ground line, and draw a A perpendicular to xy to meet Ab in A. A is the horizontal trace of the line.

Proof. Because Bb is in the vertical plane of projection and is perpendicular to xy, it is perpendicular to the horizontal plane (Theor. v.); and since it passes through the point b, it must be the projector of the point in which the given line meets the vertical plane; that is, the vertical trace of the line is in Bb; but the vertical trace of the line must be in its vertical projection; it is therefore the point B in which the lines intersect.

Similarly it may be proved that A is the horizontal trace of the given line.

Corollary. If a straight line be parallel to one of the coordinate planes, its projection on the other will be parallel to the ground line.

PROBLEM III. Figs. 17 and 18.

To determine the projections of a straight line which passes through a given point, and is parallel to a given straight line.

Let Ab, a'B be the projections of the given line AB, and pp' the projections of the given point P; it is required to draw the projections of the line which passes through P, and is parallel to AB.

Construction. Through the point p draw pC parallel to Ab, and through p' draw p'c' parallel to a'B; pC and p'c' are the projections required.

Proof. The line through P parallel to AB must have its horizontal and vertical projections respectively parallel to Ab and a B (Theor. XVI.), and passing through p and p', the projections of P; therefore pĈ and p'c' are the projections required.

Corollary. As there can only be one line of which pC and p'c' are the projections, it follows that if the vertical and horizontal projections of two straight lines are respectively parallel to one another, the lines are also parallel.

PROBLEM IV. Figs. 19 and 20.

Given the projections of two points, to determine the distance between them.

Let aa', bb' be the projections of two points A and B ; it is required to find the length of the line AB.

It is obvious from figure 19 that AB is the hypotenuse of a right-angled triangle ABC of which AC is equal to ab, the horizontal projection of the line, and BC is the difference of the heights of A and B. Hence the following:—

Construction. Through a draw a line parallel to xy, meeting bb' in the point c', and make c'a, equal to ba: b'a is the distance required.

Proof. The right-angled triangle ab'c' has the two sides ac' and b'c' equal respectively to AC and BC of the triangle ABC; therefore ab ́ is equal to AB.

Remarks. The triangle a 'b'c' may also be considered as the vertical projection of ABC when the vertical projecting plane of AB is turned about Bb, as an axis, till it comes parallel to the vertical plane of projection, when every line of the figure AabB is equal to its vertical projection (Theor. XV. Cor. 1).

It may be observed that the problem might be solved in a similar manner by constructing a right-angled triangle with its base equal to a'b', and the perpendicular equal to the difference of Aa' and Bb'. If A and B were on opposite sides of the vertical plane, the perpendicular would be the sum of the projectors.

PROBLEM V. Figs. 19, 20, 21, 22.

Given the projections of a point, and a line through the point, to lay off a given distance from the point along the line.

Let bh, b'h' be the projections of the line, and bb' the projections of the point; it is required to find the projections of a point on BH at a given distance from B.

Construction. Take any point I on the line and con