tag:blogger.com,1999:blog-6933544261975483399.post948151718524432019..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 759: Equilateral Triangle, Transversal, Trisection of sides, Congruence, AngleAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger9125tag:blogger.com,1999:blog-6933544261975483399.post-58073648218224556812019-02-25T04:55:51.310-08:002019-02-25T04:55:51.310-08:00posons AB = R et AH la hauteur issue de A.
Alors :...posons AB = R et AH la hauteur issue de A.<br />Alors : BD = R/3 ; BH = R/2 et BE = 2R/3<br />BD/BH=2/3 et BE/BA=2/3, donc BD/BH= BE/BA<br />D’où : (ED) parallèle à (AH) et x= 30°.<br />English translation :<br />let AB = R and AH be the height of A.<br />Then: BD = R / 3; BH = R / 2 and BE = 2R / 3<br />BD / BH = 2/3 and BE / BA = 2/3, so BD / BH = BE / BA<br />Hence: (ED) parallel to (AH) and x = 30 °.<br />Medhttps://www.blogger.com/profile/14862518129950779394noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-50869197677442704152015-09-20T04:53:51.866-07:002015-09-20T04:53:51.866-07:00Easily BED is a 30-60-90 Tr. Having BE = 2.BD and ...Easily BED is a 30-60-90 Tr. Having BE = 2.BD and the included angle 60. So < BED = 30 and hence x = 30.<br /><br />Sumith Peiris<br />Moratuwa <br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-14804762996092513322013-10-10T06:27:50.755-07:002013-10-10T06:27:50.755-07:00http://www.mathematica.gr/forum/viewtopic.php?f=20...http://www.mathematica.gr/forum/viewtopic.php?f=20&t=27308&p=133642Μιχάλης Νάννοςhttps://www.blogger.com/profile/02379101429577964881noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-22146420179874111022013-02-23T07:08:19.280-08:002013-02-23T07:08:19.280-08:00Let BD = AE = a . Draw DG // AB intersecting AC at...Let BD = AE = a . Draw DG // AB intersecting AC at G. now DC = BE = 2a. In tr.ABC, by BPT, GC = 2a and AG = a.<br />now tr. DGC is equilateral. so DG = 2a . In tr. FGD and tr.FAE are similar. so FA = a. Now FA = AE = a. So ∠FEA = ∠AFE = x. ∠ FEA = ∠ GDF = x. In tr. DFG ∠DFG + ∠ GDF = ∠DGC. 2x = 60° or x = 30°kalpithttps://www.blogger.com/profile/08431159710282150473noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-12064281122275777712012-11-27T08:57:32.143-08:002012-11-27T08:57:32.143-08:00Let G be the midpoint of [BE].(Draw a picture your...Let G be the midpoint of [BE].(Draw a picture yourself)<br />From BG/BA=BD/BC =1/3 follow that triangles BDG and BAC are similar; moreover DG/CA=1/3.<br />Now, in triangle BDE, median DG=BE/2- half of side of base, or that suffice that angle BDE is right. In fact, because triangle BDG is equilateral, follow that triangle EBD (in that order of vertices) is of type 30-60-90.<br />The rest is simple.<br /><br />Visit also http://ogeometrie-cip.blogspot.com/CiPhttps://www.blogger.com/profile/01324868096212244655noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-24150574468075660052012-06-02T04:39:04.319-07:002012-06-02T04:39:04.319-07:00AE = BD = a/3 where AB = BC = AC = a
So BE = CD = ...AE = BD = a/3 where AB = BC = AC = a<br />So BE = CD = 2a/3 <br />Apply Menelau's Theorem to ∆ABC<br />and transversal DEF<br />(BE/EA).(AF/FC).(CD/DB) = 1 numerically<br />Easy to see BE/EA = 2 = CD/DB<br />Follows AF/FC = 1/4, AF = a/3 = AE<br />So x = ∠AFE = ∠AEF = (1/2)∠EAC = 30°Pravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-56229525190007746802012-06-02T04:09:20.238-07:002012-06-02T04:09:20.238-07:00Draw EG ∥ DC (G on AC)
∆s GDC and AEG are equilate...Draw EG ∥ DC (G on AC)<br />∆s GDC and AEG are equilateral and <br />BEGD is a parallelogram<br />1/2 = EG/DC = FG/FC <br />G is the midpoint of FC<br />FG = GC = EB = DG<br />∆FDC is rt angled at D<br />x = 30°Pravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-44128951225054420912012-06-01T23:10:39.673-07:002012-06-01T23:10:39.673-07:00please disregard my previous solution. new Solutio...please disregard my previous solution. new Solution below include minor typo corrections. <br />Apply Menelaus’s theorem to triangle ABC with secant FED<br />DB/DC x FC/FA x EA/EB=1 => ½ x FC/FA x ½= 1<br />So FA/FC= ¼ => CA/CF= ¾<br />From A draw AH //FD ( H on CB)<br />We have CA/CF=CH/CD= ¾ => CH=3/4.CD=3/4 .2/3 .CB= ½. CB<br />AH will be an angle bisector and altitude of triangle ABC, so x= angle(HAC)= 30Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-4728157763497688882012-06-01T23:09:39.862-07:002012-06-01T23:09:39.862-07:00By Menelaus theorem on triangle ABC, (BD/DC)*(CF/F...By Menelaus theorem on triangle ABC, (BD/DC)*(CF/FA)*(AE/EB)=1.<br />Hence CF/FA = 4/1. Therefore FA:AC=1:3.<br />Since ABC is an equilateral triangle, FA = (1/3)AC = (1/3)AB = AE<br />So triangle AEF is an isosceles triangle. Angle EAF = 120, hence x = 30.WFungnoreply@blogger.com