Question:
- How do I know area of a parallelogram is well defined? That is, how do I know choice of base and relative height does not matter?
My original question had to do with triangles which is more of a headache (three choices of bases and relative heights) but I suppose I should tackle the smaller mountain first.
I am having a difficult time convincing myself that the choice of base does not matter when proving the area of a parallelogram. I know that given a parallelogram of base $b$ and height $h$ its area is $b\cdot h$. The tools I have are the following…
(P1) Every region has an area given by a positive real number.
(P2) Given a square of side length $s$, its area is $s^2$.
(P3) If the interiors of two regions do not intersect, then the area of their union is the sum of their areas.
(P4) Congruent regions have the same area.
(T1) Given a rectangle with base length $b$ and height $h$, its area is given by $b\cdot h.$
For T1, the area of a rectangle, there is no worry about ambiguity for choice of base. For instance in rectangle $FECD$ below, if I choose $b_1=DC$ then $h_1=DF$, where if I choose $b_2=DF$ then $h_2=DC$… Thus $$b_1\cdot h_1 = (DC)(DF)=(DF)(DC)=b_2\cdot h_2$$ hey thats good! No Ambiguity!
But what about for parallelograms?
The idea for the standard proof given base $DC$ and height $CE$ above is given in purple. Virtually forcing rectangle $FECD$ to show up with constructing parallel lines and $HL\cong$ on $\triangle EBC$ and $\triangle FAD$. But I have no idea how to show that the area is the same if we were to allow $DA$ the base and $DG$ the height.
I need to show that $$(EC)(DC)=(AD)(DG)$$I'm at a loss. Any guidance would be amazing. I am trying to avoid coordinate geometry and only prove this with the most basic postulates in euclidean geometry (mostly to avoid circular reasoning).
Best Answer
You have a "proof" that $Area(ECDF) = Area(ABCD)$ that requires you to show first that $\triangle AFD \cong \triangle BEC$, after which (if $\triangle AFD$ does not intersect $\triangle BEC$) you know how to show that $Area(ECDF) = Area(ABCD)$.
But how do you know that $\triangle AFD$ does not intersect $\triangle BEC$?
To complete a proof that requires that $\triangle AFD$ does not intersect $\triangle BEC$, you must first prove that $\triangle AFD$ does not intersect $\triangle BEC$. This is a relatively simple fact, but I think it takes more effort to prove it that it does to just prove that the quadrilaterals have equal area without making that assumption.
The proof I think you need is the proof requested by Why do these parallelograms have the same area. Your requirement is to prove the special case where one of the two parallelograms is a rectangle. But proving the general case also proves the special case. And the general case is no harder to show than your special case.