If a straight line be set up at right angles to two straight lines which cut one another, at their common point of section, it will also be at right angles to the plane through them.
如果一条直线垂直于两条相交直线，且垂足为它们的交点，那么这条直线也垂直于这两条直线所在的平面。
For let a straight line EF be set up at right angles to the two straight lines AB, CD, which cut one another at the point E, from E; I say that EF is also at right angles to the plane through AB, CD. For let AE, EB, CE, ED be cut off equal to one another, and let any straight line GEH be drawn across through E, at random; let AD, CB be joined, and further let FA, FG, FD, FC, FH, FB be joined from the point F taken at random <on EF>. Now, since the two straight lines AE, ED are equal to the two straight lines CE, EB, and contain equal angles, [I. 15] therefore the base AD is equal to the base CB, and the triangle AED will be equal to the triangle CEB; [I. 4] so that the angle DAE is also equal to the angle EBC. But the angle AEG is also equal to the angle BEH; [I. 15] therefore AGE, 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 to EB; therefore they will also have the remaining sides equal to the remaining sides. [I. 26] Therefore GE is equal to EH, and AG to BH.
设直线EF垂直于相交于E的AB和CD，在EF上任取一点F，作等长线段AE、EB、CE、ED，并过E任作直线GEH。
And, since AE is equal to EB, while FE is common and at right angles, therefore the base FA is equal to the base FB. [I. 4] For the same reason FC is also equal to FD. And, since AD is equal to CB, and FA is also equal to FB, the two sides FA, AD are equal to the two sides FB, BC respectively; and the base FD was proved equal to the base FC; therefore the angle FAD is also equal to 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, AG are equal to the two sides FB, BH.
连接AD、CB，以及FA、FG、FD、FC、FH、FB。由AE=ED，CE=EB，且对角相等，得AD=CB，三角形AED全等于CEB，故角DAE等于角EBC。
And the angle FAG was proved equal to the angle FBH; therefore the base FG is equal to the base FH. [I. 4] Now since, again, GE was proved equal to EH, and EF is common, the two sides GE, EF are equal to the two sides HE, EF; and the base FG is equal to the base FH; therefore the angle GEF is equal to the angle HEF. [I. 8] Therefore each of the angles GEF, HEF is right. Therefore FE is at right angles to GH drawn at random through E.
又角AEG等于角BEH，结合AE=EB，得三角形AGE全等于BEH，故GE=EH，AG=BH。由AE=EB，EF公共且垂直，得FA=FB；同理FC=FD。
Similarly we can prove that FE 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; [XI. Def. 3] therefore FE is at right angles to the plane of reference. But the plane of reference is the plane through the straight lines AB, CD. Therefore FE is at right angles to the plane through AB, CD.
由AD=CB，FA=FB，FD=FC，得角FAD等于角FBC。再由AG=BH，FA=FB，角FAG等于角FBH，得FG=FH。于是三角形GEF全等于HEF，故角GEF等于角HEF，均为直角。同理可证EF与平面内任一直线垂直，因此EF垂直于该平面。