PROPOSITION 39. THEOREM.
Equal triangles on the same base, and on the same side of it, are between the same parallels.
Let the equal triangles ABC, DBC be on the same base BC, and on the same side of it : they shall be between the same parallels.
Join AD.
AD shall be parallel to BC.
For if it is not, through A draw
AE parallel to BC, meeting BD
at E. [I. 31.
and join EC.
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 triangle ABC is equal to the triangle DBC.[Hyp.
Therefore also the triangle DBC is equal to the triangle EBC, [Axiom 1.
the greater to the less ; which is impossible.
Therefore AE is not parallel to BC.
In the same manner it can be shewn, that no other
straight line through A but AD is parallel to BC;
therefore AD is parallel to BC.
Wherefore, equal triangles &c., q.e.d.
PROPOSITION 40. THEOREM.
Equal triangles, on equal bases, in the same straight line, and on the same side of it, are between the same parallels.
Let the equal triangles ABC, DEF be on equal bases BC, EF, in the same straight line BF, and on the same side of it: they shall be between the same parallels.
Join AD.
AD shall be parallel to BF.
For if it is not, through A
draw AG parallel to BF,
meeting ED at G [I. 31.
and join GF.
