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 Math Help > Geometry > Polygons and Triangles > Rectangles, Squares > Finsler-Hadwiger Theorem

If two squares ABCD and PQRS are joined at a corner, identifying D with P, then the figure formed by joining the midpoints of AC, CQ, QS, and SA is a square.

### Proof, using the Fundamental Theorem of Directly Similar Figures

First, "directly similar" means similar in the same rotational sense, as opposed to "inversely similar", which means similar in the opposite rotational sense.

The FTDSF says if the lines connecting the corresponding vertices of two directly similar polygons are divided in equal ratios, then the resulting polygon is directly similar to the given two polygons.

By considering the vertices of the first square in order BCDA, and the second in order PQRS, with r=1/2, then the midpoints of BP, CQ, DR, and AS are the vertices of the red figure, which by FTDSF is similar to the other squares.

### Lemma 1: A parallelogram is formed by joining any quadrilateral's side midpoints

We will prove that The figure formed by joining the side midpoints of a quadrilateral is a parallelogram, and the area of the parallelogram is half the area of the quadrilateral.

Let ABCD be a quadrilateral.  The vector cross product of the diagonals gives the magnitude of the area: K = (1/2)|AC×BD|.  (You can see this because the diagonals form four small triangles.  Consider the two pieces of diagonal AC that are cut by diagonal BD.  Let vectors p represent AC, and p1 and p2 are the two pieces of this diagonal, and similarly, let vectors q1+q2=q represent BD.  Then the area vectors of the four triangles are (1/2)(p1×q1), (1/2)(p2xq1), (1/2)(p1×q2), (1/2)(p2xq2).  The sum of these four vectors is (1/2)(p×q).)

Now, let W, X, Y, Z be the midpoints of AB, BC, CD, and DA, respectively.  From an arbitrary point, O, we consider the vector difference

d1=OX-OW.  OX=(1/2)(OC-OB), and OW=(1/2)(OB-OA), so d1=(1/2)(OC-OA) = (1/2)(AC).  Similarly,
d2=OY-OZ.  OY=(1/2)(OC-OD), and OZ=(1/2)(OD-OA), d2=(1/2)(OC-OA) = (1/2)(AC),

which proves WXYZ is a parallelogram (opposite sides have the same vector).  Moreover, the area of the parallelogram is the vector cross product of two adjacent sides one of which is |(1/2)p × (1/2)q| = (1/4)|p×q|, exactly half the area of the quadrilateral ABCD.

### Area of the red square MACMCQMQSMSA is (1/4)( a2 + b2 + 2ab sinθ)

Consider the quadrilateral ACQS.  The blue squares divide it into four regions, consisting of half of square ABCD, half of square PQRS, triangle APS, and triangle CPQ.  If we let a represent the length of the side of ABCD, and b represent the length of the side of PQRS, then these areas, respectively, are (1/2)a2, (1/2)b2, (1/2)ab sinθ, and (1/2)ab sin(π-θ).

Since sin(π-θ)=sin(θ), the area of quadrilateral ACQS is (1/2)(a2+b2)+ab sinθ.  A figure formed by joining the midpoints of the sides of a quadrilateral is a parallelogram whose area is half that of the quadrilateral.  Therefore, by lemma 1, the area of the red square is half the area of quadrilateral ACQS.

The area of the red square MACMCQMQSMSA is (1/4)( a2 + b2 + 2ab sinθ)

### Internet references

Mathworld: Finsler-Hadwiger Theorem , which is a special case of the Fundamental Theorem of Directly Similar Figures

Cut-the-knot: Similar Triangles, which explains the Fundamental Theorem of Directly Similar Figures