PROPOSITION VII. THEOREM.

If the line AP is perpendicular to the plane MN, any line DE parallel to AP will be perpendicular to the same plane.

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

By the Corollary of the preceding Theorem, BC is perpendicular to the plane

M

P

B

E

APDE; therefore the angle BDE is right: but the angle EDP is right also, since AP is perpendicular to PD, and DE parallel to AP; therefore the line DE is perpendicular to the two straight lines DP, DB; therefore it is perpendicular to their plane MN.

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 in the same plane, which (4. V.) is impossible.

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 understood not to be in the same plane; otherwise the proposition (24. I.) would be already known.

PROPOSITION VIII. THEOREM.

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

rallel; hence it will not meet the plane MN; hence (Def. 2.) it is parallel to that plane.

PROPOSITION IX. THEOREM.

Two planes MN, PQ perpendicular to the same straight line AB, are parallel to each other.

For, if they can meet anywhere, let O be one of their common points, and join 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 perpen

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M

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C

diculars let fall, from the same point O, upon the same straight line; which is impossible: therefore the planes MN, PQ, cannot meet each other; therefore they are parallel.

PROPOSITION X. THEOREM.

The intersections EF, GH, of two parallel planes MN, PQ with a third plane FG, are parallel.

The line AB, which is perpendicular to the plane MN, is also perpendicular to the plane PQ, parallel to MN.

Having drawn any line BC in the plane PQ (see the fig. of Prop. 9.), along the lines AB and BC, extend a plane ABC, intersecting the plane MN in AD; the intersection AD will (10. V.) be parallel to BC; but the line AB, being perpendicular to the plane MN, is perpendicular to the straight line AD; therefore also to its parallel BC: hence the line AB be

ing perpendicular to any line BC drawn through its foot in the plane PQ, is consequently perpendicular to that plane.

PROPOSITION XII. THEOREM.

The parallels EG, FH, comprehended between two parallel planes MN, PQ, are equal.

Through the parallels EG, FH, (see the fig. of Prop. 10.) draw the plane EGHF to meet the parallel planes in EF and GH. The intersections EF, GH, (10. V.) are parallel to each other; so likewise are EG, FH; therefore the figure EGHF is a parallelogram; therefore EG=FH.

Cor. Hence it follows that two parallel planes are every where equidistant; for if EG and FH are perpendicular to the two planes MN, PQ, they will be parallel to each other, (7. V. Cor. 1.); and therefore equal, as has just been shewn.

PROPOSITION XIII. THEOREM.

If two angles CAE, DBF 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.

Make AC-BD, AE=BF; M and join CE, DF, AB, CD, EF. Since AC is equal and parallel to BD, the figure ABDC is a parallelogram (Prop. 31. I.); 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;

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A

B

D

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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 parallel to BDF, drawn through the point A, were to meet the lines CD, EF, in points different from C and E, for instance in G and H; then, (12. V.), the three lines. AB, GD, FH would be equal: 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 CADB, EABF, the angles CAE, DBF, formed by the intersections of the parallel planes will be equal; for (10. V.) the intersection AC is parallel to BD, and AE to BF, therefore the angle CAE=DBF.

PROPOSITION XIV. THEOREM.

If three straight lines AB, CD, EF, not situated in the same plane, are equal and parallel, the triangles ACE, BDF formed by joining the extremities of these straight lines will be equal, and their planes will be parallel.

For, since AB (see the preceding figure), is equal and parallel to CD, the figure ABDC is a parallelogram; hence the side AC is equal and parallel to BD. For a like reason the sides AE, BF are equal and parallel, as also CE, DF; therefore the two triangles ACE, BDF, are equal; and consequently, as in the last Proposition, their planes are parallel.

PROPOSITION XV. THEOREM.

Two straight lines, included between three parallel planes, are cut proportionally.

EB:: AG: GD; in like manner, the intersections AC, GF being parallel, AG: GD :: CF: FD; the ratio AG : GD is the same in both; hence

AE: EB:: CF: FD,

PROPOSITION XVI. THEOREM.*

Let ABCD be any quadrilateral situated or not situated wholly in one plane; if two straight lines EF, GH cut the opposite sides proportionally, so that AE: EB:: DF: FC, and BG: GC:: AH: HD; then will the straight lines EF, GH, cut each other in such a manner that we shall have HM: MG :: AE : EB, and EM: MF:: AH: HD.

Draw along AD any plane AbHcD, which shall not lie along GH; through the points E, B, C, F, draw Ee, Bb, Cc, Ff, parallel to GH, to meet that plane in e,b,c,f. Because Bb, GH, Cc, are parallel, (15. III.) we have bH: Hc: BG: GC:: AH: HD; therefore (20. III.) the triangles AHb, DHc

:

are similar. Also we have Ae: eb :: AE: EB, and Df: fc:: DF FC; therefore Ae: eb:: Df: fc, or componendo, Ae: Df:: Ab: Dc; but, because the triangles AHb, DHc, are similar, we have Ab: Dc: : AH : HD; therefore Ae: Df: AH: HD; also the triangles AHb, cHD being similar, the angle HAe=HDƒ; therefore (20. III.) the triangles АHe, DHƒ are similar; therefore the angle AHe=DHf. Hence it follows that eHf is a straight line, and therefore the three parallels Ee, GH, Ff are situated in the same plane, which plane will contain the two straight lines EF, GH; therefore these latter must cut each other in a point M. Lastly, because Ee, MH, Ffare parallel, we have EM: MF:: eH: Hƒ:: AH:

HD.

And, by a similar construction in reference to the side AB, it may be shewn that HM: MG :: AE: EB.

* May be omitted, having no immediate connexion with what follows.

Ep.