Geometry in rectangular envelopes

1. GEOMETRY IN RECTANGULAR ENVELOPES

Having worked through the last few pages we have learned that:

A rectangular envelope can only be made from a quadrilateral, whose diagonal lines are perpendicular to each other, by folding it along the lines formed by joining the midpoints on each of its sides.
If the sides of the triangles, which are folded on top of the rectangle, coincide with the diagonal lines of the quadrilateral then it is a rhombus.
An envelope can be made from a square, as a square is really a special kind of rhombus.
We cannot make an envelope from any kind of parallelogram other than a rhombus.

The following window has been designed to illustrate certain geometric properties which will allow us to draw some conclusions about rectangular envelopes.

Move control points A, B, C and D to change the shape to make different quadrilaterals.

In this window quadrilateral ABCD, has diagonals AC and BD which intersect at O and parallelogram M1M2M3M4 where M1, M2, M3 and M4 are the corresponding mid-points with respect to AB, BC, CD and DA

Why is M1M2M3M4 a parallelogram?

We need to prove that side M1M2 is parallel to side M3M4 and that side M1M4 is parallel to M2M3.

The proof:

The diagonal line BD divides quadrilateral ABCD into two triangles ADB and CBD.

Line M1M4 divides sides AD and AB into two equal parts. According to Thales' theorem M1M4 is parallel to BD and therefore triangle ADB is similar to AM4M1, with a ratio of M1M4/BD=1/2.

In the same way, line M3M4 divides sides CD and CB into two equal parts and again, as stated by Thales' theorem, M2M3 is parallel to BD.

As we have shown that M1M4//BD and M2M4//BD, thenM1M4//M2M3.

By following a similar type of reasoning we also find that M1M2//M3M4

Which conditions are satisfied when M1M2M3M4 is a right-angled parallelogram?

If sides M1M4 and M2M3 are perpendicular to sides M1M2 and M3M4 then M1M2M3M4 is a right-angled parallelogram.

As we saw in the section above, sides M1M4 and M2M3 are parallel to diagonal BD and sides M1M2 and M3M4 are parallel to diagonalAC. Consequently, if diagonal AC and BD are perpendicular in the quadrilateral ABCD then M1M2M3M4 is a right-angled parallelogram.

Why can an envelope be made when the diagonal lines which intersect in the quadrilateral ABCD are perpendicular?

Here we are interested in the reflection of points A, B, C and D about the lines M1M4, M1M2, M2M3 and M3M4.

The reflection of vertex A about line M1M4 is A' which shows that:

AA' is perpendicular to M1M4
AH=HA'

Triangle AM1M4 is similar to triangle ABD whose ratio, as we already know, is 1/2. As AA' is the perpendicular height of triangle ABD then A' is the base of this height, which is found on BD, the base of the triangle.

When diagonals AC and BD are perpendicular then A' coincides with the intersection point O and the triangle symmetrical to AM1M4 is OM1M4

Use the same process of reasoning with the other triangles BM1M2, CM2M3 and DM3M4

Vertices A, B, C and D only coincide with point O when the diagonal lines AC and BD are perpendicular. The envelope is made by folding the quadrilateral along the lines M1M2, M2M3, M3M4 and M4M1.

If the diagonals intersect at the central point of the quadrilateral ABCD then the shape is a rhombus.

	REVIEWING GEOMETRY WITH RECTANGULAR ENVELOPES
	Maths Workshop: Maths in an envelope