See complete Problem 133

Triangle, Angle Bisectors, Collinear Points. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Wednesday, July 2, 2008

### Elearn Geometry Problem 133

Labels:
angle bisector,
collinear,
triangle

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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.

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