Find GH perpendicular to AB, CD (Prop. IV.); make GK and GK' each equal to HP; and join PK, PK'. Each of these lines is equally inclined to AB and CD.

For, since HG, HP are equal to HG, GK, and the included angles GHP, PGK are right angles (Constr.), the base HK is equal to the base PG.

Also, since the two sides HP, HK are equal to the two KG, GP, and the

A

Κ'

base KP common, the angles KPH, GKP are equal. And these are the angles stated in the enunciation to be equal.

In the same manner it may be shown that K'P makes equal angles with AB and CD.

PROPOSITION XI.

To construct a trihedral angle equal to a given one.

This problem branches out into several cases according to the essential data by which the pattern trihedral is given. In all cases, however, it is possible, in virtue of the theorems already given, to completely construct the trihedral angle or its supplementary one from adequate data.

Suppose this done, we may copy this trihedral angle by taking the three most convenient of its six parts (three plane angles and three dihedral angles) as one data.

The student should be required to discuss the several cases in order.

SECTION III.

CONSTRUCTIONS RELATING TO THE PLANE, LINE, AND POINT.

PROPOSITION I.

The projections of a point on two of the coordinate planes are given to find its projection on the third.

Eid.

Let A be the given point, and YX, XZ, ZY the three planes of projection; there are given a,, a2 the projections of A on the planes XY, XZ to find a, on YZ.

Now in the eidograph, the three lines Aa, Aa, Aa, are perpendicular to the three planes; and hence the planes through them two and two are parallel to the coordinate planes and the parallel edges are equal, that is,

A

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Orth. Elevation and Profile.

Now if we make the plane ZY revolve about OZ till it coincides with XZ, behind OZ; then the line OY will coincide with OX, and aα, with a1a', the points a, a, describing quadrants during the revolution.

Whence from O in the orthograph as centre, describe the quadrant aa', and draw aga's perpendicular to OX, meeting the line a1a in a': this is the orthograph with respect to the elevation and profile.

3

If with YZ and XY, we proceed in a similar manner, we shall get the orthograph of the point in respect of the plan and profile.

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Orth. Plan and Profile.

PROPOSITION II.

The orthograph of a line on two planes is given to construct the orthograph of either of them with the third plane.

It is evident that since a line is given when two points in it are given, that if we take the projections of any two points in it, and employ the preceding proposition, we shall only have to draw the line through the two projections upon the third plane to resolve the problem.

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These orthographs are representative of the elevation and section, and of the plan and section respectively.

SCHOLIUM. 1. When the line is given under some particular circumstances, the construction is simplified. Three cases deserve notice from their frequent occurrence in practice.

1. When the traces of the line are given.

屋

DEE

Let a1, B2 be the traces: draw as in the figure: then the dark line is evidently the projection on the third plane.

2. When the line itself is the section of any plane whatever with a profile plane; and that profile plane can be taken as the coordinate profile plane.

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This is obviously only the particular case of the preceding, in which a, coincides with a and B, with B.

3. When the line is parallel to one of the coordinate planes.

SCHOLIUM 2. It will be seen hereafter that the same process can be opplied to any figures whatever. In fact it will be obvious that we can always construct by points as we have here done the line aß.

PROPOSITION III.

Given the traces of a line to find its projections; and, conversely, given the projections of a line to find its traces.

(1) Let a1, B2 be the traces of the line, and through a,ß, draw planes a112, a12ß, perpendicular to the planes XY, XZ, cutting the two planes as in the figure of the eidograph.

Then, since XZ and aßß, are perpendicular to XY, the line 8,ß, is perpendicular to XY, and to the line OX in it. (Pls. 11. 18). In like manner aa is perpendicular to OX. Also, the projection on the horizontal plane is aß, and that on the vertical is a. Wherefore these projections can be found without actually describing the projecting planes; since from the positions of a1, B, and the axis OX, we can find B, and α.

Again, had the vertical plane been turned to the horizontal as in the orthograph, and the points a1, B2 been given, the same processes might be performed upon the united plane that were separately performed upon the two planes in eidograph. Whence it follows, that if

we:

Draw BB, and aα, perpendicular to OX, and join a1ß1, aß2, these will respectively be the horizontal and vertical projections which were required.

(2) The converse follows from the same considerations, viz. :—

Draw the projections ab1, ab2 to meet the axis in B1, a; and then perpendicular to the axis, draw aα, to meet a,b, in a1, and 6,8, to meet ab, in ß: then a1, В are the traces of the line whose projections are 4qb, asbg.

SCHOLIUM. In the preceding figure, the traces are given for the line lying in the first region; but the process is the same for all the others. COR. 1. If the projections of two points in the line be given, the same things may be found, viz., the traces and the projections of the line passing through them: for the projections of the line will pass through the projections of the points.

The projections being found by drawing these lines on the coordinate planes; then the traces of the lines will be found as in the problem itself.

COR. 2. Hence, also, if a point and a line be given by means of their projections, the projections and traces of a line through the given point parallel to the given line can be found.

Thus, let a (or a,a,) be the given point, and be (or be1, b,c) the

given line; then lines through a, and a parallel to be, and be, will be the projections of a line through a parallel to bc.

For, when lines are parallel their projections on any plane will be parallel, and the converse: but in the present case, the projections are, by construction, parallel; and hence they are the projections of parallel lines.

PROPOSITION IV.

Given the projections of two points in a line, to construct a line equal to the distance of those points.

In the eidograph, let AB be the segment of the line a12, a line equal to which is to be constructed in orthograph, having given the projections a,a,, b1b2.

Let a BB, be the plane which projects the line horizontally, and Bb1, Aa, the lines which project the points A, B. Draw AC parallel to a, B, meeting Bb, in C.

Then snce a BB, is a right angle, and AC is parallel to a11, ACB is a right angle, and therefore given. Also, since Aa, is parallel to