# Isosceles triangles, vectors and centroids

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,
$\overrightarrow{BP}=\overrightarrow{CD}$
Similarly,
$\overrightarrow{AE}=\overrightarrow{CR}$
$\overrightarrow{AQ}=\overrightarrow{BF}$
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
$\overrightarrow{AM}=\overrightarrow{AN}$
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
$\overrightarrow{FD}+\frac{1}{2}\overrightarrow{DE}$
So
$\overrightarrow{FM} = \frac{2}{3}(\overrightarrow{FD}+\frac{1}{2}\overrightarrow{DE})$
$\overrightarrow{AM} = \overrightarrow{AF}+\frac{2}{3}(\overrightarrow{FD}+\frac{1}{2}\overrightarrow{DE})=\overrightarrow{AF}+\frac{2}{3}\overrightarrow{FD}+\frac{1}{3}\overrightarrow{DE}$
Similarly,
$\overrightarrow{AN}=\overrightarrow{AQ}+\frac{2}{3}(\overrightarrow{QP}+\frac{1}{2}\overrightarrow{PR})=\overrightarrow{AQ}+\frac{2}{3}\overrightarrow{QP}+\frac{1}{3}\overrightarrow{PR}$
We will rewrite the vector AM:
$\overrightarrow{AM} = \overrightarrow{AB}+\overrightarrow{BF}+\frac{2}{3}(\overrightarrow{FB}+\overrightarrow{BC}+\overrightarrow{CD})+\frac{1}{3}(\overrightarrow{DC}+\overrightarrow{CA}+\overrightarrow{AE})$
$\overrightarrow{AM} = \overrightarrow{AB}+\overrightarrow{BF}-\frac{2}{3}\overrightarrow{BF}+\frac{2}{3}\overrightarrow{BC}+\frac{2}{3}\overrightarrow{CD}-\frac{1}{3}\overrightarrow{CD}+\frac{1}{3}\overrightarrow{CA}+\frac{1}{3}\overrightarrow{AE}$
$\overrightarrow{AM} = \overrightarrow{AB}+\frac{1}{3}\overrightarrow{BF}+\frac{2}{3}\overrightarrow{BC}+\frac{1}{3}\overrightarrow{CD}+\frac{1}{3}\overrightarrow{CA}+\frac{1}{3}\overrightarrow{AE}$
We can rewrite this as:
$\overrightarrow{AM} = \frac{2}{3}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{BF}+\frac{1}{3}\overrightarrow{BC}+\frac{1}{3}\overrightarrow{BC}+\frac{1}{3}\overrightarrow{CD}+\frac{1}{3}\overrightarrow{CA}+\frac{1}{3}\overrightarrow{AE}$
But
$\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=0$
$\frac{1}{3}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{BC}+\frac{1}{3}\overrightarrow{CA}=0$
So
$\overrightarrow{AM} = \frac{2}{3}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{BF}+\frac{1}{3}\overrightarrow{BC}+\frac{1}{3}\overrightarrow{CD}+\frac{1}{3}\overrightarrow{AE}$
We will now make the substitutions we found earlier:
$\overrightarrow{AM} = \frac{2}{3}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AQ}+\frac{1}{3}\overrightarrow{BC}+\frac{1}{3}\overrightarrow{BP}+\frac{1}{3}\overrightarrow{CR}$
$\overrightarrow{AM} = \overrightarrow{AQ}-\frac{2}{3}\overrightarrow{AQ}+\frac{2}{3}\overrightarrow{AB}+\frac{2}{3}\overrightarrow{BP}-\frac{1}{3}\overrightarrow{BP}+\frac{1}{3}\overrightarrow{BC}+\frac{1}{3}\overrightarrow{CR}$
$\overrightarrow{AM} = \overrightarrow{AQ}+\frac{2}{3}\overrightarrow{QA}+\frac{2}{3}\overrightarrow{AB}+\frac{2}{3}\overrightarrow{BP}+\frac{1}{3}\overrightarrow{PB}+\frac{1}{3}\overrightarrow{BC}+\frac{1}{3}\overrightarrow{CR}$
$\overrightarrow{AM} = \overrightarrow{AQ}+\frac{2}{3}(\overrightarrow{QA}+\overrightarrow{AB}+\overrightarrow{BP})+\frac{1}{3}(\overrightarrow{PB}+\overrightarrow{BC}+\overrightarrow{CR})$
$\overrightarrow{AM} = \overrightarrow{AQ}+\frac{2}{3}\overrightarrow{QP}+\frac{1}{3}\overrightarrow{PR}$
$\overrightarrow{AM}=\overrightarrow{AN}$
So M = N
Hence these triangles have the same centroid.