tag:blogger.com,1999:blog-6933544261975483399.post2291152256593240199..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Elearn Geometry Problem 5Antonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger15125tag:blogger.com,1999:blog-6933544261975483399.post-67961462621610209522022-10-28T07:57:22.466-07:002022-10-28T07:57:22.466-07:00<BCD=2x
<BDA=4x
<ABD=180-7x
Consider tri...<BCD=2x<br /><BDA=4x<br /><ABD=180-7x<br /><br />Consider triangle and by sine law<br />sin3x/BD=sin(180-7x)/AD<br />AD/BD=sin7x/sin3x--------(1)<br /><br />Consider triangle BDC and by sine law again<br />sin(180-4x)/BC=sin2x/CD<br />BC/CD=sin4x/sin2x=2cos2x--------(2)<br /><br />Equating (1) and (2) as AD=BC & BD=CD<br />sin7x=2sin3xcos2x<br />sin7x=sin5x+sinx<br />sin7x-sin5x=sinx<br />2cos6xsinx=sinx<br />sinx(2cos6x-1)=0<br />cos6x=1/2<br />6x=60<br />x=10Marcohttps://www.blogger.com/profile/04632526355171968456noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-50189669575206931622020-12-09T20:30:11.136-08:002020-12-09T20:30:11.136-08:00This comment has been removed by the author.Sailendra Thttps://www.blogger.com/profile/12056621729673423024noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-8692145485200534002020-06-05T06:43:39.905-07:002020-06-05T06:43:39.905-07:00Grab a pen and a piece of paper, and illustrate al...Grab a pen and a piece of paper, and illustrate along.<br /><br />Firstly, let us calculate some angles:<br />Angle BCD = Line AE = line DE = line BD = line CD<br />> Triangles ADE and BCD are identical.<br />> Angle BAE = x<br />> Angle BDE = 2x<br /><br />Add point F to line AB and connect it to point D such that line DF is the angle bisector of angle BDE. Now, connect point F to point E. <br /><br />We can see that:<br />> Angle BDF = angle EDF = x<br />> Triangles BDF and EDF are congruent by SAS because:<br /> · Angles BDF and EDF are equal.<br /> · Line BD = line DE<br /> · They share a side.<br />> Triangle ADF is isosceles because angle DAF = angle ADF = 3x.<br />> Since triangle ADF is isosceles, line AF = line DF.<br />> Triangles FAE and EDF are congruent by SAS because:<br /> · Angle FAE = angle FDE = x<br /> · Line AE = line DE<br /> · Line AF = line DF<br /><br />Let angle AFE have a value of q such that angle AFE = angle DFE = angle DFB = q.<br /><br />Now, we solve the problem:<br />> 3q = 180°, angle DFB = q = 180°/3 = 60°<br />> Since all three angles in a triangle add up to 180°, <br />180° - 7x + x + 60° = 180°<br />180° - 6x = 120°<br />6x = 60°<br />x = 60°/6 = 10°Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-13693460688486094732019-02-12T09:09:48.304-08:002019-02-12T09:09:48.304-08:00See this video<a title="See" href="https://youtu.be/4hwo8Kur-E0" rel="nofollow"> See this video</a>rv.littlemanhttps://www.blogger.com/profile/05572092955468280791noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-85621930593269315962017-07-16T22:03:46.315-07:002017-07-16T22:03:46.315-07:00Solution:
https://drive.google.com/file/d/0B5CPfrO...Solution:<br />https://drive.google.com/file/d/0B5CPfrOPKJ1Tem42ZTRZckhjTDQ/view?usp=sharingAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-7858697501954335932017-07-14T14:27:18.880-07:002017-07-14T14:27:18.880-07:00Since BDC is isosceles, m(BDA) = 4x
Now construct ...Since BDC is isosceles, m(BDA) = 4x<br />Now construct an isosceles triangle AED such that m(EAD) = m(EDA) = 2x<br />the traingles AED and BDE are congruent implying AE = ED = DB and M(BDE) = 2*m(BAE) <br />=> on careful observation the quadrilateral BAED is similar to problem -4<br />so m(ABD) = 120-x<br />from triangle, ABC, 3x+120-x+2x+2x = 180<br />=> x = 10Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-25319012316596399272016-10-15T08:55:03.029-07:002016-10-15T08:55:03.029-07:00https://www.youtube.com/watch?v=MeEZakv8WxIhttps://www.youtube.com/watch?v=MeEZakv8WxIgeoclidhttps://www.blogger.com/profile/07989522895596673545noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-41654348935123821152016-08-25T03:55:25.279-07:002016-08-25T03:55:25.279-07:00"Note that both angles EAB and EAD are equal ..."Note that both angles EAB and EAD are equal 2x"<br /><br />What? How come? If you subtract <EAD=2x from <BAD=3x, the remainder <BAE=x, not 2x.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-47635562378021333122016-05-02T04:54:53.010-07:002016-05-02T04:54:53.010-07:00Dear Antonio - is there a proof using properties o...Dear Antonio - is there a proof using properties of 80-20-80 triangles? <br /><br />Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-66116187869032971462012-04-08T09:45:17.352-07:002012-04-08T09:45:17.352-07:00Extend BD to E such that BD = DE
Verify BCE is a ...Extend BD to E such that BD = DE <br />Verify BCE is a right angle; Angle BDA = 4x; ABD = Pi - 7x<br />Now cos 2x = BC / BE = BC / 2BD<br />=> 2 cos 2x = BC / BD = AD / BD = sin (pi - 7x) / sin3x = sin 7x / sin 3x<br />=> 2 sin 3x cos 2x = sin 7x<br />=> sin 5x + sin x = sin 7x<br />=> sin 7x - sin 5x = sin x<br />=> 2 cos 6x sin x = sin x<br />=> cos 6x = 1/2<br />=> 6x = pi/3<br />=> x = pi/18 = 10 degPravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-37485377233424474612012-04-07T05:58:10.825-07:002012-04-07T05:58:10.825-07:00Read "x = 10 degrees" instead "x=10...Read "x = 10 degrees" instead "x=100"Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-87989902349432183632012-04-06T20:55:41.469-07:002012-04-06T20:55:41.469-07:00Draw an angular bisector of BDA.
