Area of Parallelogram 증명

Theorem
The area of a parallelogram equals the product of one of its bases and the associated altitude.

Proof

There are three cases to be analysed: the square, the rectangle and the general parallelogram.

Square

From Area of Square: (ABCD)=a 2 where a is the length of one of the sides of the square.
The altitude of a square is the same as its base.
Hence the result.
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Rectangle

Let ABCD be a rectangle.

Then construct the square with side length (AB+BI) as shown in the figure above.
Note that □CDEF and □BCHI are squares.
Thus □ABCD≅□CHGF .
Since congruent shapes have the same area, (ABCD)=(CHGF) (where (FXYZ) is the area of the plane figure FXYZ ).
Let AB=a and BI=b .
Then the area of the square AIGE is equal to:

				(a+b) 2	=		a 2 +2(ABCD)+b 2
				(a 2 +2ab+b 2 )	=		a 2 +2(ABCD)+b 2
				ab	=		(ABCD)

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Parallelogram

Let ABCD be the parallelogram whose area is being sought.
Let F be the foot of the altitude from C
Also label a point E such that DE is the altitude from D (see figure above).
Extend AB to F .
Then:

				AD	≅		BC
				∠AED	≅		∠BFC
				DE	≅		CF

Thus:

△AED≅△BFC⟹(AED)=(BFC)

So:

(ABCD)=EF⋅FC=AB⋅CF

■ proofwiki