Saturday, July 25, 2009

Problem 326. Equilateral triangle, Semicircle, Equal arcs

Proposed Problem
Click the figure below to see the complete problem 326 about Equilateral triangle, Semicircle, Equal arcs.

 Problem 326. Equilateral triangle, Semicircle, Equal arcs.
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Complete Problem 326
Level: High School, SAT Prep, College geometry


  1. let O be the centre of semicircle drawn of side BC of equilateral ABC whose side length is s. AB=BC=CA=s and BO=CO=s/2and OD=OE=OB=OC=s/2 given arcBD=arcDE=arcCE so BD=DE=CE we know that equal chords makes equal angles at the centre so angleBOD=angleDOE=angleEOC=angleBOC/3=60. consider triangleBOD, angleBOD=60,BO=OD so angleOBD=angleBDO=60hence triangle OBDis a equilateral triangle so BD=DO=OB=s/2. consider triangle ABD, angleABF=angleFBD=60 so BF is angular bisector of angleABD hence BF=2AB.BD/AB+BD cos 60 = s/3 . similarlly from triangleACE, CG=S/3 Hence BF=CG=s/3. FG=BC-BF-CG=s-s/3-s/3=s/3. so BF=FG=GC=s/3. HENCE AD,AE DIVIDE BC IN EQUAL PARTS.

  2. Angle ABC = Angle BOD = 60
    OD || AB
    triangle ABF ~ triangle DOF
    AB = BC = 2OB = 2OD
    So BF = 2OF = (2/3)OB = (1/3)BC

  3. harvey.littleman@sciences.heptic.frJune 1, 2021 at 2:01 AM

    See the drawing

    Define 0 middle of BC. O is the center of semicircle OD diameter BC
    Define a=OB=OD=OC => AB=AC=BC=2a
    Arc BD=arc DE=arc EC => ∠BOD=∠DOE=∠EOC=Π/3

    OB=OD=a and ∠BOD=Π/3
    =>BD=a, ΔBOD is equilateral and BD//AC
    In the same way, DE=a, ΔDOE is equilateral and DE//BC
    and EC=a, ΔEOC is equilateral and EC//AB

    Define H the intersection of lines AB and DE
    Define I the intersection of lines AC and DE
    ∠ABC=Π/3 and ∠OBD=Π/3 => ∠HBD=Π/3
    ∠ABC=Π/3 and BC//DE => ∠BHD=Π/3
    ∠HBD=Π/3 and ∠BHD=Π/3 => ΔBHD is equilateral
    In the same way, ΔCIE is equilateral

    ΔAIE is similar to ΔACG => AI/IE=AC/CG =>3=2a/CG =>CG=2a/3
    ΔAHD is similar to ΔABF => AH/HD=AB/BF =>3=2a/BF =>BF=2a/3
    BC= 2a=BF+FG+GC=2a/3+FG+2a/3 => FG=2a/3
    Therefore BF=FG=GC