les, - plane is determined 
GEOMETRY. 
lan ints A and C lying on the plane, the whole line AC 
Pa be in the line, therefore the position of the 
: the single condition of its con- 
. taining the two straight lines AB, AC. 
‘Cor. 1. Any triangle ABC, or three points A, B, C, 
not in a strai ht ine, determine the position of a plane. 
* Cor. 2. Therefore also any two parallels AP, ED 
(Fig. 141.) determine the position of a plane, for'if a 
straight line AD meet them, the plane of the two lines 
AP, AD is that of the parallels AP, ED. 
Prop. III. Tueror. 
_.. If two planes cut each other, their common section 
is a straight line. if 
Draw a straight line joining any two points E and F 
in the’ common section: of two -planes' AB, CD, this 
Jine will be wholly in the plane AB (by the definition 
of a plane), and also whallp in the plane CD ; there- 
fore it is in both s at-once, and consequently is 
their common section. 
Prov: IV. Tueor. 
OE a straight line AP is’ ndicular to two-straight 
Jines PB, PC at the point of their intersection P, it is 
perpendicular to the plane in which these lines are. 
Through P draw any straight line PQ in the plane 
of the lines PB, PC; through Q, any point in that line, 
draw a straight line to meet PB, PC, so that BQ=QC, 
(Prob. 5. Sect. 4. Part I.) Join AC, AQ, AB; and 
because ABC is a triangle, of which the base BC is bi- 
sected at Q, therefore ° 
AB?*+AC?=2BQ?+2AQ?, (14. 4. Part. I’) 
In like manner, in the triangle PBC, we have 
7 PB* 4. PC? — 2BQ742PQ% 3.5 = 
therefore, taking equals from equals, we have — 
AB* — PB? AC*_ PC? = 2AQ* — 2PQ”, 
But because the triangles APB, APC are right angled 
at P, AB?— PB?= AP?; and AC?— PC?— AP* 
(11, 4.) ; therefore ‘ 
AP?4 AP? = 2AQ?— 2PQ*, ~ 
that is, 2A P? = 2AQ? — 2PQ%, and AP?-=AQ?—PQ?, 
and AQ?=AP?4 PQ, hence in the triangle APQ, the 
angle at P must be a right angle, (11, 12, and 13. of 4. 
Part.I-), and AP is perpendicular to any line whatever 
on the plane of the lines PB, PC, therefore it is perpen- 
dicular to the plane itself, (Def. 1.) 
Cor. 1. The shortest line that can be drawn to a 
; oy from A any point above it, is the perpendicular 
Cor. 2. Only, one perpendicular PA can be drawn 
to a given plane, from a given point P in that plane: 
For if two perpendiculars could be drawn, a plane 
might pass along them, and meet the given plane in 
PQ ; and thus two lines would be perpendicular to PQ, 
which is impossible. _ 
Cor. 3. It is also impossible to draw more than one 
straight line dicular to a plane, from a point with- 
out it. Fer if there could be two perpendiculars AP, 
AQ, the triangle APQ would have two right angles, 
which is impossible. ' 
‘Prop. V: Turor: 
If a straight line AP is perpendicular toa plane MN, 
VOL, X. PART I. 
225 
every straight line DE, 
to the ie plane. 
Let a plane pass along the 
to meet the plane MN in the line PD. In this plane 
draw a line BDC through D, perpendicular to PD, 
take equal distances DB, D€, and join AB, AC, AD, 
And because in the triangles DBP, DCP, DB= DC, 
and DP is common to both, and the angle PDB=PDC, 
therefore PB = PC (6.1. Part I.) Again, in the tris 
angles APB, APC, AP is common to both, and it has 
been shewn that PB = PC, besides the angles APB, 
APC are right angles, because AP is perpendicular to 
the e MN, therefore AB=AC, so that the triangle 
ABC is isosceles; hence BD is'perpendicular to AD 
(12, 1. Part I.) ; but BD is also perpendicular to PD, 
et oe.go haga therefore ‘BD «is perpendiculat to the 
e of the triangle APD (4:), wins consequently to 
E, which is in that plane; since then EDB isa mght 
angle, and also EDP is a right angle (1 Cor. 21. Part I.), 
the line DE is perpendicular to the two lines DP, DB; 
it is therefore perpendicular to the plane MN. 
Cor. 1. Conversely, if the two straight lines AP, 
DE are perpendicular to the same plane MN, they’ are 
parallel: For if they are not, let a line be drawn 
through D parallel to AP, this line will be perpendi- 
cular to the» plane MN ; therefore through the same 
point D, two: perpendiculars: canbe drawn to the same: 
plane, which.is impossible, (2 Cor: 4:).. 
Cor. 2. Two straight lines A and B parallel to a 
third C, thoughnot inthe same plane, are parallel to one 
another; for suppose a plane perpendicular to the line 
€, the lines A and B; which are | weer to C, will be 
perpendicular to this plane ; therefore by the preceding 
eorollary, they: will be parallel to one another. 
Pror. VI. Tueror. 
els AP, ED, soas .. 
Pig, 
i i ‘or Pla 
parallel to AP, is perpendicular ear old 
Angles. 
141. 
Two planes MN, PQ perpendicular to the same Fig. 142. 
straight line AB, are parallel to each other. 
For if they could meet, let O be one of their com- 
mon points ; join OA, OB; then, because AB is per- 
pendicular to the two planes, the angles OAB, OBA 
are right angles (Def. 1.); therefore OA, OB are 
two perpendiculars from the same point on the same 
straight line, which is impossible ; therefore the planes 
cannot meet, that is, they are parallel. 
Prop. VII. Turor. 
The intersections EF, GH of chk planes MN, ,; 
2 ig. 143, 
PQ, with a third plane FG, are el. 
For if the lines EF, GH situated in the same plane, 
be not parallel, they would meet if produced ; there- 
fore, the planes in which they are would also meet, and 
consequently would not be parallel. 
Prov. VIIL. Tueor. 
A straight line AB perpendicular to a plane MN, is Fig. 142. 
also perpendicular to any plane PQ parallel to MN. 
From B draw any straight line BC in‘the plane PQ, 
and let a plane passing through BC and AB meet the 
plane MN in AD, then AD will be parallel to BC, 
2F 
