Basic trig question to proof FA = BG

pka said:
Actually yes that is exactly the most commonly used definition for parallelogram: SEE HERE or HERE .
The operative words are pairs of parallel sides.
Click to expand...

Well, no.

The Wikipedia definition is:

In Euclidean geometry, a parallelogram is a simple (non-self-intersecting) quadrilateral with two pairs of parallel sides. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure. The congruence of opposite sides and opposite angles is a direct consequence of the Euclidean parallel postulate and neither condition can be proven without appealing to the Euclidean parallel postulate or one of its equivalent formulations.​

Congruence of sides is not considered part of the definition.

Similarly, the Wolfram definition is:

A parallelogram is a quadrilateral with opposite sides parallel (and therefore opposite angles equal).​

The "and congruent" part is not part of a minimal definition, which I think was Subhotosh's point.
 
Not sure why I would defend Subotosh but i do realize that he is right.

Just because a definition is common does not mean it is correct. I was brainwashed because I recently saw that definition in the notes of a student whom I tutor. The thing is if you can prove some part of a definition then you should remove that part from the definition.
 
Of course, none of this really matters to the OP. We need to be told what is allowed to be assumed.
 
pka said:
Actually yes that is exactly the most commonly used definition for parallelogram: SEE HERE or HERE .
The operative words are pairs of parallel sides.
Click to expand...
I beg to differ. After ascertaining pair of parallel sides (for parallelogram)within the definition, the congruence of opposite sides (and opposing angles) becomes a property which can be proven (Like in the definition of isosceles triangles - we do not include congruence of base angles).
 
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