Elements of Geometry and Trigonometry

Cor. It is evident likewise, that BC is perpendicular to the plane APD, since BC is at once perpendicular to the two straight lines AD, PD.

Scholium. The two lines AE, BC, afford an instance of two lines which do not meet, because they are not situated in the same plane. The shortest distance between these lines is the straight line PD, which is at once perpendicular to the line AP and to the line BC. The distance PD is the shortest distance between them, because if we join any other two points, such as A and B, we shall have AB>AD, AD>PD; therefore AB>PD.

The two lines AE, CB, though not situated in the same plane, are conceived as forming a right angle with each other, because AE and the line drawn through one of its points parallel to BC would make with each other a right angle. In the same manner, the line AB and the line PD, which represent any two straight lines not situated in the same plane, are supposed to form with each other the same angle, which would be formed by AB and a straight line parallel to PD drawn through one of the points of AB.

PROPOSITION VII. THEOREM.

If one of two parallel lines be perpendicular to a plane, the other will also be perpendicular to the same plane.

Let the lines ED, AP, be parallel; if AP is perpendicular to the plane NM, M then will ED be also perpendicular to it.

Through the parallels AP, DE, pass a plane; its intersection with the plane MN will be PD; in the plane MN draw BC perpendicular to PD, and draw AD.

By the Corollary of the preceding Theorem, BC is perpendicular to the plane APDE; therefore the angle BDE is a right angle; but the angle EDP is also a right angle, since AP is perpendicular to PD, and DE parallel to AP (Book I. Prop. XX. Cor. 1.); therefore the line DE is perpendicular to the two straight lines DP, DB; consequently it is perpendicular to their plane MN (Prop. IV.).

Cor. 1. Conversely, if the straight lines AP, DE, are perpendicular to the same plane MN, they will be parallel; for if they be not so, draw through the point D, a line parallel to AP, this parallel will be perpendicular to the plane MN; therefore

through the same point D more than one perpendicular might be erected to the same plane, which is impossible (Prop. ÏV. Cor. 2.).

Let AB be parallel to CD of the plane NM; then will it be parallel to the plane M

NM.

Cor. 2. Two lines A and B, parallel to a third C, are parallel to each other; for, conceive a plane perpendicular to the line C; the lines A and B, being parallel to C, will be perpendicular to the same plane; therefore, by the preceding Corollary, they will be parallel to each other.

The three lines are supposed not to be in the same plane; otherwise the proposition would be already known (Book I. Prop. XXII.).

PROPOSITION VIII. THEOREM.

If a straight line is parallel to a straight line drawn in a plane, it will be parallel to that plane.

For, if the line AB, which lies in the plane ABDC, could meet the plane MN, this could only be in some point of the line CD, the common intersection of the two planes: but AB cannot meet CD, since they are parallel; hence it will not meet the plane MN; hence it is parallel to that plane (Def. 2.).

PROPOSITION IX. THECREM.

Two planes which are perpendicular to the same straight line, are parallel to each other.

Let the planes NM, QP, be perpendicular to the line AB, then will they be parallel.

For, if they can meet any where, let O be one of their common points, and draw OA, OB; the line AB which is perpendicular to the plane MN, is perpendicular to the straight line OA drawn through its foot in that plane; for the same reason AB is perpendicular to BO; therefore OA and OB are two perpendiculars let fall from the same point Q, upon the same straight line; which is impossible (Book I. Prop. XIV.); therefore the planes MN, PQ, cannot meet each other; consequently they are parallel.

Let the parallel planes NM, QP, be intersected by the plane EH; then will the lines of intersection EF, GH, be parallel.

For, if the lines EF, GH, lying in the same plane, were not parallel, they would meet each other when produced; therefore, the planes MN, PQ, in which those lines lie, would also meet; and hence the planes would not be parallel.

PROPOSITION X. THEOREM.

If a plane cut two parallel planes, the lines of intersection will be parallel.

PROPOSITION XI. THEOREM.

If two planes are parallel, a straight line which is perpendicular to cne, is also perpendicular to the other.

Let MN, PQ, be two parallel planes, and let AB be perpendicular to NM; then will it also be perpendicular to QP.

PROPOSITION XII. THEOREM.

Let MN, PQ, be two parallel planes, and FH, GE, two parallel lines: then will EG=FH

For, through the parallels EG, FH, draw the plane EGHF, intersecting the parallel planes in EF and GH. The intersections EF, GH, are parallel to each other (Prop. X.); so likewise are EG, FH; therefore the figure EGHF is a parallelogram; consequently, EG=FH.

Having drawn any line BC in the plane PQ, through the lines AB and BC, draw a plane ABC, intersecting the plane MN in AD; the intersection AD will be parallel to BC (Prop. X.); but the line AB, being perpendicular to the plane MN, is perpendicular to the straight line AD; therefore also, to its parallel BC (Book I. Prop. XX. Cor. 1.): hence the line AB being perpendicular to any line BC, drawn through its foot in the plane PQ, is consequently perpendicular to that plane (Def. 1.).

The parallels comprehended between two parallel planes are equal.

Cor. Hence it follows, that two parallel planes are every where equidistant: for, suppose EG were perpendicular to the plane PQ; the parallel FH would also be perpendicular to it (Prop. VII.), and the two parallels would likewise be perpendicular to the plane MN (Prop. XI.); and being parallel, they will be equal, as shown by the Proposition.

PROPOSITION XIII. THEOREM.

If two angles, not situated in the same plane, have their sides parallel and lying in the same direction, those angles will be equal and their planes will be parallel.

Let the angles be CAE and DBF.
Make AC BD, AE=

BF; and draw CE, DF, AB, CD, EF. Since AC is equal and parallel to BD, the figure ABDC is a parallelogram; therefore CD is equal and parallel to AB. For a similar reason, EF is equal and parallel to AB; hence also CD is equal and parallel to EF; hence the figure CEFD is a parallelogram, and the side CE is equal and parallel to DF; therefore the triangles CAE, DBF, have their corresponding sides equal; therefore the angle CAE-DBF.

Again, the plane ACE is parallel to the plane BDF. For suppose the plane drawn through the point A, parallel to BDF, were to meet the lines CD, EF, in points different from C and E, for instance in G and H; then, the three lines AB, GD, FH, would be equal (Prop. XII.): but the lines AB, CD, EF, are already known to be equal; hence CD=GD, and FH=EF, which is absurd; hence the plane ACE is parallel to BDF.

Cor. If two parallel planes MN, PQ are met by two other planes CABD, EABF, the angles CAE, DBF, formed by the intersections of the parallel planes will be equal; for, the intersection AC is parallel to BD, and AE to BF (Prop. X.); therefore the angle CAE=DBF.

PROPOSITION XIV. THEOREM.

If three straight lines, not situated in the same plane, are equal and parallel, the opposite triangles formed by joining the extremities of these lines will be equal, and their planes will he parallel

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