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.

**Proof**

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,

Similarly,

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

So

Similarly,

We will rewrite the vector AM:

We can rewrite this as:

But

So

We will now make the substitutions we found earlier:

So M = N

Hence these triangles have the same centroid.