Monday, October 18, 2010

Problem 530: Cyclic Quadrilateral, Diagonal, Diameter, Perpendicular, Congruence, Math Education

Geometry Problem
Click the figure below to see the complete problem 530 about Cyclic Quadrilateral, Diagonal, Diameter, Perpendicular, Congruence, Math Education.

Complete Problem 530

Level: High School, SAT Prep, College geometry

1. AC meet BD on G, draw OM perpendicular to BD
▲OMG ~ ▲GFC => MF/GM = R/OG
▲OMG ~ ▲AEG => EM/GM = R/OG
=> EM = MF
=> BE = FD

2. Let M and N are the projection of O over BC and AD.
Note 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

3. triangle ABE ~ triangle ACD
(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.

4. Problem 530
Draw 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