Proof that by joinning the midpoints of the sides of a rectangle we obtaina rhombus

[Tristan Pouliot]

Hi there,
In order to prove that a rhombus is created, my teacher wants us to prove two things: that vector EF= HG (already done) and that ||EF||= ||FG|| . Can you help me figure out the second condition?

 

[pka]

Consider any convex quadrilateral \(ABCD\) where \(DEFG\) are the midpoints of the sides. (Just as they are in your diagram).
We can prove that \(DEFG\) is a parallelogram.
\(\begin{gathered} \overrightarrow {AB} + \overrightarrow {BC} = \overrightarrow {AC} \hfill \\ \frac{1}{2}\overrightarrow {AB} + \frac{1}{2}\overrightarrow {BC} = \frac{1}{2}\overrightarrow {AC} \hfill \\ \overrightarrow {EF} = \frac{1}{2}\overrightarrow {AC} \hfill \\ \end{gathered} \)
Likewise \(\overrightarrow {HG} = \frac{1}{2}\overrightarrow {AC} \) using \(\overrightarrow {AD} + \overrightarrow {DC} = \overrightarrow {AC}\)
Therefore \(\overrightarrow {EF}\| \overrightarrow {HG} \) and \(\left\| {\overrightarrow {EF} } \right\| = \left\| {\overrightarrow {HG} } \right\|\).
Which proves that \(DEFG\) is a parallogram.

 

[Steven G]

The proof is not that hard.
Do you see that you have 4 congruent triangles? Can you label all the angles in your diagram? There are only three different ones. Once you do this you should see the what to do next.

So please update your triangle by labeling all the angles and then post back.

 

[Deleted member 4993]

Hi there,
In order to prove that a rhombus is created, my teacher wants us to prove two things: that vector EF= HG (already done) and that ||EF||= ||FG|| . Can yoView attachment 21895u help me figure out the second condition?

You say that you have proven

Vector EF= vector FG

Now the 2nd part is what??