## Quadrilateral Parallelogram

#### Problem

Three points, A, B, and C, are chosen at random such that OABC forms a quadrilateral. The midpoints of each edge, P, Q, R, and S, are joined.

Prove that the quadrilateral PQRS will always be a parallelogram.

#### Solution

We shall prove this result by consideration of vectors.

Let OA = 2** a**, OB = 2

**, and OC = 2**

`b`**.**

`c`So AB = AO + OB = -2** a** + 2

**and CB = CO + OB = -2**

`b`**+ 2**

`c`**.**

`b`The position vectors of P, Q, R, and S respectively will be:

OP = (1/2)OA =

OS = (1/2)OC =

OQ = OA + (1/2)AB = 2

OR = OC + (1/2)CB = 2

`a`OS = (1/2)OC =

`c`OQ = OA + (1/2)AB = 2

**+ -**`a`**+**`a`**=**`b`**+**`a``b`OR = OC + (1/2)CB = 2

**+ -**`c`**+**`c`**=**`b`**+**`b``c`Using these we can obtain vectors for each side of the quadrilateral PQRS:

PQ = PO + OQ = -

SR = SO + OR = -

PS = PO + OS = -

QR = QO + OR = -(

**+**`a``a`+**=**`b``b`SR = SO + OR = -

**+**`c`**+**`b`**=**`c``b`PS = PO + OS = -

**+**`a``c`QR = QO + OR = -(

**+**`a`**) +**`b`**+**`b`**= -**`c`**+**`a``c`As PQ is parallel with and equal to SR, similarly with PS and QR, we prove that the quadrilateral PQRS is a parallelogram.

Was it necessary to consider PS and QR to prove the result?

Problem ID: 250 (02 Dec 2005) Difficulty: 3 Star