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Some others might be added; but it may be stated here generally, that any one condition being given, the corresponding one follows, either by direct or ex absurdo deduction, at once.
PROPOSITION XII. A plane is drawn perpendicular to a given line at its middle point :
then (1.) Any point in the plane is equidistant from the extremities of the
line; and (2.) All points equidistant from the extremities of the line are situated
in that plane. (1.) Let AB be a straight line, C its middle point, and MN a plane
drawn through C perpendicular to AB: then any point P in MN will be equidistant from A and B.
For join PA, PB, PC.
Then since AB is perpendicular to the plane MN, meeting it in C, and CP is in the same plane, the angles ACP, BCP, are right angles, and therefore equal to one another (Prop. 11.). Wherefore the sides AC, CP are equal to BC, CP and the included angles equal; and (Euc. 1. 4) hence AP is equal to PB.
(2.) Let Q be any point equidistant from A and B: it will lie in the plane MN.
For join QA, QB; and, if possible, let Q be not in the plane MN, but on one side of it-suppose the same side with B. Then QA will cut MN in some point, P. Join PB.
Then, by the preceding case, PA is equal to PB; and hence adding PQ to both, AQ is equal to BP and PQ together. But BP and PQ together are greater than BQ (Euc. 1. 20); and hence AQ is greater than BQ, which is absurd.
PROPOSITION XIII. If a line be perpendicular to a plane, and from its intersection with
the plane a line be drawn perpendicular to any line in the plane ; then the plane through the perpendiculars will be perpendicular to the line in the plane. Let AB be perpendicular to the plane MN, and let BC be drawn
perpendicular to the line DE in that plane: then the plane through AB, BC will be perpendicular to DE.
For in DĒ take on opposite sides of C the equal segments CF, CG; and draw FB, GB to B, and FA, GA to any other point A in the perpendicular.
Then since BC, CF are equal to BC, CG each to each, and the included angles BCF, BCG equal (being right angles by hypothesis) BF is equal to BG (Euc. I. 4.).
Again, since F and G are equidistant from B, the lines AF, AG are equal (Prop. xi.). Wherefore the sides AC, CF are equal to the sides AC, CG, and the base AF to the base AG; and hence the angle ACF is equal to its adjacent angle ACG. Whence AC is perpendicular to DE, but BC is also perpendicular to DE by hypothesis ; and hence DE is perpendicular to both BC and CA; and hence to the plane BH drawn through them (Prop. 11.). And this plane is that drawn through the perpendiculars AB, BC; since AC is in the same plane with them.
Cur. If the same hypothesis be made, all lines drawn from C to meet AB are perpendicular to DE.
This is proved above, and is often given as the principle theorem ; whilst that which has been enunciated is sometimes made a corollary from it.
PROPOSITION XIV. (1.) If two straight lines be perpendicular to a plane, they are parallel
to one another; and (2.) If two lines be parallel, the plane which is perpendicular to one
of them will be perpendicular to the other.
(1.) Let AB, CD be perpendicular to the plane MN: they will be parallel to one another.
For join BD; draw a line from D to any point A in AB; and in the plane MN the line EDF perpendicular to BD.
Then since AB is perpendicular to MN, and BD to the line EF in MN, the plane ABD is perpendicular to EF (Prop. XIII.); that is DB and DA are each perpendicular to EF. But CD is also perpendicular to the plane MN, and hence CD is perpendicular EF ; that is, EF is perpendicular to DB, DA, DC at the point D. Wherefore the three lines DC, DA, DB being perpendicular to EF at the point D, they are in one plane (Prop. 111.). But AB is in the same plane with DA, DB, and therefore with CD; or the two lines perpendicular to the plane MN are themselves in one plane.
Again, since AB, CD are perpendicular to the plane MN, the angles ABD, CDB are right angles. Whence these lines being in
the same plane ABDC, and perpendicular to the same line BD in it, they are parallel.
(2.) Let AB, CD be parallel, and the plane MN be perpendicular to one of them AB, it will be perpendicular to the other CD.
Since AB, CD are parallel, they are in one plane. Let this intersect MN in BD, and draw ·EF through D perpendicular to BD.
Then, since AB is parallel to CD, and ABD a right angle (AB being perpendicular to MN) the angle BDC is also a right angle, or CD is perpendicular to the line BD in the plane MN.
Again, since AB is perpendicular to MN and BD to a line EF in MN; the line EF is perpendicular to the plane ABD (Prop. XIII.); that is to the plane ABDC. Whence CD is perpendicular to EF; and it has been proved to be perpendicular to DB. It is hence
perpendicular to the plane MN which contains the lines BD, EF (Prop. 11.).
If a plane cut two parallel planes, it makes equal dihedral angles
with them. Let the plane MN cut
the parallel planes PQ, RS in MQ, DS; the dihedral angles PMQN, RDSN will be equal.
