Question

bisectors of interior angles of a parallelogram form a rectangle

Answer 1

Suppose ABCD is a parallelogram labelled cyclically. Then, consider the bisectors of angles ABC and BCD, with points E and F, distinct from B and C, on them respectively. Are BE and CF parallel? If they were, then BC would be a traversal across parallel lines, giving us, by cointerior angles: 

angle EBC + angle BCF = 180 degrees ... (1) 

Since BE and CF are bisectors of angles ABC and BCD respectively, we have: 

angle EBC = (1/2) angle ABC ... (2) 
angle BCF = (1/2) angle BCD ... (3) 

Substituting (2) and (3) into (1), we get: 

(1/2) angle ABC + (1/2) angle BCD = 180 degrees 
angle ABC + angle BCD = 360 degrees 

But, since AB is parallel to CD, and BC traverses it, we must have: 

angle ABC + angle BCD = 180 degrees 

which is a contradiction. Thus, BE and CF are not parallel. 

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Since BE and CF are not parallel, they must intersect at some point G. I claim that angle BGC is a right angle. Again, since AB is parallel to CD, we have: 

angle ABC + angle BCD = 180 degrees 
(1/2) angle ABC + (1/2) angle BCD = 90 degrees 
angle CBG + angle GBC = 90 degrees ... (remember that these are bisectors) 

By the angle sum of a triangle: 

angle CBG + angle GBC + angle BGC = 180 degrees 
90 degrees + angle BGC = 180 degrees 
angle BGC = 90 degrees 

as required.