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PRO P. V. THEOREM.

If two right lines be parallel to each other, and one of thein 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, since the fide AB is equal to the side ED (by Hyp.), the side common to each of the triangles BAD, BED, and the angles ABD, BDE right angles (by Const. and Hyp.), the side ad will be equal to the side EB (1. 4.)

Again, since the sides AD, de are equal to the sides EB, BA, and the side Ae is common to each of the tri. angles EAD, EBD, the angle ADE will be equal to the angle ABE (1. 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 intersects them, they will be all in the same plane (VII. 4. Cor.); and the

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angle ABD being a right angle, the angle cds will also be a right angle (1.25.)

But since 'es is at right angles to DB, DA, it is also at right angles to the plane which passes through them (VII. 3.); and consequently to DC (VII. Def. 2.)

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

PRO P. VI. THEOREM.

If two right lines be parallel to the faine right line, though not in the same planc 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 same plane with it; then will as 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 passes 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 also be perpendicular to that plane (VII. 5.)

And, And, in like manner, it may be proved that the line cp is also perpendicular to the plane HGK.

But when two right lines are perpendicular to the same plane, they are parallel to each other (VII. 4.); whence the line AB is parallel to the line cd, as was to be shewn.

PRO P. VII.

VII. THEOREM.

If two right lines that meet each other, be parallel to two other right lines that meet each other, though not in the same plane with them, the angles contained by those lines will be equal.

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Let the two right lines AB, BC, which meet each other in the point B, be parallel to the two right lines de, EF which meet each other in the point E ; then will the angle ABC be equal to the angle der.

For make ba, BC, ED, EF all equal to each other (1.3.), and join AD, CF, BE, Ac and DF.

Then, because BA is equal and parallel to ED (by Hyp.), AD will be equal and parallel to Be (I. 29.)

And, for the same reason, CF will also be equal and parallel to BE.

But lines which are equal and parallel to the same line, though not in the same plane with it, are equal and parallel to each other (I. 26. and VII. 6.); whence AD is equal and parallel to CF.

And since lines which join the corresponding extremes of two equal and parallel lines are also equal and parallel (I. 29.), Ac will be equal and parallel to Dr.

The three sides of the triangle ABC are, therefore, equal to the three sides of the triangle der, each to each; whence the angle ABC is equal to the angle Def (I. 7.), as was to be Thewn.

PRO P. VIII. PROBLEM.

To draw a right line perpendicular to a given plane, from a given point in the plane.

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Let A be the given point, and bc the given plane; it is required to draw a right line from the point A that shall be perpendicular to the plane Bc.

Take any point E above the plane Bc, and join EA; and through a draw AF, in the plane Bc, at right angles with EA (I. 11.); then if Ea be also at right angles with any other line which meets it in that plane; the thing required is done.

But if not, in the plane Bc, draw AG at right angles to AF (I. 11.); and in the plane eg, which pasies through the points E, A, G, make ah perpendicular to

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AG (I. 11.), and it will also be perpendicular to the plane BC, as was required.

For, since the right line FA is perpendicular to each of the right lines AE, AG (by Const.), it will also be perpendicular to the plane eg which passes through those lines (VII. 3.)

And because a right line which is perpendicular to a plane is perpendicular to every right line which meets it in that plane (Def. 2.), FA will be perpendicular to AH.

But AG is also perpendicular to Ah (by Const.); whence Ah, being perpendicular to each of the right lines FA, AG, it will also be perpendicular to the plane Bc (VII. 3), as was to be shewn.

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To draw a right line perpendicular to a given plane, from a given point above it.

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Let A be the given point, and BG the given plane ; it is required to draw a right line from the point A that shall be perpendicular to the plane BG.

Take any right line bc, in the plane BG, and draw AD perpendicular to bc (I. 11.); then if it be also perpendicular to the plane BG, the thing required is done.

But if not, draw de, in the plane BG, at right angles to BĆ (I.11), and make AF perpendicular to De (I. 12.);

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