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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
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
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)
BD= BE ( AB=BC ,(1/2)AB=(1/2)BC)
Thus triangle ODB is congruence to OEB by RHS.
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.
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