« ForrigeFortsett »
Book XI. Let AB, CD be two parallel straight lines, and let one of thene
AB be at right angles to a plane; the other CD is at right angies to the same plane.
Let AB, CD, meet the plane in the points B, D, and join BD. therefore AB, CD, BD are in one plane. In the plane, to which AB is at right angles, draw DE at right angles to BD, and make DE equal to AB, and join BE, AE, AD. And because AB
is perpendicular to the plane, it is perpendicular to every stright 2. 3. Def.s1. line which meets it, and is in that plane". therefore each of the
angles ABD, ABE, is a right angle, and because the straight line
BD meets the parallel straight lines AB, CD, the angles ABD, b. 29. 1.
CDB are together equal o to two right angles. and ABD is a right
EDB, because each of them is a right 0. 4. .
angle; therefore the base AD is equal o
D IDA; wherefore the angle ABE is equal d. 8. 1.
a to the angle EDA. and ABE is a right
is perpendicular e to the plane which passes thro' BD, DA, and 1.3.Def.14. fhall : make right angles with every straight line meeting it in that
plane. but DC is in the plane passing thro' BD, DA, because all three are in the plane in which are the parallels AB, CD, wherefore ED is at right angles to DC; and therefore CD is at right angles to DE. but CD is also at right angles to DB; CD then is at right angles to the two straight lines DE, DB in the point of their intersection D; and therefore is at right angles to the plane paffing thro' DE, DB, which is the fame plane to which AB is at right angles. Therefore if two straight lines, &c. Q. E. D.
e. 4. Ir.
PROP. IX. THEO R.
to the same straight line, and not in the same plane with it, are parallel to one another.
a. 4. Il.
Let AB, CD be each of them parallel to EF, and not in the fame plane with it; AB shall be parallel to CD.
In EF take any point G, from which draw, in the plane paffing thro' EF, AB, the straight line GH at right angles to EF; and in the plane passing thro' EF, CD, draw GK at right angles to the faine EF. and because EF is per- A H pendicular both to GH and GK, EF
B is perpendicular to the plane HGK passing thro' them. and EF is parallel
G to AB; therefore AB is at right an
F gles b to the plane HGK. for the fame reafon, CD is likewise at right angles to the plane HGK. therefore CK D AB, CD are each of them at right angles to the plane HGK. but if two straight lines be at right angles to the fame plane, they Mall be parallelo to one another. therefore AB is parallel to CD. c. 6. 11, Wherefore two straight lines, &c. Q. E. D.
b. 8. 11.
PROP. X. THEO R.
IF two straight lines meeting one another be parallel
to two others that meet one another, and are not in the same plane with the first two; the first two and the other two shall contain equal angles.
Let the two straight lines AB, BC which meet one another be parallel to the two straight lines DE, EF that meet one another, and are not in the faine plane with AB, BC. The angle ABC iş equal to the angle DEF.
Take BA, BC, ED, EF all equal to one another; and join AD, CF, BE, AC, DF. because BA is equal and parallel to ED, there;
Book XI. fore AD is both equal and parallel to
BE. For the same reafon, CF is equal 2. 33. I.
and parallel to BE. Therefore AD and
C to BE. Bat straight lines that are paral
iel to the fame straight line, and not in b. 9. 11. the fame plane with it, are parallel b to
one another. Therefore AD is parallel 6. 1. Ax. 1. to CF; and it is equal to it, and AC, DI join them towards the fame parts ; D
F and therefore, a AC is equal and parallel
to Dr. and because AB, BC are equal to DE, EF, and the base d. 8. I.
AC to the base DI's the angle ABC is equal d to the angle DEF.
PRO P. XI. PROB.
o draw a straight line perpendicular to a plane, from a given point above it.
Let A be the given point above the plane BII; it is required to draw froin the point A a straight line perpendicular to the plane BH.
In the plane draw any straight line BC, and from the point A draw a AD perpurdicular to LC. If then AD bu als perpendi
cular to the plane DH, the thing required is alrcady done, but if it b. 11.1.
he not, from the point D draw b
right angles to ED and DA, BC is d. 4. 11. at right angles to the plane pal.
fing thro' ED, DA. And GII is 3 D
the same plane; wherefore GH is at right angles to the plane thro' ( 3. Def.ir. ED, DA, and is perpendicular r to every straight line meeting it in
that pline. But AF, which is in the plane thro’ED, DA meets it,
C. 31. I.
e. 8. 11.
therefore GH is perpendicular to AF, and consequently AF, is Book Xi.
PROP. XII. PROB.
To crect a ftraight line at right angles to a given
plane, from a point given in the plane.
2. II. II.
Let A be the point given in the plane ; it is required to erect a
Froin any point B above the plane
b. 3. 1. A draw 6 AD parallel to BC. because therefore AD, CB are two parall straight A lines, and one of them BC is at right angles to the given plane, the other AD ish also at right angles to it. therefore a straight line has been erect. c. 8.11, ed at right angles to a given plane from a point given in it. Which was to be done.
PRO P. XIII. THEOR.
*ROM 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 per, pendicular to a plane from a point above the plane.
For, 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; and let a plane pafs thro' BA, AC; the common section of this with the given plane is a straight a line
Book XI. passing through A. let DAE be their common fection. therefore
the straight lines AB, AC, DAE are in one plane. and because
D А E which is imposible. Also, from a point above a plane there can
be but one perpendicular to that plane; for if there could be two, b.6. Ir. they would be parallel b to one another, which is absurd. There
fore from the same point, &c. 0. E. D.
PROP. XIV. THEOR.
LANES to which the same straight line is perpendicular, are parallel to one another.
Let the straight line AB be perpendicular to each of the planes CD, EF; these planes are parallel to one another.
If not, they hall meut one onother when produced ; let them meet; their common section shall be a straight line GH, in which take any point K, and join AK, BK. then be
cause AB is perpendicular to the plane 4.3. Def. 11. EF, it is perpendicular a to the straight C
H line BK which is in that plane. there
F fore ADK is a right angle. for the fame A
B reason, BAK is a right angle; wherefore the two angles AEK, BAK of the triangle ABK are equal to two right
E b. 17. I. angles, which is imposible b. therefore
D the planes CD, EF though produced do c.8. Def.11. not meet one another ; that is, they are parallel Therefore
planes, &c. Q: E. D.