Geometry Problem
Click the figure below to see the complete problem 530 about Cyclic Quadrilateral, Diagonal, Diameter, Perpendicular, Congruence, Math Education.
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Complete Problem 530
Level: High School, SAT Prep, College geometry
Monday, October 18, 2010
Problem 530: Cyclic Quadrilateral, Diagonal, Diameter, Perpendicular, Congruence, Math Education
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AC meet BD on G, draw OM perpendicular to BD
ReplyDelete▲OMG ~ ▲GFC => MF/GM = R/OG
▲OMG ~ ▲AEG => EM/GM = R/OG
=> EM = MF
=> BE = FD
Let M and N are the projection of O over BC and AD.
ReplyDeleteNote that M and N are the midpoints of BC and AD and MO=1/2 AB , NO=1/2CD
Triangle DFC similar to triangle OMC ( angle COM= angle CDF)
And DF= CD.OM/CO= ½.CD.AB/CO
Triangle ABF similar to triangle AON ( Angle AON= angle ABE)
And BE=AB.ON/AO=1/2. AB.CD/AO
so DF=BE
Peter Tran
triangle ABE ~ triangle ACD
ReplyDelete(angle D = angle E = 90, angle ABE = angle ACD)
which implies AB/AC = BE/CD or AB. CD = AC. BE
similarly triangle CDF ~ triangle CAB
which implies CD/CA = DF/AB or AB. CD = AC. DF
from the above two relations we get
AB. CD = AC. DF = AC. BE which implies that
BE = DF
Q. E. D.
Problem 530
ReplyDeleteDraw DM perpendicular in AB(the point M belongs to the AB).Si DM intersects AE at N, then the point N is orthocenter the triangle ABD.Then BN is perpendicular in AD or
BN//CD and DN//BC so the BCDN is parallelogram.So BN=CD and <BNE=<FCD (NE//FC ).
Then triangle BNE= triangle FCD.Therefore BE=FD.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE