Hello all, i had a question.
, you: )

Given: an equilateral triangle abc, medians ap, bq, & cr. their point of concurrency is o, which is the centroid of the triangle.
to prove that: centroid o is the circumcentre of the triangle abc.
if we prove that centroid o is the circumcentre of the triangle, then it automatically becomes the centre of the circumcircle.
proof: since ap is median, so p is mid point of bc.
ie, bp = pc.
ab = ac ( as triangle abc is equilateral)
ap=ap ( common side)
hence triangle abp is congruent to acp( by sss congruence criterion)
=> angle apb = angle angleapc ( corresponding parts of congruent triangles)
but their sum = 180°
so, each angle has to be 90°.
that shows that ap is perpendicular bisector of bc.
similarly, pyove that bq & cr are perpendicular bisectors of ac & ab respectively.
so now, the point of concurrency ‘o' of these perpendicular bisectors becomes circumcentre of the triangle. ( as circumcentre is the point of concurrency of 3 perpendicular bisectors of the sides of the triangle). and this centre is also the centre of circum circle.
this way centroid o coincides with circumcentre o
[hence proved]

Ibelieve it's a. 2.2 - x = -10.85 (plz tell me if it's right : ) )