Equal triangles which are on the same base and on the same side are also in the same parallels.
同底同侧的相等三角形，其顶点在同一条与底边平行的直线上。
等积三角形 ABC、DBC 同底 BC、同侧；要证 AD 与 BC 平行。设作 AE 平行 BC 交 BD 于 E，由面积等推 E=D。
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Let ABC, DBC be equal triangles which are on the same base BC and on the same side of it; [I say that they are also in the same parallels.] And [For] let AD be joined; I say that AD is parallel to BC. For, if not, let AE be drawn through the point A parallel to the straight line BC, [I. 31] and let EC be joined.
设同底同侧两个相等三角形的顶点不在同一条平行于底边的直线上。
Therefore the triangle ABC is equal to the triangle EBC; for it is on the same base BC with it and in the same parallels. [I. 37] But ABC is equal to DBC; therefore DBC is also equal to EBC, [C.N. 1] the greater to the less: which is impossible.
过其中一个顶点作平行线，另一顶点若不在其上，则会形成同底但不同平行线间的三角形。
Therefore AE is not parallel to BC. Similarly we can prove that neither is any other straight line except AD; therefore AD is parallel to BC.
由 euclid-elements/book1-prop-037 的反向比较，可推出一个三角形大于另一个。
Therefore etc.
这与已知相等矛盾，所以顶点在同一平行线上。