From: quadratic on
How does one prove that in any convex quadrilateral, the quadrilateral
formed by the midpoints of the sides is a parallelogram?
From: alainverghote on
Good evening,

It is simple algebra:
summits in this order A(xa,ya),B(xb,yb),C(xc,yc),
D(xd,yd);
midpoint E of AB (xa+xb)/2,(ya+yb)/2
... F BC ....
G ,H
You'll get EF=GH ; FG=HE

Alain
From: Achava Nakhash, the Loving Snake on
On May 4, 6:30 am, quadratic <quadra...(a)juno.com> wrote:
> How does one prove that in any convex quadrilateral, the quadrilateral
> formed by the midpoints of the sides is a parallelogram?

Draw a diagonal of the quadrilateral. This diagonal, together with
two of the sides of the quadrilateral, form a triangle. Connecting
the midpoints of those sides yields a line segment parallel to the
diagonal and half its length by very elementary geometry always taught
in high school. Hence the line segment connecting the midpoints of
the triangle formed by this diagonal is also parallel to the diagonal
and half its length. In particular these two midpoint connectors are
parallel to each other. We see that the the other two midpoint
connectors are parallel to each other as well, so the four of them
form a parallelogram. You can even find the area of this
parallelogram from what I have just said.

Regards,
Achava
From: alainverghote on
Bonjour Achava,

Yes, it is a theorem taught in 'colleges'
just before Thales (14 old pupils) ,
Of course the algebric way is longer


Alain
From: Ken Pledger on
In article
<39b1b326-77bc-45d6-842b-5907ff7567ff(a)d19g2000yqf.googlegroups.com>,
quadratic <quadratic(a)juno.com> wrote:

> How does one prove that in any convex quadrilateral, the quadrilateral
> formed by the midpoints of the sides is a parallelogram?


You can probably track down several ways of proving it by
searching for "Varignon's Theorem". Incidentally, the quadrilateral
needn't be convex, and it needn't even be in one plane.

Ken Pledger.