M1C1B1)+m(AHC1)=90-A-@+90-@=180-A-2@<br />Therefore in Tr.AGQ, m(AGQ)=2@=m(CGC1) -----------------(2)Sailendra Thttps://www.blogger.com/profile/12056621729673423024noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-40400292011500375452020-10-20T11:02:09.748-07:002020-10-20T11:02:09.748-07:00No , Alpha =∠(ADA1)
beta=∠(CBA)No , Alpha =∠(ADA1)<br />beta=∠(CBA)Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-6111668620215216282020-10-12T14:32:14.201-07:002020-10-12T14:32:14.201-07:00arent alpha and beta are same angles?arent alpha and beta are same angles?Anonymoushttps://www.blogger.com/profile/17821589001449105440noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-74911992322258409302019-09-10T22:59:13.835-07:002019-09-10T22:59:13.835-07:00Let BE extended meet M1B1 at Y and A1B1 extended a...Let BE extended meet M1B1 at Y and A1B1 extended at X<br /><br />alpha1 = alpha since BDB1E is concyclic<br /><br />If < BAC = p, then < BXD = 90-alpha<br />So < B1YX = 180-2.alpha-2p = < MBC = 90- p since BM//BM1<br /><br />Hence 2.alpha + p = 90<br /><br />Now consider quadrilateral AGA1D<br /><br />180-p + 180- theta + 2.alpha + p = 360<br /><br />Therefore Theta = 2.alpha<br /><br />Sumith Peitis<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-51656759195832474832015-07-06T21:02:28.305-07:002015-07-06T21:02:28.305-07:00http://s7.postimg.org/ccpzrcdez/pro_1128.png
Obse...http://s7.postimg.org/ccpzrcdez/pro_1128.png<br /><br />Observe that quadrilateral DBEB1 is cyclic so α = α 1<br />Let B1M1 cut BE extension at N<br />Let b=∠ (MBC) => ∠ (MCB)= β<br />Since BM//B1M1 => ∠ (M1Nx)= β<br />And ∠ (CFE)= 180- β - α<br />And ∠ (M1B1C1)= β - α =∠ (M1C1B1)<br />In triangle GFC1 θ = 180-(180- β - α)-( β - α)= 2. α<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com