tag:blogger.com,1999:blog-6933544261975483399.post6389501382582516308..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Elearn Geometry Problem 4Antonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger19125tag:blogger.com,1999:blog-6933544261975483399.post-5642956368599722082021-02-06T20:48:40.831-08:002021-02-06T20:48:40.831-08:00For easy typing, I use z instead of "alpha&qu...For easy typing, I use z instead of "alpha" for the following<br />angle CDB=90-z<br />Consider triangle CDB<br />sin2z/BD=sin(90-z)/CD<br />CD=BDcosz/sin2z<br />CD=BD/2sinz<br /><br />Consider triangle BAD<br />angle ABD=x-angle CBD=x-90+z<br />sinz/BD=sin(x-90+z)/AD<br />AD=BDsin(x+z-90)/sinz<br />AD=-BDcos(x+z)/sinz<br /><br />CD=AD<br />BD/2sinz=-BDcos(x+z)/sinz<br />1/2=-cos(x+z)<br />x+z=120<br />x=120-zMarcohttps://www.blogger.com/profile/04632526355171968456noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-58376977295784061742020-12-09T17:31:07.224-08:002020-12-09T17:31:07.224-08:00Let angle bisector of m(BCD) meet AB at E.
Connect...Let angle bisector of m(BCD) meet AB at E.<br />Connect ED and since given AD=DC=BC=>ED is perpendicular to AC, AEC is an isosceles and BCDE is a kite => AE=EC and EB=ED<br />Observe that the triangles DAE, DCE and BCE are congruent (SAS) <br />Hence m(AED)=m(DEC)=m(CEB)=180/3=60 =>m(DBE)=30<br />Now its easy to see x=30+90-alpha=120-alphaSailendra Thttps://www.blogger.com/profile/12056621729673423024noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-42792086050668504012020-05-18T05:10:04.727-07:002020-05-18T05:10:04.727-07:00Grab and pen and a piece of paper, and illustrate ...Grab and pen and a piece of paper, and illustrate accordingly.<br /><br />Add point E to the figure and connect it to line C. Do so in such a way that line CE is the angle bisector of angle BCD. Next, connect point D to point E. That is all for the illustrations. <br /><br />We can see that:<br /><br />> Triangles CDE and CBE are congruent because<br />~ They are share line CE <br />~ Line CD = line BC<br />~ Angle DCE = angle BCE<br /><br />> Triangles CDE and ADE are congruent because<br />~ Due to two equal opposite angles and equal adjacent lengths, quadrilateral ADCE is a concave kite.<br />~ They share side DE<br />~ This is a special case where SSA works, but only because they make up a kite. <br /><br />> Since triangles ADE, CDE and CBE are congruent, angle AED = angle CED = angle CEB.<br />> Angle AED = angle CED = angle CEB = 180 / 3 = 60<br /><br />Now, we solve the problem:<br /><br />Just focus on triangle CBE. The three angles in a triangle add up to 180.<br />x + a + 60 = 180<br />x + a = 120<br />x = 120 - aAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-14255490789112455492017-07-13T10:26:16.872-07:002017-07-13T10:26:16.872-07:00Construct point E such that E lies on AB and EC is...Construct point E such that E lies on AB and EC is an angle bisector of <DCB. Hence, <DCE=<BCE.<br />Construct line ED.<br />Triangle DCE is congruent to BCE. Hence <EDC=x<br />Since <EAD=<ECD and <DAC=<DCA,<EAC=<ECA <br />Hence, EA=EC<br />Hence triangle EAD and ECD are congruent. <br />Hence <EDA=x <br />By angle sum of quadrilateral 3x+3a=360°<br />So x+a=120°<br /> x=120°-a<br />Proved Web Adminhttps://www.blogger.com/profile/06383500595320162829noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-38093698113583215592016-10-15T08:54:39.861-07:002016-10-15T08:54:39.861-07:00https://www.youtube.com/watch?v=2XwhJStz4bAhttps://www.youtube.com/watch?v=2XwhJStz4bAgeoclidhttps://www.blogger.com/profile/07989522895596673545noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-89441036992569779162015-10-11T03:10:35.535-07:002015-10-11T03:10:35.535-07:00Let the bisector of < DCB meet AB at E and BD a...Let the bisector of < DCB meet AB at E and BD at F. Draw altitude DG of Tr. ABD.