Joining the midpoints of the consecutive sides of a quadrilateral

  • Thread starter Thread starter lamerali
  • Start date Start date
lamerali
Messages
62
Reaction score
0

Homework Statement



Use Cartesian vectors in two-space to prove that the line segments joining midpoints of the consecutive sides of a quadrilateral form a parallelogram.


Homework Equations





The Attempt at a Solution



the only thing I can think of is; the only way joining the midpoints of the consecutive sides of a quadrilateral will form a parallelogram is if the quadrilateral IS a parallelogram in the first place...I'm pretty sure this is not the correct answer so any guidance is greatly appreciated...THANKS!

lamerali
Calculus and vectors
 
Physics news on Phys.org
The best approach is to draw the picture, including the mid-point line segments. Denote one side of the parallelogram by \mathbf{a} the other by \mathbf{b}. Then express the purportedly parallel line segments in terms of a and b.
 
This is a theorem from Euclidean geometry. If the proposition is given to you for a general quadrilateral, there is no requirement for it to have its opposite sides be parallel. So you're not allowed to argue starting from a parallelogram. (The proposition is surprising because it doesn't seem like it ought to be true, yet it is!)

For setting up the problem, you're allowed to make life easier for yourself by placing one of the vertices of the quadrilateral at the origin (0,0) and lay one of its sides along, say, the x-axis, so a second vertex is (a, 0). From there, the other two vertices will just be at some other two points on the plane, (b,c) and (d,e). Now find the coordinates of the midpoints of the four sides and calculate the components of the vectors linking midpoints of consecutive sides. What do you notice about the components of the vectors on opposite sides of the new quadrilateral they form?
 
Oh man, I misread quadrilateral as parallelogram.
 
Wow, that actually works…
Thank you guys!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top