tag:blogger.com,1999:blog-6933544261975483399.post8800805879512043496..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1079: Quadrilateral, Isosceles Triangle, Diagonals, Double AngleAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-83390398599080408452015-02-22T22:58:00.044-08:002015-02-22T22:58:00.044-08:00AD subtends an angle of 2α at B while it subtends ...AD subtends an angle of 2α at B while it subtends an angle of α at D. Moreover, B lies on the perpendicular bisector of AD since AB=BD=a. Hence, B is the circumcentre of Tr. ADC whose circumradius is a= BA=BD=BC.Ajithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-14863766803594573712015-02-08T08:49:56.855-08:002015-02-08T08:49:56.855-08:00Fie BP perpendicular pe AC ,P situat pe AC,daca no... Fie BP perpendicular pe AC ,P situat pe AC,daca notam cu M intersectia dreptelor AC cu BP =>BMDC-patrulater insciptibil, <BAM=<MDB=<ACB=.AB=BC=aion radunoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-23502964850359419802015-02-07T14:45:14.208-08:002015-02-07T14:45:14.208-08:00Draw the circle with center B and radii BA . This ...Draw the circle with center B and radii BA . This circle is the locus of every point which see the segment AD with angle ACD therefor C lies on the circle, therefore CB = BA = aAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-51740497197156598202015-02-07T11:35:42.137-08:002015-02-07T11:35:42.137-08:00Fie BPperpendicular pe AC ,psituat pe AC,daca nota...Fie BPperpendicular pe AC ,psituat pe AC,daca notam cu M intersectia dreptelor AC cu BP =>BMCE-patrulater insciptibil,AB=BC=aion radunoreply@blogger.com