From: quadratic on 4 May 2010 09:30 How does one prove that in any convex quadrilateral, the quadrilateral formed by the midpoints of the sides is a parallelogram?
From: alainverghote on 4 May 2010 09:07 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 4 May 2010 13:31 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 4 May 2010 10:03 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 4 May 2010 17:02
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. |