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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:

Attachment 27448

For question 1, I have written down the following:

AB // DC => Angle BAC + Angle ADC = 180^{o}. α + δ = 90^{o}. => Angle ASD = 90^{o}

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 90^{o} 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!

Re: Bisections of parallelogram form a rectangle

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

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

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

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

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

$\displaystyle \Rightarrow \measuredangle RSP=90^o$

Do you think you can continue from here?

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?

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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).