Proposed Problem

Click the figure below to see the complete problem 345 about Equal circles, Tangents, Concurrent lines, Hexagon, Semiperimeter.

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Complete Problem 345

Level: High School, SAT Prep, College geometry

## Sunday, August 23, 2009

### Problem 345. Equal circles, Tangents, Concurrency, Hexagon, Semiperimeter

Labels:
circle,
concurrent,
congruence,
hexagon,
semiperimeter,
tangent

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1. Connect BC3 and CB3

ReplyDeletenote that BC3=CB3 and angle( BC3E)= angle (CB3E)=90 , angle (BEC3)= angle (B3EC)

Triangle BEC3 congruence with triangle B3EC ( case ASA)

so BE=CE

Similarly we also have BH=HC >> HE is the perpendicular bisector of BC

With the same logic DG, FM are the perpendicular bisectors of AB and AC

So 3 perpen. bisectors DG, EH, FM are concurrent at the center of circumcircle of triangle ABC

2. from part 1 we have DA=DB, BE=CE, CF=AF

so s= half of perimeter of polygon ADBECF = DA+EB+CF=DB+EC+FA

Peter Tran