# Thread: Bisections of parallelogram form a rectangle

1. ## Bisections of parallelogram form a rectangle

Hello everyone,

I am having trouble writing down the exact proofs of these two questions:

1. The four bisections of a parallelogram form a rectangle.
2. If the bisections of quadrileteral ABCD form a rectangle, ABCD is a parallelogram.

First off, here's the analysis figure:

For question 1, I have written down the following:

AB // DC => Angle BAC + Angle ADC = 180o. α + δ = 90o. => Angle ASD = 90o
The same can now be done with Angle CQB. However, Angles R and P are not in triangles, so I'm stuck there.

For question 2, I obviously give the reverse proof, starting from Angle ASD being 90o and then reading backwards from question 1, but I have the same problem with angles P and R as I had with question 1.

If anyone could help me with this, thanks a lot!

2. ## Re: Bisections of parallelogram form a rectangle

$\measuredangle BAD=\measuredangle DAS + \measuredangle SAB = 2 \measuredangle DAS$

$\measuredangle ADC=\measuredangle ADS + \measuredangle SDC = 2 \measuredangle ADS$

$AB || DC \Rightarrow \measuredangle BAD+\measuredangle ADC=180^o \Rightarrow 2 \measuredangle DAS+2 \measuredangle ADS=180^o \Rightarrow$

$\Rightarrow \measuredangle DAS+\measuredangle ADS=90^o$

$\measuredangle DAS+\measuredangle ADS+\measuredangle DSA=180^o \Rightarrow \measuredangle DSA =180^o-90^o=90^o\Rightarrow$

$\Rightarrow \measuredangle RSP=90^o$

Do you think you can continue from here?

3. ## Re: Bisections of parallelogram form a rectangle

Alright thank you very much. When I have proven that angles RSP and BQC are 90 degrees, can I then say that the other 2 angles must be 90 degrees because otherwise a quadrileteral is impossible?

4. ## Re: Bisections of parallelogram form a rectangle

You're welcome!

Uhm, no. But you can easily prove the other two angles are right angles (triangles DPC and ARB).