Kitap 11
14
For the same reason the angle BAK is also right.
14
Thus, in the triangle ABK, the two angles ABK, BAK are equal to two right angles: which is impossible. [I. 17]
14
Therefore the planes CD, EF will not meet when produced; therefore the planes CD, EF are parallel. [XI. Def. 8]
14
Therefore planes to which the same straight line is at right angles are parallel. Q. E. D.
PROPOSITION 15.
15
If two straight lines meeting one another be parallel to two straight lines meeting one another, not being in the same plane, the planes through them are parallel.
15
For let the two straight lines AB, BC meeting one another be parallel to the two straight lines DE, EF meeting one another, not being in the same plane; I say that the planes produced through AB, BC and DE, EF will not meet one another.
15
For let BG be drawn from the point B perpendicular to the plane through DE, EF [XI. 11], and let it meet the plane at the point G; through G let GH be drawn parallel to ED, and GK parallel to EF. [I. 31]
15
Now, since BG is at right angles to the plane through DE, EF, therefore it will also make right angles with all the straight lines which meet it and are in the plane through DE, EF. [XI. Def. 3]
15
But each of the straight lines GH, GK meets it and is in the plane through DE, EF; therefore each of the angles BGH, BGK is right.
15
And, since BA is parallel to GH, [XI. 9] therefore the angles GBA, BGH are equal to two right angles. [I. 29]
