Proposed Problem

Click the figure below to see the complete problem 385.

See also:

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

Subscribe to:
Post Comments (Atom)

http://s43.radikal.ru/i100/0911/e1/52e881bd9a75.jpg

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

Part 1:

ReplyDeleteExtend BE to meet AC at F and BD to meet AC at G.

AD is the perpendicular bisector of tr. ABG so BD = DG and likewise

BE is the perpendicular bisector of tr. CBF so BE = EF.

That means triangle BFG is similar to tr. BDE by SAS (in a 1:2 ratio)

and DE is parallel to AC.

Part 2:

DE = 1/2 FG from the ratio of the triangles

FG = AC - (AF + GC)

From the perpendicular bisectors AB = AG and BC = CF

Putting that together:

DE = 1/2 ( AC - ([AC - BC] + [AC - AB]) = 1/2(b -([b-a] +[b-c])

= 1/2(a + c - b) = 1/2 (a + b + c - 2b) = s - b