Prove that in an equilateral triangle centroid and circumcentre coincide
Answers
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]
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An equilateral triangle ABC in which D, E and F are the mid - points of sides BC, CA and AB respectively.
The centroid and circum center are coincident.
Draw medians AD, BE and CF.
Let G be the centroid of ∆ABC i.e. the point of intersection of AD , BE and CF. In triangles BEC and BFC. we have,
∠B = ∠C
BC = BC
BF = CE
[ ∵ AB = AC → ½AB = ½AC → BF = CE ]
∆BEC ≅ ∆BFC
⇒BE = CF ........................(I)
Similarly, ∆CAF ≅ ∆CAD
⇒ CF = AD......................(2)
From (I) and (2) , we get
AD = BE = CF
⇒⅔AD = ⅔BE = ⅔CF
⇒GA = GB = GC
⇒G is equidistant from the vertices
⇒G is the circum center of ∆ABC
Hence, the centroid and circum center are coincident.
