In an equilateral triangle, prove that the centroid & the circumcentre of the triangle coincide.

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See the attached figure.

(ABC are triangle vertices, DEF are midpoints on triagle sides and O is the intersection of medians)

In triangles ABD and ACD

BD= CD (by median definition)

AD= AD (common segment)

angles ABC= ACB (equlateral triangle)

results triangles ABD=ACD

results angle BAD=CAD (=60/2=30 degree)

By smilarity...

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See the attached figure.

(ABC are triangle vertices, DEF are midpoints on triagle sides and O is the intersection of medians)

In triangles ABD and ACD

BD= CD (by median definition)

AD= AD (common segment)

angles ABC= ACB (equlateral triangle)

results triangles ABD=ACD

results angle BAD=CAD (=60/2=30 degree)

By smilarity all angles BAD=CAD=CBE=BCF=ACF=BCF(=30 deg)

We know angle ABC=ACD=BAC (=60 deg) because the triangle is eqilateral

Results angles ADC=ADB=BEA=BEC=CFA=CFB (=90 deg)

Now the triagle is equilateral it means all medians are equal

AD=BE=CF

Now we demonstrate triangles OBD=OCD

OD=OD (common degment)

CD=BD (by definition of median)

angle ODC=ODB (=90 deg)

Results OB=OC and by similarity

OB=OC=OE=OA=OF

It means circumcenter = centroid (qvod erat demonstratum)

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