# 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 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)

Let ABC be an equilateral triangle. Also let D,E and F are resp. mid points of side AB ,BC and AC.Join A to E,B to F and C to D and intersect these at G. G is centroid of triangle ABC.Since triangle is an equilateral therefore AE=CD=BF.Also CD will perpendicular to AB ,AE perpendicular to BC and BF perpendicular to AC.

In triangle ODB and OEB.

angle ODB=angle OEB = 90 (Right angle)

OB=OB (common)

BD= BE ( AB=BC ,(1/2)AB=(1/2)BC)

**Thus triangle ODB is congruence to OEB by RHS.**

So

**OD=OE**

similarly we can prove **OD=OF**

**Thus O is incentre of triangle ABC ,which centroid.**

To prove O as circumcentre.

consider triangle OAD and triangle OBD

please add to proof.

AD = BD ( D is mid point of AB)

angle ODA= angle ODB =90 (proved)

OD = OD (common)

Thus triangle OAD is congruence to triangle OBD by SAS.

**Therefore OA = OB**

**Similarly ,we can prove OA = OC**

**Therefore O is circumcentre of triangle ABC.**

**Thus O is centroid ,incentre and circumcentre of triangle ABC.**