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

Labels:
circle,
congruence,
cyclic quadrilateral,
diagonal,
diameter,
perpendicular

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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