Then, because the two sides AE, ED are equal to the two sides BE,EC, each to each, [Construction.
and that they contain equal angles AED, BEC; [I. 15.
therefore the base AD is equal to the base BC, and the angle DAE is equal to the angle EBC. [I. 4.
And the angle AEG is equal to the angle BEH; [I. 15.
therefore the triangles AEG, BEH have two angles of the one equal to two angles of the other, each to each;
and the sides EA, EB adjacent to the equal angles are equal to one another; [Construction.
therefore EG is equal to EH, and AG is equal to BH. [I. 26.
And because EA is equal to EB, [Construction.
and EF is common and at right angles to them, [Hypothesis.
therefore the base AF is equal to the base BF. [I. 4.
For the same reason CF is equal to DF.
And since it has been shewn that the two sides DA, AF are equal to the two sides CB, BF, each to each,
and that the base DF is equal to the base CF;
therefore the angle DAF is equal to the angle CBF. [I. 8.
Again, since it has been shewn that the two sides FA, AG are equal to the two sides FB, BH, each to each,
and that the angle FAG is equal to the angle FBH;
therefore the base FG is equal to the base FH. [I. 4.
Lastly, since it has been shewn that GE is equal to HE, and EF is common to the two triangles FEG, FEH;
and the base FG has been shewn equal to the base FH;
therefore the angle FEG is equal to the angle FEH. [I. 8.
Therefore each of these angles is a right angle. [I, Defn. 10.
In like manner it may be shewn that EF makes right angles with every straight line which meets it in the plane passing through AB, CD.
Therefore EF is at right angles to the plane in which are AB, CD. [XI. Definition 3.
Wherefore, if a straight line &c. q.e.d.
