Then the triangle ABC is equal to the triangle GEF,
because they are on equal bases BC, EF, and between
the same parallels, [I. 38.
But the triangle ABC is equal to the triangle DEF. [Hyp.
Therefore also the triangle DEF is equal to the triangle
GEF, [Axiom 1.
the greater to the less ; which is impossible.
Therefore AD is not parallel to BF,
In the same manner it can be shewn that no other straight line through A but AD parallel to BF; therefore AD is parallel to BF,
Wherefore, equal triangles &c. q.e.d.
PROPOSITION 41. THEOREM.
If a parallelogram and a triangle he on the same base and between the same parallels, the parallelogram, shall be double of the triangle.
Let the parallelogram ABCD and the triangle EBC be
on the same base BC, and between the same parallels
BC, AE : the parallelogram ABCD shall be double of the triangle EBC.
Join AC.
Then the triangle ABC
is equal to the triangle EBC
because they are on the same
base BC, and between the same
parallels BC, AE. [I. 37.
But the parallelogram ABCD
is double of the triangle ABC,
because the diameter AC bisects the parallelogram. [I . 34.
Therefore the parallelogram ABCD is also double of the
triangle EBC.
Wherefore, if a parallelogram &c. q.e.d.
