Online Geometry theorems, problems, solutions, and related topics.
Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 873.
Reflect BD along AD to B'D Since AD is the angle bisector, A-C-B' collinear As DB = DC = DB', D is the center of circle(BCB')So angleBDB' = 2*angleBCA = 96.AngleBDA = AngleB'DA = (1/2)*AngleBDB' = (1/2)*96 = 48 = angleBCAHence B,D,C,A concylic.AngleBAC + angleBDC = 180(5x + 5x) + (48 + x) = 180x = 12
There is another solution, namely x=42/5, which occurs when the reflection of BD is CD.
Could you please give your detail solution?
http://img23.imageshack.us/img23/8498/problem8731.pngLet circumcircle of ABC cut AD at D’We have D’B=D’C ( chords of congruent angles) and DB=DCSo DD’ is a perpendicular bisector of BC or D coincide to D’.To satisfy requirements of the problem, DD’ cannot be a perpendicular bisector of BC So D coincide to D’Quadrilateral ABDC is cyclic so 48+5x+5x+x=180 => x=12
Is there any other way?
Yes there is another wayDraw perpendiculars DM & DN to AB & AC which can easily be proved to be equal. Hence Tr. BMD & Tr. CDN are congruent. So < MBD = < DCN = 6x which shows that ABCD is cyclic. Hence < BCD = 5x and adding the angles in Tr.,ACD which add upto 180 and solving we get x = 12Sumith PeirisMoratuwaSri Lanka
Problem 873The point D is the intersection of the perpendicular bisector BC and bisector <CAB. So by the theorem of South Pole the ABDC is cyclic, then <ABC=x,<DCB=<DAB=5x=<BCD.But <ABD+<ACD=180 or x+5x+48=5x=180 or x=12.APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE