tag:blogger.com,1999:blog-6933544261975483399.post6120060308187580112..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 830: Quadrilateral, Triangle, Angles, 30 degrees, CongruenceAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-6933544261975483399.post-67757133013365989402016-08-21T09:00:58.056-07:002016-08-21T09:00:58.056-07:00Problem 830
Form the equilateral triangle ADE ...Problem 830<br />Form the equilateral triangle ADE ( the point C is inside the triangle ADE).So EC is bisector<br />Perpendicular the AD and <AEC=<CED=30(AC=CD, AE=ED) is <BDE=<ADE-<ADB=60-34=26.<br />But AD=BD=AE=ED(<ABD=180-73-34=73) then the point D is circumcenter the triangle ABE.<br />So <AEB=(<ADB)/2=34/2=17 and <BAE=(<BDE)/2=26/2=13. Then <EBC=<EBA-<CBA=(180-17-13)-(73+30)=47=<BEC (=17+30=47) therefore BC=CE.But BD=DE so triangleBCD=triangleECD.<br />Therefore x=<BDC=(<BDE)/2=26/2=13.<br />APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE<br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-27184798881646592212015-06-20T02:14:57.178-07:002015-06-20T02:14:57.178-07:00Draw CF and DE perpendicular to AD and BCrespectiv...Draw CF and DE perpendicular to AD and BCrespectively, Then easily FD equal to DE and CFD and CED Tr.s are congruent.Hence < CDE = x + 34. so 2x + 34=60. So x=13.<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-22458511578663068432012-12-18T07:11:29.626-08:002012-12-18T07:11:29.626-08:00Geometric Solution of the problem 830 by Michael T...Geometric Solution of the problem 830 by Michael Tsourakakis from Greece<br />See the image:http://img820.imageshack.us/img820/2807/geogebra.png<br />solution<br />CM is perpendicular of AD. Because CA=CD , CM is mediator of the AD.<br />AB and MC meet of G. So triangle AGD is isosceles .So , angle GDA=73<br />E is point of the GM with AED is equilateral triangle.<br />Angle GDE= angle EAG =73-60=13, therefore , angle EDB=73-34-13=26<br />Additional, angle DAB=73 and angle BDA=34 ,so , angle ABD=73,therefore DA=DB ,so ,DB=DE and triangle BDE is isosceles.Therefore,2 angle DBE=1800-260=1540 so, angle DBE= angle DEB=77.But angle CBD=30, therefore, angle EBC=47 <br />But, angle BEC=77- angle CED=77-30=47.Therefore , triangle BCE is isosceles and CB=CE .But DB=DE .So , DC is mediator of the BE ,therefore, DE is bisector of the angle BDE=26.So,x = angle CDB=13<br /><br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-66680323672590811452012-12-16T04:10:05.953-08:002012-12-16T04:10:05.953-08:00Nice proof, Peter!Nice proof, Peter!Ajithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-1813973242580242502012-12-14T23:17:43.683-08:002012-12-14T23:17:43.683-08:00http://img577.imageshack.us/img577/4899/problem830...http://img577.imageshack.us/img577/4899/problem830.png<br /><br />See drawing per attached sketch<br />Extend BC and from D draw DF ⊥to BC<br />We have ∠BDF=60 and DF=1/2. BD<br />In isosceles triangle ACD draw altitude CE<br />We have DE=1/2.AD=1/2.BD=DF<br />CE and CF are tangent lines from point C to circle centered D with radius=DE=DF ( see Sketch)<br />∠EDF=60+34=94<br />∠FCD=90-1/2. 94=43<br />So in triangle BCD, external angle FCD=30+x => x=43-30=13 <br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com