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

Labels:
arc,
equilateral,
semicircle,
triangle

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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