*40. Projective theorems and metrical theorems. Linear construction.* This theorem is the connecting link between the general protective theorems which we have been considering so far and the metrical theorems of ordinary geometry. Up to this point we have said nothing about measurements, either of line segments or of angles. Desargues’s theorem and the theory of harmonic elements which depends on it have nothing to do with magnitudes at all. Not until the notion of an infinitely distant point is brought in is any mention made of distances or directions. We have been able to make all of our constructions up to this point by means of the straightedge, or ungraduated ruler. A construction made with such an instrument we shall call a _linear_ construction. It requires merely that we be able to draw the line joining two points or find the point of intersection of two lines. *41. Parallels and mid-points.* It might be thought that drawing a line through a given point parallel to a given line was only a special case of drawing a line joining two points. Indeed, it consists only in drawing a line through the given point and through the "infinitely distant point" on the given line. It must be remembered, however, that the expression "infinitely distant point" must not be taken literally. When we say that two parallel lines meet "at infinity," we really mean that they do not meet at all, and the only reason for using the expression is to avoid tedious statement of exceptions and restrictions to our theorems. We ought therefore to consider the drawing of a line parallel to a given line as a different accomplishment from the drawing of the line joining two given points. It is a remarkable consequence of the last theorem that a parallel to a given line and the mid-point of a given segment are equivalent data. For the construction is reversible, and if we are given the middle point of a given segment, we can construct _linearly_ a line parallel to that segment. Thus, given that _B_ is the middle point of _AC_, we may draw any two lines through _A_, and any line through _B_ cutting them in points _N_ and _L_. Join _N_ and _L_ to _C_ and get the points _K_ and _M_ on the two lines through _A_. Then _KM_ is parallel to _AC_. _The bisection of a given segment and the drawing of a line parallel to the segment are equivalent data when linear construction is used._ *42.* It is not difficult to give a linear construction for the problem to divide a given segment into _n_ equal parts, given only a parallel to the segment. This is simple enough when _n_ is a power of _2_. For any other number, such as _29_, divide any segment on the line parallel to _AC_ into _32_ equal parts, by a repetition of the process just described. Take _29_ of these, and join the first to _A_ and the last to _C_. Let these joining lines meet in _S_. Join _S_ to all the other points. Other problems, of a similar sort, are given at the end of the chapter.