<br /><br />Tr. s BCF & DCF are congruent and these are also congruent with Tr. ADG. So DG = DF = FB hence in right Tr. BGD BD = 2DG which means that < ABD = 30 and since < DBC = 90-@ the result follows<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-52247278573766314232015-08-01T01:47:31.657-07:002015-08-01T01:47:31.657-07:00Draw an angle bisector of angle BCD and let it me...Draw an angle bisector of angle BCD and let it meet at the point E on the side AB<br />Then connect E and D.<br />Since In Tri ACD , angle ACD = angle CAD <br /> angle ECD = angle EAD = alpha<br /> CE = AE ( Isosceles tri AEC )<br />Since In Tri BCE and Tri ADE <br /> BC = AD ( given )<br /> angle BCE = angle EAD = alpha<br /> CE = AE (proved)<br /> Tri BCE congruent to Tri ADE. This implies x = angle CBE = angle ADE<br />Since in Tri CDE and in Tri ADE <br /> CE = AE ( proved )<br /> CD = AD ( given )<br /> angle ECD = angle EAD<br /> Tri CED congruent Tri ADE , angle CDE = angle ADE = x<br />Since Tri BCE , Tri CDE , Tri ADE are congruent to each other<br /> angle BEC = angle CED = angle AED<br /> and BEC + CED + AED =180 degree<br /> 3(AED) = 180 degree<br /> AED = 60 degree<br />IN Tri CEA , angle CEA = 2(AED) angle = 120 degree. Since it Is isosceles <br /> angle ECA = angle EAC = 60 degree<br />IN Tri CDA , angle DCA = angle DAC = 60 - alpha , and angle CDA = 120 + 2(alpha)<br />AND FINALLY <br /> ANGLE CDA = 360 - (ANGLE EDC + ANGLE EDA)<br /> ANGLE CDA = 360 - (ANGLE EDC + ANGLE EDC) ( EDC = EDA)<br /> ANGLE CDA = 360 - ANGLE 2(EDC)<br /> ANGLE CDA = 360 - 2X (EDC = X)<br /> 120 + 2(ALPHA) = 360 - 2X (ANGLE CDA = 120 + 2(ALPHA))<br /> 2X - 360 = - 2(ALPHA) - 120<br /> 2X = 360 - 120 - 2(ALPHA)<br /> = 240 - 2(ALPHA)<br /> = 2( 120 - ALPHA)<br /> THEREFORE X = 120 - ALPHA Anonymoushttps://www.blogger.com/profile/12136684908469824769noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-82400293139959253572012-09-11T02:47:56.491-07:002012-09-11T02:47:56.491-07:00hallo my dear brothers and sisters ...... here is ...hallo my dear brothers and sisters ...... here is a simple solution for this problem<br />lets start<br />join A,C. join B,D.<br />draw an angle bisector of angle DCB.it intersects DB at the point F and<br />also intersects AB , at the point E.<br />now consider triangle DBC<br />CD=CB (given)<br />angle DCF=angle FCB=alpha<br />angle FDC = angle FBC = (180- 2 alpha)/2=90-alpha<br />so angle DFC = angle BFC = 90<br />so angle EFD = angle EFB = 90<br />now consider two triangle DFC and BFC<br />cf is common <br />angle DCF=angle FCB=alpha<br />angle DCF=angle FCB=alpha<br />triangle CDF is congruent with triangle CBF<br />so FD = FB<br /><br />now consider two triangle DEF and EFB<br />EF is common <br />FD = FB<br />angle EFD = angle EFB = 90<br />two triangle DEF and EFB are congruent with each other<br />so angle BEF = angle FED............... (1)<br />consider triangle ADC<br />AD = DC<br />this is an isosceles triangle<br />so angle DAC = angle DCA<br />consider triangle AEC<br />angle EAC = angle EAD + angle DAC<br />angle ECA = angle ECD + angle DCA<br />angle EAD = angle ECD = alpha<br />so angle EAC = angle ECA<br />so triangle AEC is isosceles triangle<br />so AE = EC<br />now consider two triangle AED and CED<br />AE=CE<br />ED is common<br />AD=DC<br />triangle AED and CED are congruent<br />so angle AED = angle FED ..........(2)<br />from (1) and (2)<br />angle BEF = angle FED = angle AED = 180/3 =60<br />so angle EBF = 90 - 60 = 30<br />ANGLE ABC= angle EBF + angle FBC<br />= 30 +90 - alpha = 120 - alphaas it ishttps://www.blogger.com/profile/12504861338709342832noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-37443807669257671852012-05-15T20:12:32.267-07:002012-05-15T20:12:32.267-07:00Join BD.
Let E be the circumcentre of ∆ABD.
