For let a straight line / E F / be set up at right angles to the two straight lines / A B /, / C D /, which cut one another at the point E, from E; I say that / E F / is also at right angles to the plane through / A B /, / C D /.
For let / A E /, / E B /, / C E /, / E D / be cut off equal to one another, and let any straight line / G E H / be drawn across through E, at random; let / A D /, / C B / be joined, and further let / F A /, / F G /, / F D /, / F C /, / F H /, / F B / be joined from the point F taken at random <on / E F / >.
Now, since the two straight lines / A E /, / E D / are equal to the two straight lines / C E /, / E B /, and contain equalangles, [G.40]
therefore the base / A D / is equal to the base / C B /, and the triangle< A E D > will be equal to the triangle < C E B >; [G.24] so that the angle < D A E ) is also equal to the angle < E B C ).
[G.40]But the angle < A E G ) is also equal to the angle < B E H );
therefore < A G E >, < B E H > are two triangles which have two angles equal to two angles respectively, and one side equal to one side, namely that adjacent to the equal angles, that is to say, / A E / to / E B /; therefore they will also have the remaining sides equal to the remaining sides. [G.51]
Therefore / G E / is equal to / E H /, and / A G / to / B H /.
And, since / A E / is equal to / E B /, while / F E / is common and at right angles,
[G.24]therefore the base / F A / is equal to the base / F B /.
For the same reason / F C / is also equal to / F D /.
And, since / A D / is equal to / C B /, and / F A / is also equal to / F B /, the two sides / F A /, / A D / are equal to the two sides / F B /, / B C / respectively; and the base / F D / was proved equal to the base / F C /;
[G.29]therefore the angle < F A D ) is also equal to the angle < F B C ).
And since, again, / A G / was proved equal to / B H /, and further / F A / also equal to / F B /, the two sides / F A /, / A G / are equal to the two sides / F B /, / B H /. And the angle < F A G ) was proved equal to the angle < F B H );
[G.24]therefore the base / F G / is equal to the base / F H /.
Now since, again, / G E / was proved equal to / E H /, and / E F / is common, the two sides / G E /, / E F / are equal to the two sides / H E /, / E F /; and the base / F G / is equal to the base / F H /;
[G.29]therefore the angle < G E F ) is equal to the angle < H E F ).
Therefore each of the angles < G E F ), < H E F ) is right.
Therefore / F E / is at right angles to / G H / drawn at random through E.
Similarly we can prove that / F E / will also make right angles with all the straight lines which meet it and are in the plane of reference.
But a straight line is at right angles to a plane when it makes right angles with all the straight lines which meet it and are in that same plane; [G.505]
therefore / F E / is at right angles to the plane of reference.
But the plane of reference is the plane through the straight lines / A B /, / C D /.
Therefore / F E / is at right angles to the plane through / A B /, / C D /.
