tag:blogger.com,1999:blog-6933544261975483399.post5345501059869818956..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 12: Triangle, Cevian, Angles, CongruenceAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger10125tag:blogger.com,1999:blog-6933544261975483399.post-77545744215329623862022-10-30T02:17:29.190-07:002022-10-30T02:17:29.190-07:00A better solution after 1.5 years of thinking lol
...A better solution after 1.5 years of thinking lol<br /><ABD=115-x<br />By considering triangle ABD with sine law<br />sin(115-x)/AD=sinx/BD<br />AD/BD=sin(115-x)/sinx--------(1)<br />By considering triangle BCD with sine law<br />sin115/BC=sin50/BD<br />BC/BD=sin115/sin50---------(2)<br />Equating (1) & (2)<br />sin(115-x)/sinx=sin115/sin50<br />sin115sinx=sin(115-x)sin50<br />cos(115-x)-cos(115+x)=cos(65-x)-cos(165-x)<br />cos(115-x)+cos(65-x)=cos(65-x)-cos(165-x)<br />cos(115-x)=-cos(165-x)<br />cos(115-x)+cos(165-x)=0<br />2cos(140-x)cos50=0<br />cos(140-x)=0<br />140-x=90<br />x=50Marcohttps://www.blogger.com/profile/04632526355171968456noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-77904945484027856442021-02-16T12:10:20.963-08:002021-02-16T12:10:20.963-08:00E on AC such that BC=CE. So AE=CD and m(AEB)=m(CDB...E on AC such that BC=CE. So AE=CD and m(AEB)=m(CDB) and BE=CD => AEB is congruent to CDB => x=50 degrees. <br /><br />Additionally E is the power of a point wrt circumcircle of BDC. Sailendra Thttps://www.blogger.com/profile/12056621729673423024noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-50861182038470152772021-02-10T22:12:26.620-08:002021-02-10T22:12:26.620-08:00Consider triangle ABC
sinx/BC=sin50/AB
BC=ABsinx/s...Consider triangle ABC<br />sinx/BC=sin50/AB<br />BC=ABsinx/sin50<br /><br />Consider triangle ABD<br />sin65/AB=sin(115-x)/AD<br />AD=ABsin(115-x)/sin65<br /><br />BC=AD<br />sinx/sin50=sin(115-x)/sin65<br />sin65/sin50/sin(115-x)/sinx<br />It is easy to see that x=50Marcohttps://www.blogger.com/profile/04632526355171968456noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-69626615859097478762020-06-12T05:26:42.014-07:002020-06-12T05:26:42.014-07:00This solution is clearer when illustrated, so grab...This solution is clearer when illustrated, so grab a pen and a piece of paper.<br /><br />Angle BDC = 180° - 50° - 15° = 115°<br />Angle ADB = 180° - 115° = 65° <br /><br />Add point E to the figure and connect it to points A, B and E such that a duplicate of triangle BCD is formed with angle ADE = 15°. Now, we shall prove that point E sits on line AB.<br /><br />Focus on triangle BDE. It is isosceles because line DE = line BD. Angle BDE = 65° - 15° = 50°. Angle BED = (180° - 50°)/2 = 65°. Angle AED = angle BDC = 115° because triangles ADE and BDC are identical. <br />Angle AED + angle BED = 180°, which is the angle measure of a straight line. Therefore, points A, E and B are collinear.<br /><br />Since points A, E and B are collinear, angle BAC = angle EAC <br />So, angle BAC = angle BCD because triangles ADE and BCD are congruent.<br />Therefore, angle BAC = 50°<br /><br />Angle x = 50°Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-53032066164820909372016-10-15T08:58:37.406-07:002016-10-15T08:58:37.406-07:00https://www.youtube.com/watch?v=L_AY8lZynqkhttps://www.youtube.com/watch?v=L_AY8lZynqkgeoclidhttps://www.blogger.com/profile/07989522895596673545noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-37984904072578610692015-01-09T00:27:39.788-08:002015-01-09T00:27:39.788-08:00Since I am not able to see all links above, I post...Since I am not able to see all links above, I post my solution, hoping it is not among the previous ones, although it is very simple: Take E - reflection of C in BD, DE=CD and easy angle chase gives <ADE=50, i.e. triangles ADE and BCD are congruent (s.a.s.), hence AE=BD and <DAE=15, therefore triangles ABE and BAD are congruent (s.s.s.) with <BAE=<ABD. From this we get x+15=115-x, or x=50.<br />Best regards,<br />Stan FulgerAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-72169767074223529722012-04-12T09:35:34.632-07:002012-04-12T09:35:34.632-07:00Solution is uploaded to the following link:
https...Solution is uploaded to the following link:<br /><br />https://docs.google.com/open?id=0B6XXCq92fLJJZWplWFJKTGMxUEkAnonymoushttps://www.blogger.com/profile/07812499400423119847noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-25099563218517605662012-03-03T09:36:33.186-08:002012-03-03T09:36:33.186-08:00Solution by cristian to this problem:
http://yout...Solution by cristian to this problem:<br /><br />http://youtu.be/mhh1YO0LLIIEder Contreras Ordeneshttps://www.blogger.com/profile/08915905860162020945noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-37618619867597066352011-08-02T23:44:09.751-07:002011-08-02T23:44:09.751-07:00Denote angle ABD=α=115º-x. Suppose α>65º then A...Denote angle ABD=α=115º-x. Suppose α>65º then AD>AB, and since AD=BC, BC>AB.Now in triangle ABC we get x>50º, that is 115º-α>50º, or α<65º in contradiction to our assumption α>65º. So α=65º and x=50º.davidnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-36885354332195348522009-05-03T23:58:00.000-07:002009-05-03T23:58:00.000-07:00Locate a point E on AC such that BC=CE. Join B to ...Locate a point E on AC such that BC=CE. Join B to E. Now it's easy to see that AE=CD and angle AEB = 115 = angle BDC which makes BE=BD. Hence tringles ABE & CBD are SAS congruent; thus x=50.<br />Ajit: ajitathle@gmail.comAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.com