For since the planes PQ, RS are parallel, the edges MQ, DS of the dihedral angles are parallel (Prop. I. Chap. 1.). Draw then a profile plane TV to one of the angles through A, to cut the system of planes in ABV, AE, BF.
Then since PQ, RS are parallel, and TV is perpendicular to one of them PQ, it is perpendicular to the other RS. It is hence also a pro
Q file plane of both the dihedral angles.
Again, since PQ, RS are parallel planes cut by the profile plane TV, the lines AE, BF are parallel (Prop. I. Chap. 1.), and AV is common to both the profile angles. Wherefore the angle EAV is equal to the angle FBV; and hence the profile angles are equal. Wherefore, finally, the dihedral angles are equal.
Cor. 1. If the parallel planes be produced, the alternate dihedral angles will be equal; and the two interior dihedral angles are together equal to two dihedral right angles. For this is the case with their profile angles (Euc. 1. 29) and hence with the dihedral angles themselves.
Cor. 2. The converse, when enunciated in an unrestricted manner, is not true. But put thus, it is true:
If two planes meet a third plane in parallel lines, and make with it equal dihedral angles in the same direction, those planes will be parallel. The italics indicate the restrictions alluded to.
Cor. 3. The opposite dihedral angles of a parallelopiped are equal. (Prop. xx. Chap. 1.).
PROPOSITION XVI. If a line be perpendicular to a plane, every plane drawn through it
will be perpendicular to that plane. Let the line AB be perpendicular to the plane MN: then any plane PQ drawn through AB will be perpendicular to MN.
For let the plane PQ cut MN in BQ, and in MN draw the line BC perpendicular to BQ; and also draw the plane AC through AB, BC.
Then since AB is perpendicular to MN, it is perpendicular to BQ; and BC is perpendicular to BQ by construction. Whence ABC is the profile angle of the dihedral angle PQBN (Def. 4).
But AB being perpendicular to MN, ABC is a right angle (Prop. 11.); and the profile angle of the planes being a right angle, then the dihedral angle is a right angle (Def. 4). That is, the plane PQ is perpendicular to MN.
PROPOSITION XVII. If two planes be perpendicular to one another, every line drawn in one
plane perpendicular to the common section will be perpendicular to the other plane ; and every line drawn from a point in one plane perpendicular to the other plane lies wholly in the former, and is perpendicular to the common section. (1.) Let the plane PQ be perpendicular to the plane MN; and let
N the line AB be drawn in one plane PQ perpendicular to MQ the common section of the planes: then AB will be perpendicular to the other plane MN.
For in the plane MN, draw BC perpendicular to the common section MQ. Then since AB, BC are perpendicular to MQ, the plane through them is the profile plane, and ABC the profile angle. But the dihe
dral angle PQMN is a right angle (Hyp.), and hence ABC is a right angle (Prop. x. Scholium.).
But AB is perpendicular to MQ (Hyp.), and hence it is perpendicular to the two lines BC, BM in the plane MN: that is, to the plane MN itself.
(2.) All lines drawn from points, as R, in the plane PQ perpendicular to the plane MN, will be wholly in the plane PQ.
For, if possible, let some one of them RS be without the plane PQ. From R draw RQ perpendicular to the line QM.
Then by the preceding case RQ is perpendicular to MN. But by admission RS is also perpendicular to MN: that is, two lines RQ, RS can be drawn from the same point R perpendicular to the plane MN. This is impossible (Prop. iv.); and hence the perpendicular to MN falls wholly in the plane PQ.
(3.) All perpendiculars from points in PQ to the plane MN are perpendicular to the line MQ.
For if possible let AH be drawn in the plane PQ perpendicular to MN, yet not perpendicular to MQ; and draw AB perpendicular to MQ.
Then reasoning as in the last case, the conclusion follows.
(1.) If two planes which intersect be perpendicular to a third plane,
their intersection will also be perpendicular to this third plane; and (2.) If a plane be perpendicular to the intersection of two other planes,
it will be perpendicular to each of those planes.
(1.) Let the two planes PQ, RS which intersect in RQ, be both perpendicular to the plane MN; then will RQ be perpendicular to the plane MN.
For, from any point A in the intersection AQ, draw a line AP in PQ perpendicular to AQ; and similarly a line SB in the plane RS perpendicular to QS.
Then (Prop. xvii.), AP, SB are perpendicular to. MN; and hence (Prop. XIV.) they are parallel to one another.
Also, since the planes PQ, RS which intersect in RQ are drawn through the parallel lines AP, BS, the intersection QR is parallel to both of them (Prop. 11. Chap. 1.).
And, finally, since AP is perpendicular to MN, and QR parallel to AP, it follows that QR is also perpendicular to MN.
(2.) Let the plane MN be perpendicular to QR the intersection of the two planes PQ, RS: then it will be perpendicular to the planes PQ, RS themselves.
For, since RQ is perpendicular to MN, the planes PQ, RS passing through it are also perpendicular to MŃ (Prop. xvi.); that is, MN is perpendicular to PQ and RS.