A somewhat more complicated form of this instrument may also be made by pupils in manual training, as is shown in this illustration from Bion's great treatise. The principle involved may be taken up in class, even if the instrument is not used. It is evident that, unless the workmanship is unusually good, this form of parallel ruler is not as accurate as the common one illustrated above. The principle is sometimes used in iron gates. [Illustration: PARALLEL RULER OF THE EIGHTEENTH CENTURY N. Bion's "Traite de la construction ... des instrumens de mathematique," The Hague, 1723] THEOREM. _Two parallelograms are congruent if two sides and the included angle of the one are equal respectively to two sides and the included angle of the other._ This proposition is discussed in connection with the one that follows. THEOREM. _If three or more parallels intercept equal segments on one transversal, they intercept equal segments on every transversal._ These two propositions are not given in Euclid, although generally required by American syllabi of the present time. The last one is particularly useful in subsequent work. Neither one offers any difficulty, and neither has any interesting history. There are, however, numerous interesting applications to the last one. One that is used in mechanical drawing is here illustrated. [Illustration] If it is desired to divide a line _AB_ into five equal parts, we may take a piece of ruled tracing paper and lay it over the given line so that line 0 passes through _A_, and line 5 through _B_. We may then prick through the paper and thus determine the points on _AB_. Similarly, we may divide _AB_ into any other number of equal parts. Among the applications of these propositions is an interesting one due to the Arab Al-Nair[=i]z[=i] (_ca._ 900 A.D.). The problem is to divide a line into any number of equal parts, and he begins with the case of trisecting _AB_. It may be given as a case of practical drawing even before the problems are reached, particularly if some preliminary work with the compasses and straightedge has been given. Make _BQ_ and _AQ'_ perpendicular to _AB_, and make _BP_ = _PQ_ = _AP'_ = _P'Q'_. Then [triangle]_XYZ_ is congruent to [triangle]_YBP_, and also to [triangle]_XAP'_. Therefore _AX_ = _XY_ = _YB_. In the same way we might continue to produce _BQ_ until it is made up of _n_ - 1 lengths _BP_, and so for _AQ'_, and by properly joining points we could divide _AB_ into _n_ equal parts. In particular, if we join _P_ and _P'_, we bisect the line _AB_. [Illustration] THEOREM. _If two sides of a quadrilateral are equal and parallel, then the other two sides are equal and parallel, and the figure is a parallelogram._