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PROP, I. THEO Ř E M.
The common section of any two planes is a right line.
Let AB, cd be two planes, whose common section is EF ; then will Ef be a right line.
For if not, let Fge bé a right line, drawn in the plane AB; and fke another right line, drawn in the plane cd.
Then, since the lines FGE, FKE are in different planes, they must fall wholly without each other.
But the line FGE, having the same extremities with the line Fke, will coincide with it: whence they coincide and fall wholly without each other, at the same time, which is absurd.
The lines FGE, FKE cannot, therefore, be right lines; and consequently the line EF, which lies in each of the planes, must be a right line, as was to be shewn.
SCHOLIUM. One part of a right line cannot be in a plane, and another part out of it. For since the line can be produced in that plane, the part out of the plane, and the part produced would have different directions, which is absurd.
PRO P. II. THEOREM.
Any three right lines which mutually intersect each other, are all in the same plane.
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 same 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 must also be in it; or otherwise its parts would not lie in the same direction.
And, because the points b, c are also in this plane, the whole line Bc must likewise be in it; for the same reason,
But the line AB is in the plane AD, by hypothesis ; whence the three lines AB, BC, CA are all in the fame plane, as was to be shewn.
COR. Any two right lines which intersect each other, are both in the fame plane; and through any three points a plane may be extended.
If a right line be perpendicular to two other right lines, at their point of intersection, it will also be perpendicular to thę plane which passes through those lines.
Let the right line AB be perpendicular to each of the two right lines BC, BD, at their point of intersection B; then will it also be perpendicular to the plane which passes through those lines.
For make BD equal to BC; and, in the plane which passes through those lines, draw any right line Bes and join the points cD, AD, AE and Ac:
Then because the side bc is equal to the side BD (by Const.), and the perpendicular AB is common to each of the triangles ABC, ABD, the side as will also be equal to the fide AC (I. 4.)
And fince the triangles CAD, CBD are isosceles, the rectangle of c£, ED, together with the square of EB, is equal to the square of DB; and the rectangle of cE, ED together with the square of ea, is equal to the square of AD (II. 20.)
From each of these equals, take away the rectangle of ÇE, ED which is common, and the difference of the
squares of EB, EA will be equal to the difference of the squares of DB, AD.
But the difference of the squares of DB, ad is equal to the square of AB (II. 14. Cor.); whence the difference of the squares of EB, EA will also be equal to the square of AB; and consequently AB is perpendicular to be, as was to be shewn.
COROLL. If a right line be perpendicular to three other right lines, at their point of intersection, those lines will be all in the same plane.
For if either of them, as be, were above or below the plane which passes through the other two, the angle ABE would be less or greater than a right angle.
If two right lines be perpendicular to the same plane, they will be parallel to each other.
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, since 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 side AB, is equal to the side ed (by Conf.), the side de common to each of the triangles BAD, BED, and the angles ABD, Bde right angles (by Hyp. and Const.), the side AD will also be equal to the side EB (I. 4.)
Again, since the sides AD, DE are equal to the sides EB, BA, and the side ae is common to each of the triangles EBA, EDA, the angle ADE will also 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, those lines, together with the line AB, will be all in the same plane (VII. 3. Cor.)
Since, therefore, the lines AB, BD, DC are all in the same 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 shewn.
Cor. Any two parallel right lines AB, CD, are in the same plane ; and any right line da, which intersects those parallels, is in the same plane with them.