There will be no circles involved today.
I discovered something quite different in flavour:
Extend each side of the triangle ABC in both directions.
D is the point on BC such that AB = AD.
E is the point on AC such that BC = BE.
F is the point on AB such that CA = CF.
P is the point on BC such that AC = AP.
Q is the point on AB such that CB = CQ.
R is the point on AC such that BA = BR.
Triangles PQR and DEF have the same centroid.
Here the centroid is labelled G.
I will prove this result using vectors.
Let’s examine the following configuration first:
I have taken the triangle XYZ and put Y’ and Z’ on YZ such that XY = XY’ and XZ = XZ’.
Construct K, the foot of the perpendicular from X to YZ.
As XZ’Z and XYY’ are isosceles, K is the midpoint of Z’Z and of YY’.
So KZ’ = KZ and KY = KY’
So YZ’ = KY – KZ’ = KY’ – KZ = Y’Z
So YZ’ = Y’Z
Now we will apply this to our original diagram:
BP = CD; in fact, as they lie along the same line,
Now suppose the centroid of DEF is M and the centroid of PQR is N.
We will show that M and N are the same point by proving that
To do this we will use the fact that the centroid divides each median in the ratio 1:2.
The vector from F to the midpoint of DE is
We will rewrite the vector AM:
We can rewrite this as:
We will now make the substitutions we found earlier:
So M = N
Hence these triangles have the same centroid.