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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the problem 745 details.
The slution is uploaded to the following link:https://docs.google.com/open?id=0B6XXCq92fLJJNkJFWExzWE1NSW8
The solution is uploaded to the following link:https://docs.google.com/open?id=0B6XXCq92fLJJM0w2aWEyS3lMNEk
http://img715.imageshack.us/img715/7306/problem745.pngSince ∠ACD=90 => AD is a diameterSince AB=BD => BO perpendicular to ADLet BE cut AD at F and r = radius of the circleSince EF//CD => AE/AC=AF/AD => AF=10/7 .rNote that ∠EAF=∠OBF= alphaIn ∆ BOF , tan(alpha)=OF/OB=3/7In ∆CAD, tan(alpha)=x/7=3/7So x=3
B is the midpoint of arc AD and BE is perpendicular to chord AC. Hence by the Broken Chord Theorem, 5 = 2 + x or x = 3If you wish to prove the theorem all you need to do is extend AC to F such that CF =CD=x. Now /_ABD = /_ACD = 2*/_AFD and B being the midpoint of arc AD lies on the perpendicular bisector of AD. Therefore B is the circumcentre of triangle ADF and BE is perpendicular to AF so AE = EF or 5 = 2 + x or x = 3Ajit
http://img18.imageshack.us/img18/6838/p745resuelto.png(look at the picture)Note that ABCD is cyclic, then 7*BD=BD*x+2sqrt(2)*BD*sqrt(2)<=>x=3Greetings
AD is a diameter so < BDA = 45 = < BCA. Hence BE = EC = 2 and so applying Pythagoras to Tr. ABE, AB^2 = 29 from which we have AD^2 = 58. So x^2 = 58 - 7^2 = 9 and thus x = 3Sumith PeirisMoratuwaSri Lanka
Another proof without using Pythagoras...Extend EC to F such that CF = 3. So Tr. ABF is isoceles and hence B is the centre of circle ADF. Hence < AFD = 1/2 of < ABD = 45 and so x = CF = 3Sumith PeirisMoratuwaSri Lanka
Still another method much simpler than both previous proofs...,Find G on AE such that GE = 2. Then prove that Tr.s AGB & BDC are congruent SAA and the result follows. Sumith PeirisMoratuwaSri Lanka
Problem 745From the theorem of broken chord (Archimedes) is arc AB=arc BD then AE=EC+CD.So 5=2+x .Therefore x=3APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE