prove that the centre of a circle coincides with the centroid of an equilateral triangle inscribed in it

Asked by Tejas Kulkarni | 14th Feb, 2014, 02:06: PM

Expert Answer:

Dear Student,
 
Soln:
 
Given: An equilateral triangle ABC in which D,E and F are the mid-points of the sides BC,CA and AB respectively.
 
To Prove: The centroid and the ircumcenter coinsident.
 
Construction: Draw medians AD, BE and CF.
 
Proof: Let G be the centroid of triangleABC  i.e. the point of intersection of  AD, BE and CF.
In triangleBCE and triangleBFC, we have
angleB=angleC=60o,
BC=BC
 
and,     BF=CE               (since AB=ACrightwards double arrow1 halfAB=1 halfACrightwards double arrowBF=CE)
triangleBCE approximately equal to triangleBFC
BE=CF              ....(i)
Similarly, triangleCAF approximately equal to triangleCAD
rightwards double arrowCF=AD      ........(ii)
From (i) and (ii) we get,
 AB=BE=CF
rightwards double arrow2 over 3AD=2 over 3BE=2 over 3CF
rightwards double arrowGA=GB=GC
rightwards double arrowG is equidistance from vertices
rightwards double arrowG is the circumcenter of  triangleABC
Hence, the centroid and circumcenter are cincident.


Answered by  | 14th Feb, 2014, 06:15: PM

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