tag:blogger.com,1999:blog-6933544261975483399.post6054242389355171799..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1290 Triangle, Internal Angle Bisector, Median, Parallel, Measurement, Metric RelationsAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger9125tag:blogger.com,1999:blog-6933544261975483399.post-65681224684266912442016-12-01T13:29:47.140-08:002016-12-01T13:29:47.140-08:00G point CF meet BD, F on AB (Tr BFC isoceles)
MN ...G point CF meet BD, F on AB (Tr BFC isoceles)<br />MN is also middle line of AFC, a way to prove equality of tr MNE and DCB<br />Thanksc.t.e.onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-91140940794700732372016-12-01T12:47:17.225-08:002016-12-01T12:47:17.225-08:00To c.t.e.o
Where are points F and G?
MN is not ne...To c.t.e.o<br />Where are points F and G? <br />MN is not necessary, use MG. ThanksAntonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-50900359454046102802016-12-01T08:26:25.092-08:002016-12-01T08:26:25.092-08:00Draw CF Perpendic to BD, MN //CF
=> MN = 1/2 A...Draw CF Perpendic to BD, MN //CF <br />=> MN = 1/2 AE = 2.5 (MG middle line of AFC)<br />=> MG=EB (MGBE parallelogram)c.t.e.onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-3696984020767857142016-11-30T06:53:28.406-08:002016-11-30T06:53:28.406-08:00Let AB = c, BC = a and AC = b
we have DC = ab/a+c ...Let AB = c, BC = a and AC = b<br />we have DC = ab/a+c , AD = bc/a+c and AM = b/2<br />Since triangles AME and ADB are similar <br />AE/AM = AB/AD<br />=> c-x/(b/2) = c/(bc/a+c)<br />=> 2c-2x = a+c<br />=> x = c-a/2 = 2.5Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-7483064985698416192016-11-30T00:27:51.880-08:002016-11-30T00:27:51.880-08:00Extend ME to cut BC at F, construct the parallelog...Extend ME to cut BC at F, construct the parallelogram CFEG; due to AM=CM, ME passes through the midpoint of AG and, as it is the angle bisector of <AEG, tr. AEG is isosceles and AE=GE(=CF). With BF=BE, we are easily done.Stan Fulgerhttps://www.blogger.com/profile/07420210614111479737noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-3184451852559669292016-11-29T01:47:05.482-08:002016-11-29T01:47:05.482-08:00Problem 1290
Draw AP //BD//ME . P is on AB extende...Problem 1290<br />Draw AP //BD//ME . P is on AB extended.Is <BCP=<DBC=<DBA=<BPC, then <br />BP=BC and AM=BC so AE=EP or AB-BE=BD+BE or AB-BC=2.BE or x=2.5.<br />MANOLOUDIS APOSTOLIS4 HIGH SCHOOL OF KORYDALLOS PIRAEUS-GREECE<br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-61194765614396314932016-11-28T23:59:30.943-08:002016-11-28T23:59:30.943-08:00Extend AB to point F, such that BF=BC
Angle AFC= ...Extend AB to point F, such that BF=BC <br />Angle AFC= Angle ABC/2, hence FC is parallel to ME. <br />Since M is midpoint of AC,E must be midpoint of AF, AE=(AB+BC)/2. BE=AB-AE=(AB-BC)/2 <br />BE=5/2 <br /> Pradyumna Agashehttps://www.blogger.com/profile/10300531209692781145noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-48621857030292112702016-11-28T20:47:35.477-08:002016-11-28T20:47:35.477-08:00With the usual notation for triangle ABC,
DC = ab...With the usual notation for triangle ABC,<br /><br />DC = ab/(a+c) and so <br />MD = b/2 - ab/(a+c) = (b/2)(c-a)/(c+a)<br /><br />Now ME//BD so <br />(c-x)/x = (b/2)/MD from which<br />c/x = 1 + (b/2)/MD <br />= 1 + (c+a)/(c-a) = 2c/(c-a)<br /><br />Hence x = (c-a)/2 = 5/2 = 2.5<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-89659281839362194792016-11-28T18:13:48.645-08:002016-11-28T18:13:48.645-08:00https://goo.gl/photos/SxYFtYyAsxydH45r7
Let ME ext...https://goo.gl/photos/SxYFtYyAsxydH45r7<br />Let ME extended meet BC at F <br />Draw AN //ME . N is on BC extended ( see Sketch)<br />Observe that ∠BEF=∠BFE= u => triangle EBF is isosceles<br />∠BAN=∠ANB=u => triangle ABN is isoceles<br />Since M is the midpoint of AC => F is the midpoint of NC=> FN=FC<br />AB=BN=FN+FB=FC+x<br />BC=FC-FB =FC-x<br />So AB-BC= FC+x-(FC-x)=2x= 5 => x= 2.5 <br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com