Proposition 35.
Parallelograms which are on the same base and in the same parallels are equal to one another.
Let
ABCD,
EBCF be parallelograms on the same base
BC and in the same parallels
AF,
BC;
I say that
ABCD is equal to the parallelogram
EBCF.
For, since
ABCD is a parallelogram,
AD is equal to BC. [I. 34]
For the same reason also
EF is equal to BC, so that AD is also equal to EF; [C.N. 1] and
DE is common;
therefore the whole AE is equal to the whole DF. [C.N. 2]
But
AB is also equal to
DC; [
I. 34] therefore the two sides
EA,
AB are equal to the two sides
FD,
DC respectively,
and the angle FDC is equal to the angle EAB, the exterior to the interior; [I. 29] therefore the base EB is equal to the base FC, and the triangle EAB will be equal to the triangle FDC. [I. 4]
Let
DGE be subtracted from each; therefore the trapezium
ABGD which remains is equal to the trapezium
EGCF which remains. [
C.N. 3]
Let the triangle
GBC be added to each; therefore the whole parallelogram
ABCD is equal to the whole parallelogram
EBCF. [
C.N. 2]
Therefore etc.
Q. E. D.
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