第11卷命题 6 · 同平面垂线互相平行

For let the two straight lines AB, CD be at right angles to the plane of reference; I say that AB is parallel to CD. For let them meet the plane of reference at the points B, D, let the straight line BD be joined, let DE be drawn, in the plane of reference, at right angles to BD, let DE be made equal to AB, and let BE, AE, AD be joined. Now, since AB is at right angles to the plane of reference, it will also make right angles with all the straight lines which meet it and are in the plane of reference.

设直线AB、CD垂直于参考平面，分别交平面于B、D。连接BD，在平面内作DE垂直于BD，且使DE等于AB，连接BE、AE、AD。

[XI. Def. 3] But each of the straight lines BD, BE is in the plane of reference and meets AB; therefore each of the angles ABD, ABE is right. For the same reason each of the angles CDB, CDE is also right. And, since AB is equal to DE, and BD is common, the two sides AB, BD are equal to the two sides ED, DB; and they include right angles; therefore the base AD is equal to the base BE.

因为AB垂直于平面，所以AB与平面内所有相交直线成直角，故角ABD和角ABE为直角。同理，角CDB和角CDE也为直角。

[I. 4] And, since AB is equal to DE, while AD is also equal to BE, the two sides AB, BE are equal to the two sides ED, DA; and AE is their common base; therefore the angle ABE is equal to the angle EDA. [I. 8] But the angle ABE is right; therefore the angle EDA is also right; therefore ED is at right angles to DA. But it is also at right angles to each of the straight lines BD, DC; therefore ED is set up at right angles to the three straight lines BD, DA, DC at their point of meeting; therefore the three straight lines BD, DA, DC are in one plane.

由于AB等于DE，BD公共，且夹角为直角，所以AD等于BE（I.4）。又AB等于DE，AD等于BE，AE公共，故角ABE等于角EDA（I.8）。角ABE为直角，所以角EDA也为直角，即ED垂直于DA。

[XI. 5] But, in whatever plane DB, DA are, in that plane is AB also, for every triangle is in one plane; [XI. 2] therefore the straight lines AB, BD, DC are in one plane. And each of the angles ABD, BDC is right; therefore AB is parallel to CD.

ED同时垂直于BD、DA、DC，因此BD、DA、DC共面（XI.5）。而AB也在该平面内（XI.2），故AB、BD、DC共面。角ABD和角BDC均为直角，所以AB平行于CD（I.28）。