Draw DAE=2x, so t...Draw an angular bisector of BDA.<br />Draw DAE=2x, so that the angular bisector of BDA meets it at E.<br />So, AE=ED (EAD=EDA=2x)<br />But ∆EAD and ∆BDC are congruent under A.A.S case)<br />So, ED=BD<br />Now, Draw a Circle having the center D and Radius DC<br />Obviously it passes through E, B and C<br />Let AC and the circle meet at G<br />So, GD=BD (Radius)<br />Now connect BG<br />H be the point of intersection of BG and DE<br />Since ∆BDG is an Isosceles DH is perpendicular to BG<br />So, HGD=90-2x<br />Let F be the reflex of E over AB<br />As AF=AE, FAB=EAB=x, AB=AB<br />∆AFB and ∆ AEB are congruent<br />So FB=EB<br />Also, ∆AFE and ∆BDE are congruent (BD=AF, BDE=FAE=2x, DE=AE)<br />So, BE=FE<br />Eventually EB=EF=BF<br />So, ∆BFE is an equilateral<br />Therefore FBE=600<br />Since FE and AB are perpendicular to each other<br />ABE=300<br />Since, Ang.ABC=ABE+EBD+DBC,<br />ABC=300+900-x+2x<br />ABC=1200+x<br />But GBC=90Deg as it is an angle on semicircle (Note that GC is a diameter)<br />So, ABG=120+x-90<br />ABG=30+x<br />Since BGD=BAG+ABG<br />900-2x=3x+300+x<br />600=6x<br />So, x=100<br />Note: Since ABD = 120-x the proof for the “PROBLEM 4 is inbuilt”Anonymoushttps://www.blogger.com/profile/07812499400423119847noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-68656118410088413192011-09-04T13:21:34.963-07:002011-09-04T13:21:34.963-07:00Video solution (Spanish version) by Eder Contreras...Video solution (Spanish version) by Eder Contreras and Cristian Baeza at <a href="http://www.gogeometry.com/geometria/p005_solucion_problema_video_triangulo_angulo_ceviana_trazo_auxiliar.htm" rel="nofollow">Geometry problem 5</a>. Thanks Eder and Cristian.Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-7970089108506243342010-07-02T02:46:48.431-07:002010-07-02T02:46:48.431-07:00http://ahmetelmas.files.wordpress.com/2010/05/cozu...http://ahmetelmas.files.wordpress.com/2010/05/cozumlu-ornekler.pdfAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-76534327534868580442009-05-09T16:58:00.000-07:002009-05-09T16:58:00.000-07:00Triangle CDB is isosceles since BD=DC, therefore ...Triangle CDB is isosceles since BD=DC, therefore angle BCD=angle DBC=2x.<br /><br />Let point E be a point inside ABD triangle so that AED triangle is congruent to CDB triangle, where AE=ED=BD=DC and as it's given AD=BC. Note that both angles EAB and EAD are equal 2x.<br />Angle ADB = 4x=>angle EDB=2x. Triangle EDB is isosceles since ED=BD.<br />Angle BAE = angle BAD - angle EAD = 3x - 2x = x.<br /><br />Let's reflect point E over AB calling it F. Triangles AFB and AEB are congruent. AF=AE, FB=EB. Angle FAE = 2 * angle BAE = 2x.<br /><br />FAE and BDE are congruent isosceles triangles since AF=AE=DE=DB and angle FAE=angle EDB=2x<br />=>FE=EB, but FB=EB => triangle FBE is equilateral => angle FBE=60=>AEB=30.<br /><br />Angle EBD = (180 - angle EDB) / 2 = 90 - x.<br />Angle ABC = 180 - angle BAC - angle BCA = 180 - 5x.<br />But angle ABC = angle ABE + angle EBD + angle DBC, or<br />180 - 5x = 30 + (90 - x) + 2x=> 180 - 5x = 120 +x => x=10Anonymousnoreply@blogger.com