Proposed Problem

Click the figure below to see the complete problem 385.

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Complete Problem 385

Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

## Friday, November 13, 2009

### Problem 385. Triangle, Angle bisector, Perpendicular, Parallel, Semiperimeter

Labels:
angle bisector,
parallel,
perpendicular,
semiperimeter,
triangle

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ReplyDeleteDraw FG the altitude of Tr. BFC. Since BF bisects < B, < FBD =90 - A/2 - B/2 = C/2 = < FED since BEFD is cyclic and is also = < ECA. Hence DE is parallel to AC

ReplyDeleteSo < BED = 90 - C/2 = < EBG considering Tr. BEC. Since BEDG is cyclic it follows that the chords subtending these = angles are also =. Hence DE = BG = s-b

Sumith Peiris

Moratuwa

Sri Lanka