Friday, November 13, 2009

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

Proposed Problem
Click the figure below to see the complete problem 385.

 Problem 384. Triangle, Angle bisector, Perpendicular, Parallel, Semiperimeter.
See also:
Complete Problem 385
Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

3 comments:

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

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  2. Draw 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

    So < 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

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  3. Part 1:

    Extend 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

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