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PROP. III. THEOREM.

If a right line be perpendicular to two other right lines, at their point of interfection, it will alfo 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 thofe 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 also be equal to the fide AC (I. 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 thefe equals, take away the rectangle of CE, ED which is common, and the difference of the

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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 also be equal to the square of and confequently AB is perpendicular to BE, as was to be fhewn.

AB;

COROLL. If a right line be perpendicular to three other right lines, at their point of interfection, thofe 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.

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Let the right lines AB, CD be each of them perpendi

cular to the plane FG, then will thofe 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 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, 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 thofe 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.

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Let AB, CD be two parallel right lines, one of which, AB, is perpendicular to the plane FG; then will the other CD be also 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 EB:

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 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 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

angle ABD being a right angle, the angle CDB will also be a right angle (I.25.)

But fince ED is at right angles to DB, DA, it is also at right angles to the plane which paffes through them (VII. 3.); and confequently to DC (VII. Def. 2.)

The line DC is, therefore, perpendicular to each of the lines DE, DB; whence it is alfo perpendicular to the plane FG (VII. 3.), as was to be fhewn.

PROP. VI. THEOREM.

If two right lines be parallel to the fame right line, though not in the fame plane with it, they will be parallel to each other.

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Let the right lines AB, CD be each of them parallel to the right line EF, though not in the fame plane with it; then will AB be parallel to CD.

For take any point G in the line EF, and draw the right lines GH, GK, each perpendicular to EF (I. 11.), in the planes AF, ED of the proposed parallels :

Then fince the right line EF is perpendicular to the two right lines GH, GK, at their point of intersection G, it will also be perpendicular to the plane HGK which paffes through those ines (VIII. 3.)

And because the lines AB, EF are parallel to each other (by Hyp.), and one of them, EF, is perpendicular to the plane HGK, the other, AB, will alfo be perpendicular to that plane (VII. 5.)

And,

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