|If a straight line is set up at right angles to two straight lines which cut one another at their common point of section, then it is also at right angles to the plane passing through them.|
For let a straight line EF be set up at right angles to the two straight lines AB and CD at E, the point at which the lines cut one another.
I say that EF is also at right angles to the plane passing through AB and CD.
|Cut off AE, EB, CE, and ED equal to one another. Draw any straight line GEH across through E at random. Join AD and CB, and join FA, FG, FD, FC, FH, and FB from a point F taken at random on EF.||XI.2|
|Now, since the two straight lines AE and ED equal the two straight lines CE and EB and contain equal angles, therefore the base AD equals the base CB, and the triangle AED equals the triangle CEB, so that the angle DAE equals the angle EBC.||I.15
|But the angle AEG also equals the angle BEH, therefore AGE and BEH 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, AE equals EB. Therefore they also have the remaining sides equals to the remaining sides, that is, GE equals EH, and AG equals BH.||I.15
|And, since AE equals EB, while FE is common and at right angles, therefore the base FA equals the base FB.||I.4|
|For the same reason, FC equals FD.|
|And, since AD equals CB, and FA also equals FB, the two sides FA and AD equal the two sides FB and BC respectively, and the base FD was proved equal to the the base FC, therefore the angle FAD also equals the angle FBC.||I.8|
|And since, again, AG was proved equal to BH, and further, FA also equal to FB, the two sides FA and AG equal the two sides FB and BH, and the angle FAG was proved equal to the angle FBH, therefore the base FG equals the base FH.||I.4|
|Again, since GE was proved equal to EH, and EF is common, the two sides GE and EF equal the two sides HE and EF, and the base FG equals the base FH, therefore the angle GEF equals the angle HEF.||I.8|
|Therefore each of the angles GEF and HEF is right.
Therefore FE is at right angles to GH drawn at random through E.
Similarly we can prove that FE also makes 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, therefore FE is at right angles to the plane of reference.||XI.Def.3|
|But the plane of reference is the plane through the straight lines AB and CD.
Therefore FE is at right angles to the plane through AB and CD.
|Therefore if a straight line is set up at right angles to two straight lines which cut one another at their common point of section, then it is also at right angles to the plane passing through them.|
|Q. E. D.|
After the preceding three dubious proofs, this one is a relief. It is a little long, but it is clear.
Near the beginning of the proof, proposition XI.2 is needed to conclude that the two lines AB and CD determine a plane. The line GH is to be any line that passes through E and lies in that plane. Then, by XI.2 again, the lines AD and BC lie in this plane.
Next proposition: XI.5
© 1997, 2002