See complete Problem 133
Triangle, Angle Bisectors, Collinear Points. Level: High School, SAT Prep, College geometry
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Wednesday, July 2, 2008
Elearn Geometry Problem 133
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AE is an interior bisector of angle A
ReplyDeleteBE/EC = AB/AC
CD is an interior bisector of angle C
AD/DB = AC/BC
AE is an exterior bisector of angle B
FA/FC = AB/BC
therefore
BD/DA * AF/FC * CE/EB = BC/AC * AB/BC * AC/BC = 1
therefore
D, E, F are collinear points
Magdy Essafty
- P is on AC such that BP bisects <B.
ReplyDelete- It is known that P is the harmonic conjugate of F w/r AC.
In conclusion, F-E-D must be colinear.
Considering Pappu's hexagon Theorem A,D,B and A,C,F are two sets of collinear points. As B,E,C are collinear D,E,F should be collinear to form the pappus line AE.
ReplyDelete