Proposed Problem
Click the figure below to see the complete problem 326 about Equilateral triangle, Semicircle, Equal arcs.
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Complete Problem 326
Level: High School, SAT Prep, College geometry
Saturday, July 25, 2009
Problem 326. Equilateral triangle, Semicircle, Equal arcs
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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.
ReplyDeleteAngle ABC = Angle BOD = 60
ReplyDeleteOD || AB
triangle ABF ~ triangle DOF
AB = BC = 2OB = 2OD
So BF = 2OF = (2/3)OB = (1/3)BC
See the drawing
ReplyDeleteDefine 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
HD=DE=EI=a
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