Draw the line _AB_. (Problem V.) Let _ab_ be the line so drawn. Find _V_ and _V′_, the vanishing-points respectively of the lines _AC_ and _BC_. (Problem III.) From _a_ draw _aV_, and from _b_, draw _bV′_, cutting each other in _c_. Then _abc_ is the triangle required. If _AC_ is the line originally given, _ac_ is the line which must be first drawn, and the line _V′b_ must be drawn from _V′_ to _c_ and produced to cut _ab_ in _b_. Similarly, if _BC_ is given, _Vc_ must be drawn to _c_ and produced, and _ab_ from its vanishing-point to _b_, and produced to cut _ac_ in _a_. PROBLEM VII. TO DRAW ANY RECTILINEAR QUADRILATERAL FIGURE, GIVEN IN POSITION AND MAGNITUDE, IN A HORIZONTAL PLANE. [Illustration: Fig. 19.] Let _ABCD_ (Fig. 19.) be the given figure. Join any two of its opposite angles by the line _BC_. Draw first the triangle _ABC_. (Problem VI.) And then, from the base _BC_, the two lines _BD_, _CD_, to their vanishing-points, which will complete the figure. It is unnecessary to give a diagram of the construction, which is merely that of Fig. 18. duplicated; another triangle being drawn on the line _AC_ or _BC_. COROLLARY. It is evident that by this application of Problem VI. any given rectilinear figure whatever in a horizontal plane may be drawn, since any such figure may be divided into a number of triangles, and the triangles then drawn in succession. More convenient methods may, however, be generally found, according to the form of the figure required, by the use of succeeding problems; and for the quadrilateral figure which occurs most frequently in practice, namely, the square, the following construction is more convenient than that used in the present problem. PROBLEM VIII. TO DRAW A SQUARE, GIVEN IN POSITION AND MAGNITUDE, IN A HORIZONTAL PLANE. [Illustration: Fig. 20.] Let _ABCD_, Fig. 20., be the square. As it is given in position and magnitude, the position and magnitude of all its sides are given. Fix the position of the point _A_ in _a_. Find _V_, the vanishing-point of _AB_; and _M_, the dividing-point of _AB_, nearest _S_. Find _V′_, the vanishing-point of _AC_; and _N_, the dividing-point of _AC_, nearest _S_. Draw the measuring-line through _a_, and make _ab′_, _ac′_, each equal to the sight-magnitude of _AB_. (For since _ABCD_ is a square, _AC_ is equal to _AB_.) Draw _aV′_ and _c′N_, cutting each other in _c_. Draw _aV_, and _b′M_, cutting each other in _b_. Then _ac_, _ab_, are the two nearest sides of the square. Now, clearing the figure of superfluous lines, we have _ab_, _ac_, drawn in position, as in Fig. 21. [Illustration: Fig. 21.] And because _ABCD_ is a square, _CD_ (Fig. 20.) is parallel to _AB_. And all parallel lines have the same vanishing-point. (Note to Problem III.) Therefore, _V_ is the vanishing-point of _CD_. Similarly, _V′_ is the vanishing-point of _BD_. Therefore, from _b_ and _c_ (Fig. 22.) draw _bV′_, _cV_, cutting each other in _d_.