PROPOSITION 35. THEOREM.
Parallelograms on the same base, and between the same parallels, are equal to one another.
Let the parallelograms ABCD, EBCF be on the same
base BC, and between the same parallels AF, BC : the paral-
lelogram ABCD shall be equal to the parallelogram EBCF.
If the sides AD, EF of
the parallelograms ABCD,
EBCF, opposite to the base
BC, be terminated at the same
point D, it is plain that each of
the parallelograms is double of
the triangle BDC; [I. 34.
and they are therefore equal to one another. [Axiom 6.
But if the sides AD, EF, opposite to the base BC
of the parallelo-
grams ABCD,
EBCF be not
terminated at
the same point,
then, because
ABCD is a par-
allelogram AD is equal to BC ; [I. 34.
for the same reason EF is, equal to BC ;
therefore AD is equal to EF; [Axiom 1.
therefore the whole, or the remainder, AE is equal to the
whole, or the remainder, DF. [Axioms 2, 3.
And AB is equal to DC ; [I. 34.
therefore the two sides EA, AB are equal to the two sides FD,DC each to each;
and the exterior angle FDC is equal to the interior and
opposite angle EAB ; [I. 29.
therefore the triangle EAB is equal to the triangle
FDC. [I. 4.
Take the triangle FDC from the trapezium ABCF,
and from the same trapezium take the triangle EAB,
and the remainders are equal ; [Axiom 3.
that is, the parallelogram ABCD is equal to the parallelo-gram EBCF.
Wherefore, parallelograms on the same base &c. q.e.d.
