tag:blogger.com,1999:blog-6933544261975483399.post1360853583113923549..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Elearn Geometry Problem 238: Square, Midpoints, Lines, CongruenceAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger11125tag:blogger.com,1999:blog-6933544261975483399.post-47461685280756718332023-03-12T09:02:41.463-07:002023-03-12T09:02:41.463-07:00Solution 2
Triangles ABE, AFD & BCF are congr...Solution 2<br /><br />Triangles ABE, AFD & BCF are congruent SAS<br />So < BFC = < AEB and hence EGCF is concyclic<br />BF is thus perpendicular to AE and so ADFG is also concyclic<br />Therefore < BFC = < GAD = < AFD = < AGD<br />So AD = GD and a = b<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-49346144258053786622023-03-12T03:07:21.035-07:002023-03-12T03:07:21.035-07:00Triangle BGE ~ Triangle BCF
BG:GE=2:1
GE=c, BG=2c
...Triangle BGE ~ Triangle BCF<br />BG:GE=2:1<br />GE=c, BG=2c<br />----------------------------------------<br />By Pyth. Theorem (Triangle BEG)<br />BE^2=BG^2+GE^2=5c^2<br />BE^2=(a/2)^2<br />=>a^2=20c^2<br />----------------------------------------<br />By Pyth. Theorem (Triangle BAE)<br />BE^2+AB^2=AE^2<br />AE^2=1.25a^2=25c^2<br />AE=5c<br />AG=AE-GE=4a<br />-----------------------------------------<br />cos0Marcohttps://www.blogger.com/profile/04632526355171968456noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-32773873618928004392020-10-07T07:14:38.684-07:002020-10-07T07:14:38.684-07:00See the drawing
- ΔABE is congruent to ΔBCF by r...See the <a href="http://sciences.heptic.fr/2020/10/07/gogeometry-problem-238/" rel="nofollow"><b>drawing</b></a><br /><br /><br />- ΔABE is congruent to ΔBCF by rotation of pi/2 => ∠ AGF=90°<br />- Define H as the middle of AB and K as the middle of AG<br />- By construction HD//BF<br />- AH=HB=>AK=KG<br />- ∠ AGF=90° => ∠ AKD=90°<br />- ΔDKG is congruent to ΔDKA (SAS)<br />- =>DG=DA<br />- Therefore <b>a=b</b><br />rv.littlemanhttps://www.blogger.com/profile/15820037721044128612noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-30541049937680558872020-09-30T11:24:53.170-07:002020-09-30T11:24:53.170-07:00Draw DLM, L on AG, M midpoint of AB. Tr AML~ Tr GL...Draw DLM, L on AG, M midpoint of AB. Tr AML~ Tr GLD with coefficient 0.5c.t.e.o.https://www.blogger.com/profile/16937400830387715195noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-68328084859477372442020-09-30T01:04:57.488-07:002020-09-30T01:04:57.488-07:00See the drawing
- Triangle ABE is congruent to tr...See the <a href="https://drive.google.com/file/d/1qo8sdIG3r5RkOpVitChn0_iYp7X8goyx/view?usp=sharing" rel="nofollow"><b>drawing</b></a><br /><br />- Triangle ABE is congruent to triangle BCF by rotation of pi/2 => < AGF=90°<br />- Define H as the middle of AB and K as the middle of AG<br />- By construction HD//BF<br />- AH=HB=>AK=KG<br />- < AGF=90° => < AKD=90°<br />- triangle ADG has its height in the middle of AG<br />- => ADG is isosceles in D => DA=DG<br />- <b>Therefore a=b</b><br /><br />rv.littlemanhttps://www.blogger.com/profile/15820037721044128612noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-37123539966259398232020-09-29T12:53:07.175-07:002020-09-29T12:53:07.175-07:00- b is the length of chord GD intercepted by angle...- b is the length of chord GD intercepted by angle <DAG in circle ADFG of diameter AF<br />- a is the length of chord AB intercepted by angle <AEB in circle ABE of diameter AE<br />-but AE = AF and <AEB = <DAG hence b = a<br />Greghttps://www.blogger.com/profile/14941282981782772132noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-38499528050903393032020-06-01T08:32:23.545-07:002020-06-01T08:32:23.545-07:00ABE,BCF,ADF are congruent right triangles.
So <...ABE,BCF,ADF are congruent right triangles.<br /><br />So < BAE = < FBC = < FAD = p say. <br /><br />AE & BF are hence perpendicualar and AGFD is concyclic and hence < GAF = 90-2p & so < GFA = 2p = < ADG<br /><br />Therefore < GAD = < AGD = 90-p and so b = GD = AD = a<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-52132878745284889032013-10-10T06:59:25.189-07:002013-10-10T06:59:25.189-07:00http://www.mathematica.gr/forum/viewtopic.php?f=20...http://www.mathematica.gr/forum/viewtopic.php?f=20&t=32650&p=151226Μιχάλης Νάννοςhttps://www.blogger.com/profile/02379101429577964881noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-65483268978865105322011-02-05T09:38:23.412-08:002011-02-05T09:38:23.412-08:00We observe that AGFD is a cyclic quadrilateral.
So...We observe that AGFD is a cyclic quadrilateral.<br />So, angle GFC = GAD.<br />also, angle BFC = AFD and AFD = AGD.<br />=> GAD = AGD. => GD = AD = a.Ramprakash.Knoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-72734285923878666282009-04-15T20:28:00.000-07:002009-04-15T20:28:00.000-07:00Here's another by analytical geom.: Let A be(0...Here's another by analytical geom.: Let A be(0,0),B:(0,a),C:(a,a)& D:(a,0. So E:(a/2,a) & F:(a,a/2)and BF:x/2a + y/a = 1, AE:y = 2x. This makes G:(2a/5,4a/5) and GD^2=b^2=(2a/5-a)^2+(4a/5)^2 = a^2. Hence a=b.<br />I want to know how this can be solved by calculus. Ajit: ajitathle@gmail.comAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-10635927312166167912009-02-02T03:08:00.000-08:002009-02-02T03:08:00.000-08:00It is easy to prove that the measure of AGF is 90(...It is easy to prove that the measure of AGF is 90(traingles ABE and BCF are congruents). Let H be the intersection of BF and AD. FD= AB/2, FD is parallel to AB then, AD=DH, so GD is median of ortogonal triangle AGH. So GD = AH /2 = AD = AB.<BR/>There is another solution with calculus.macsim gabriel florinhttps://www.blogger.com/profile/08034437321820989899noreply@blogger.com