第11卷命题 8 · 平行线与平面垂直的传递性

Let AB, CD be two parallel straight lines, and let one of them, AB, be at right angles to the plane of reference; I say that the remaining one, CD, will also be at right angles to the same plane. For let AB, CD meet the plane of reference at the points B, D, and let BD be joined; therefore AB, CD, BD are in one plane. [XI. 7] 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, therefore AB is also at right angles to all the straight lines which meet it and are in the plane of reference; [XI. Def. 3] therefore each of the angles ABD, ABE is right.

设AB、CD为两条平行直线，AB垂直于参考平面，交于点B、D，连接BD，则AB、CD、BD共面。

And, since the straight line BD has fallen on the parallels AB, CD, therefore the angles ABD, CDB are equal to two right angles. [I. 29] But the angle ABD is right; therefore the angle CDB is also right; therefore CD is at right angles to BD. 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 the angle ABD is equal to the angle EDB, for each is right; therefore the base AD is equal to the base BE. And, since AB is equal to DE, and BE to AD, the two sides AB, BE are equal to the two sides ED, DA respectively, and AE is their common base; therefore the angle ABE is equal to the angle EDA.

在参考平面内作DE垂直于BD，取DE等于AB，连接BE、AE、AD。因AB垂直于参考平面，故角ABD和角ABE均为直角。

But the angle ABE is right; therefore the angle EDA is also right; therefore ED is at right angles to AD. But it is also at right angles to DB; therefore ED is also at right angles to the plane through BD, DA. [XI. 4] Therefore ED will also make right angles with all the straight lines which meet it and are in the plane through BD, DA. But DC is in the plane through BD, DA, inasmuch as AB, BD are in the plane through BD, DA, [XI. 2] and DC is also in the plane in which AB, BD are.

由平行线性质，角ABD与角CDB互补，而角ABD为直角，故角CDB为直角，即CD垂直于BD。通过三角形全等证明角EDA为直角，从而ED垂直于AD。

Therefore ED is at right angles to DC, so that CD is also at right angles to DE. But CD is also at right angles to BD. Therefore CD is set up at right angles to the two straight lines DE, DB which cut one another, from the point of section at D; so that CD is also at right angles to the plane through DE, DB. [XI. 4] But the plane through DE, DB is the plane of reference; therefore CD is at right angles to the plane of reference.

由ED同时垂直于BD和AD，得ED垂直于平面BD、DA。因DC在该平面内，故ED垂直于DC，即CD垂直于DE。结合CD垂直于BD，得CD垂直于平面DE、DB，即参考平面。