PROP. II. THEOREM.

Any three right lines which mutually interfect each other, are all in the fame plane.

E

Let AB, BC, CA be three right lines, which interfect each other in the points A, B, C ; then will those lines be in the fame plane.

For let any plane AD pass through the points A, B, and be turned round that line, as an axis, till it pass through the point c.

Then, because the points A, C are in the plane AD, the whole line AC muft alfo be in it; or otherwise its parts would not lie in the fame direction.

And, because the points B, C are also in this plane, the whole line BC muft likewife be in it; for the fame reason.

But the line AB is in the plane AD, by hypothefis; whence the three lines AB, BC, CA are all in the fame plane, as was to be shewn.

COR. Any two right lines which interfect each other, are both in the fame plane; and through any three points a plane may be extended,

PROP. III. THEOREM.

If a right line be perpendicular to two other right lines, at their point of interfec tion, it will also be perpendicular to the plane which paffes through thofe lines.

Let the right line AB be perpendicular to each of the two right lines BC, BD, at their point of interfection B ; then will it also be perpendicular to the plane which paffes through thofe lines.

For make BD equal to BC; and, in the plane which paffes through those lines, draw any right line BE; and join the points CD, AD, AE and AC:

Then because the fide BC is equal to the fide BD (by Conft.), and the perpendicular AB is common to each of the triangles ABC, ABD, the fide AD will alfo be equal to the fide AC (1. 4.)

And fince the triangles CAD, CBD are ifofceles, the rectangle of CE, ED, together with the fquare of EB, is equal to the fquare of DB; and the rectangle of CE, ED together with the fquare of EA, is equal to the fquare of AD (II. 20.)

From each of these equals, take away the rectangle of CE, ED which is common, and the difference of the

fquares of EB, EA will be equal to the difference of the fquares of DB, ad.

But the difference of the fquares of DB, AD is equal to the fquare of AB (II. 14 Cor.); whence the difference of the fquares of EB, EA will alfo be equal to the fquare of AB; and confequently AB is perpendicular to BE, as was to be fhewn,

COROLL. If a right line be perpendicular to three other right lines, at their point of interfection, those lines will be all in the fame plane.

For if either of them, as BE, were above or below the plane which paffes through the other two, the angle ABE would be lefs or greater than a right angle.

PROP. IV. THEOREM.

If two right lines be perpendicular to the fame plane, they will be parallel to each other.

D

Let the right lines AB, CD be each of them perpendicular to the plane FG, then will those lines be parallel to each other.

For join the points D, B ; and, in the plane FG, make DE perpendicular to DB, and equal to AB (I. 11. 3.); and join the points AE, AD.

Then, fince the right lines AB, CD are perpendicular to the plane FG (by Hyp.), the angles ABD, ABE, CDB and CDE will be right angles (VII. Def. 2.)

And because the fide AB, is equal to the fide ED (by Conft.), the fide DB common to each of the triangles BAD, BED, and the angles ABD, BDE right angles (by Hyp. and Conft.), the fide AD will also be equal to the fide EB (I. 4.)

Again, fince the fides AD, DE are equal to the fides EB, BA, and the fide AE is common to each of the triangles EBA, EDA, the angle ADE will alfo be equal to the angle ABE (I, 7.), and is, therefore, a right angle.

And, because the line ED is at right angles with each of the three lines DA, DB, DC, thofe lines, together with the line AB, will be all in the fame plane (VII. 3. Cor.)

Since, therefore, the lines AB, BD, DC are all in the fame plane, and the angles ABD, CDB are each of them a right angle, the line AB will be parallel to the line CD (I. 23.), as was to be fhewn,

COR. Any two parallel right lines AB, CD, are in the fame plane; and any right line DA, which interfects those parallels, is in the fame plane with them.

PROP.

PROP. V. THEOREM.

If two right lines be parallel to each other, and one of them be perpendicular to a plane; the other will also be perpendicular to that plane.

Let AB, CD be two parallel right lines, one of which,. AB, is perpendicular to the plane FG; then will the other CD be alfo perpendicular to that plane.

For join the points D, B; and, in the plane FG, make DE perpendicular to DB, and equal to BA (I. 11. 3.); and join AE, AD and ER:

Then, because AB is perpendicular to the plane FG (by Hyp.) the angles ABD, ABE will be right angles (VII. Def. 2.)

And, fince the fide AB is equal to the fide ED (by Hyp.), the fide DB common to each of the triangles BAD, BED, and the angles ABD, BDE right angles (by Conft. and Hyp.), the fide AD will be equal to the side EB (I. 4.)

Again, fince the fides AD, DE are equal to the fides EB, BA, and the fide AE is common to each of the triangles EAD, EBD, the angle ADE will be equal to the angle ABE (I. 7.), and is, therefore, a right angle.

And, because the right lines AB, CD are parallel to each other (by Hyp.), and the line AD interfects them, they will be all in the fame plane (VII. 4. Cor.); and the