PROPOSITION 36. THEOREM.
Parallelograms on equal bases, and between the same parallels are equal to one another.
Let ABCD, EFGH be parallelograms on equal bases BC, FG and between the same parallels AH, BG: the parallelogram ABCD shall be equal to the parallelogram EFGH.
Join BE, CH.
Then, because BC
is equal to FG, [Hyp.
and FG to EH,[I.34.
BC is equal to
EH; [Axiom 1.
and they are parallels, - [Hypothesis.
and joined towards the same parts by the straight lines
BE, CH.
But straight lines which join the extremities of equal and
parallel straight lines towards the same parts are them-
selves equal and parallel. [I. 33.
Therefore BE, CH are both equal and parallel.
Therefore EBCH is a parallelogram. [Definition.
And it is equal to ABCD, because they are on the same
base BC, and between the same parallels BC,AH. [I. 35.
For the same reason the parallelogram EFGH is equal
to the same EBCH.
Therefore the parallelogram ABCD is equal to the par-
allelogram EFGH. [Axiom l.
Wherefore, parallelograms &c. q.e.d.
PROPOSITION 37. THEOREM.
Triangles on the same base, and between the same parallels, are equal.
Let the triangles ABC,
DBC be on the same base
BC, and between tie same
parallels AD,BC the tri-
angle ABC shall be equal
to the triangle DBC.
Produce AD both ways
to the points E, F ; [Post. 2.