∠DEB =...Join BD.<br />Let E be the circumcentre of ∆ABD.<br />∠DEB = 2∠DAB = 2α.<br />∆s DEB, DCB are congruent (Each isosceles, common base BD, <br />equal vertical angles, each 2α)<br />Follows EA = ED = DC = DA, ∆EAD is equilateral, and ∠AED = 60°<br />So ∠ABD = (1/2)∠AED = 30°<br />Observe that ∠DBC = 90° - α<br />Hence x = 30° + (90° - α) = 120° - αPravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-90799771104491121882012-05-15T10:33:33.952-07:002012-05-15T10:33:33.952-07:00Proof:
Join AC
Through D draw a line perpendicular...Proof:<br />Join AC<br />Through D draw a line perpendicular to AC and meet AC at M and meet AB at E.<br />∵ AD = DC<br />∴ EM is the perpendicular bisector of AC<br />∴ AE = EC and ∠AEM = ∠CEM<br />∴ ∠EAD = ∠ECD = α<br />∵ DC = CB<br /> ∠DCE = ∠BCE = α<br /> Δ DCE ≡ Δ BCE<br /> ∠CEM = ∠CEB<br />∵ ∠AEM = ∠CEM = ∠CEB = 60o<br />∵ In ΔBCE, x + α = 120oAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-81586152627313883432012-04-08T02:11:26.935-07:002012-04-08T02:11:26.935-07:00solution is uploaded to the following link
https:...solution is uploaded to the following link<br /><br />https://docs.google.com/open?id=0B6XXCq92fLJJajJOTFRfNTlSTDZGdUlVZEJfSml4UQAnonymoushttps://www.blogger.com/profile/07812499400423119847noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-65644502259911666292011-09-04T13:18:58.107-07:002011-09-04T13:18:58.107-07:00Video solution (Spanish version) by Eder Contreras...Video solution (Spanish version) by Eder Contreras and Cristian Baeza at <a href="http://www.gogeometry.com/geometria/p004_solucion_problema_video_cuadrilatero_angulos_lados_iguales_trazo_auxiliar.htm" rel="nofollow">Geometry problem 4</a>. Thanks Eder.Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-59319231070520006952010-07-02T02:45:46.053-07:002010-07-02T02:45:46.053-07:00http://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-52425911475176389492010-03-13T10:08:31.927-08:002010-03-13T10:08:31.927-08:00Let the bisector of <DCB intersect AB at X. Now...Let the bisector of <DCB intersect AB at X. Now since <XCA = <XAD = a, AD = CD and XD is common, by SAS, AXD is congruent to XCD. Hence <DXC = <DXA. Since X is on the perpendicular bisector of BD it also follows that <DXC = <BXC. Therefore since <DXC = <DXA = <BXC and all three angles are supplementary, <BXC = 60. Hence <XBD = 30. Now, <XBC = <XBD + <DBC = 30 + 90 - a = 120 - a.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-42940238207137587682009-11-17T15:20:18.756-08:002009-11-17T15:20:18.756-08:00http://mate-facil.co.cc/35/
By TiNohttp://mate-facil.co.cc/35/ <br />By TiNoAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-25479594723799391062009-05-11T17:32:00.000-07:002009-05-11T17:32:00.000-07:00This comment has been removed by a blog administrator.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-70149528607130947932009-05-05T07:04:00.000-07:002009-05-05T07:04:00.000-07:00Let point E be a reflection of point D over AB.
Th...Let point E be a reflection of point D over AB.<br />Then AEB and ADB triangles are congruent <br />Angle EAB = angle DAB = alpha => angle EAD = 2 alpha => triangles AED and CBD are congruent (AE = AD = DC = BC) => ED = DB<br />But BE = DB (because AEB and ADB are congruent) => EBD is an equilateral triangle.<br /><br />Angle EBD = 60 => angle ABD = 30 (since ABD = ABE)<br />Angle DBC = 90 - alpha (since DCB is an isosceles triangle) <br /> <br />x = angle ABD + angle DBC = 30 + (90 - alpha) = 120 - alpha.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-30454827982156685592009-02-18T15:30:00.000-08:002009-02-18T15:30:00.000-08:00It is possible to provide a proof using only eleme...It is possible to provide a proof using only elementary geometry!Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-59295918761198053352008-09-16T01:35:00.000-07:002008-09-16T01:35:00.000-07:00Trazamos el segmento BD El triangulo BDC es isosc...Trazamos el segmento BD <BR/><BR/>El triangulo BDC es isosceles => angulo(DBC)=angulo(BDC)=(180-2A)/2= 90-A<BR/><BR/>Tomamos AD=DC=BC=1 <BR/><BR/>En el triangulo BDC ; 1/sen(90-A)=BD/sen(2A) => BD=2 sen(A)<BR/><BR/>En el tringulo ABD ; 1/sen(ABD)=2sen(A) / sen (A) => sen(ABD)= 1/2 => angulo (ABD) =30<BR/><BR/>Angulo(x)=angulo(DBC)+angulo(ABD)=90-A + 30= 120 - AAnonymousnoreply@blogger.com