PROPOSITION XIII.

THEOREM. From the same point in a given plane, there cannot be two straight lines at right angles to the plane, upon the same side of it: and there can be but one perpendicular to a plane from a point above the plane.

For,

DEMONSTRATION. if it be possible, let the two straight lines AB, AC be at right angles to a given plane, from the same point A in the plane, and upon the same side of it. Let a plane pass through BA, AC; the common section of this with the given plane is a straight line passing through A (a): let DAE be their common section: therefore the straight lines AB, AC, DAE are in one plane: and because CA is at right angles to the given plane, it makes

right angles with every straight line meeting it in that plane (6): but DAE, which is in that plane, meets CA; therefore CAE is a right angle: for the same reason, BAE is a right angle; wherefore the angle CAE is equal to the angle BAE (c); and they are in one plane, which is impossible. Also, from a point above a plane, there can be but one perpendicular to that plane; for if there could be two, they would be parallel to one another (d); which is absurd.

PROPOSITION XIV.

THEOREM.-If the same straight line (AB) is perpendicular to each of two planes (CD, EF), they are parallel to one

another.

DEMONSTRATION. If not, they shall meet one another when produced: let them meet; their common section is a straight line

GH, in which take any point K, and join AK, BK. Then, because AB is perpendicular to the plane EF, it is perpendicular to the straight line BK, which is in that plane (a); therefore ABK is a right angle: for the same reason BAK is a right angle; wherefore the two angles ABK, BAK of the triangle ABK, are equal to two right angles; which is impossible (6): therefore the planes CD, EF, though produced, do not meet one another; that is, they are parallel (c.

PROPOSITION XV.

THEOREM.-If two straight lines (AB, BC) meeting one another, be parallel to two other straight lines (DE, EF) which meet one another, but are not in the same plane with the first two, the plane which passes through these is parallel to the plane passing through the others.

angle; therefore also

GBA is a right angle, and GB perpendicular to BA: for the same reason, GB is perpendicular to BC; since therefore the straight line GB stands at right angles to the two straight lines BA, BC, that cut one another in B; GB is perpendicular to the plane through BA, BC (f): and it is perpendicular to the plane through DE, EF (g); therefore BG is perpendicular to each of the planes through AB, BC, and DE, EF: but planes to which the same straight line is perpendicular, are parallel to one another (h); therefore the plane through AB, BC, is parallel to the plane through DE, EF.

PROPOSITION XVI.

THEOREM.-If two parallel planes (AB, CD) be cut by another plane (EF, GH), their common sections (EF, GH), with it are parallels.

DEMONSTRATION. For, if it is not, EF, GH shall meet if produced either on the side of FH, or EG. First, let them be produced on the side of FH, and meet in the point K: therefore, since EFK is in the plane AB, every point in EFK is in that

plane (a): and K is a point in EFK; therefore K is in the plane AB: for the same reason, K is also in the plane CD; wherefore the planes AB, CD, produced, meet one another: but they do not meet, since they are parallel by the hypothesis; therefore the straight lines EF, GH do not meet when produced on the side of FH: in the same manner it may be

proved, that EF, GH do not meet when produced on the side of EG. But straight lines which are in the same plane, and do not meet, though produced either way, are parallel; therefore EF is parallel to GH.

PROPOSITION XVII.

THEOREM.-If two straight lines be cut by parallel planes, they shall be cut in the same ratio.

DEMONSTRATION. Let the straight lines AB, CD be cut by the parallel planes GH, KL, MN, in the points A, E, B; C, F, D: as AE is to EB, so shall CF be to FD.

H

Join AC, BD, AD, and let AD meet the plane KL in the point X; and join EX, XF. Because the two parallel planes KL, MN are cut by the plane EBDX, the common sections EX, BD are parallel (a): for the same reason, because the two parallel planes GH, KL are cut by the plane AXFC, the common sections AC, XF are parallel: and because EX is parallel to BD, a side of the triangle ABD; as AE to EB, so is AX to XD (6): again, because XF is parallel to AC, a side of the triangle ADC; as AX to XD, so is CF to FD; and it was proved, that AX is to XD, as AE to EB; therefore, as AE to EB, so is CF to FD (c).

ROPOSITION XVIII.

THEOREM.-If a straight line (AB) be at right angles to a plane (CK), every plane which passes through it shall be at right angles to that plane.

DEMONSTRATION. Let any plane DE pass through AB, and let CE be the common section of the planes DE, CK; take any point F in CE, from which draw FG, in the plane DE, at right angles to CE (a): and because AB is perpendicular to the plane CK, therefore it is also perpendicular to straight line in that plane meeting it (b), and consequently it is perpendicular to CE; wherefore ABF is a right angle; but GFB

every

is likewise a right angle (c); therefore AB is parallel to FG (d); and AB is at right angles to the plane CK; therefore FG is also at right angles to the same plane (e). But one plane is at right angles to another plane, when the straight lines drawn in one of the planes at right angles to their common section, are also at right angles to the other plane (f); and any straight line FG in the plane DE, which is at right angles to CE, the common section of the planes, has been proved to be perpendicular to the other plane CK; therefore the plane DE is at right angles to the plane CK. In like manner it may be proved, that all planes which pass through AB, are at right angles to the plane CK.

PROPOSITION XIX.

THEOREM.-If two planes (AB, CD) which cut one another be each of them perpendicular to a third plane, their common section (BD) shall be perpendicular to the same plane.