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they will meet; and let O be a point common to both. From draw the lines OA and OB: then, since OA lies in the plane MN, it will be perpendicular to BA at A (D. 1). For a like reason, OB will be perpendicular to AB at B: hence, the triangle OAB will have two right angies, which is impossible; consequently, the planes cannot meet, and are therefore parallel; which was to be proved.

PROPOSITION X. THEOREM.

If a plane intersect two parallel planes, the lines of intersection will be parallel.

Let the plane EH intersect the parallel planes MN and PQ, in the lines EF and GII: then will EF and GH be parallel.

For, if they are not parallel, they will meet if sufficiently prolonged, because they lie in the same plane; but if the lines meet, the planes MN and PQ, in which they lie, will also meet ; but this is impossible, because these planes are parallel: hence,

M

P

G

F

N

the lines EF and GH cannot meet; they are, therefore, parallel; which was to be proved.

PROPOSITION XI. THEOREM.

If a straight line is perpendicular to one of two parallel planes, it is also perpendicular to the other.

Let MN and PQ be two parallel planes, and let the line AB be perpendicular to PQ then will it also be perpendicular to MN.

B

For, through AB pass any plane; its intersections with MN and PQ will be parallel (P. X.); but, its intersection with PQ is perpendicular to AB at B (D. 1); hence. its intersection with MN is also perpendicular to AB at A (B. I., P. XX., C. 1): hence, AB is perpendicular to every line of the plane MN through A, and is, therefore, perpendicular to that plane; which was to be proved.

P

A

M

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Parallel straight lines included between parallel planes, are equal

Let EG and FI be any two parallel lines included between the parallel planes MN and PQ then will they be equal.

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G

N

are equal (B. I., P. XXVIII.); which was to be proved. Cor. 1. The distance between two parallel planes is mea sured on a perpendicular to both; but any two perpendiculars between the planes are equal: hence, parallel planes are everywhere equally distant.

Cor. 2. If a straight line GH is parallel to any plane MN, then can plane be passed through GĦ parallel to MN: hence, if a straight line is parallel to a plane, all of its points are equally distant from that plane.

PROPOSITION XIII. THEOREM

If two angles, not situated in the same plane, have their sides parallel and lying in the same direction, the angles will be equal and their planes parallel.

Let CAE and DBF be two angles lying in the planes MN and PQ, and let the sides AC and AE be respectively parallel to BD and BF, and lying in the same direction then will the angles CAE and DBF be equal, and the planes MN and PQ will be parallel.

Take any two points of AC and AE, as C and E, and

make BD equal to AC, and

BF to AE; draw CE, DF,
AB, CD, and EF

1°. The angles CAE and DBF will be equal.

P

H

E

D

F

B

For, AE and and BF being parallel and equal, the figure ABFE is a parallelogram (B. I., P. XXX.); hence, EF is parallel and equal to AB. For a like reason, CD is parallel and equal to AB: hence, CD and EF are parallel and equal to each other, and consequently, CE and DF are also parallel and equal to each other. The triangles CAE and DBF have, therefore, their corresponding sides equal, and consequently, the corresponding angles CAE and DBF are equal; which was to be proved.

PQ,

2. The planes of the angles MN and PQ are parallel. For, if not, pass a plane through A parallel to and suppose it to cut the lines CD and EF in G and

H.

Then will the lines GD and HF

be equal respect

ively to AB (P. XII.), equal to CD, and IIF to EF; which is impossible: hence, the planes MN and PQ must be parallel; which was to be proved.

and consequently, GD will be

Cor. If two parallel planes MN and PQ, are met by two other planes AD and AF the angles CAE and DBF, formed by their intersections, will be equal.

PROPOSITION XIV. THEOREM.

If three straight lines, not situated in the same plane, are equal and parallel, the triangles formed by joining the extremities of these lines will be equal, and their planes parallel.

Let AB, CD, and EF be equal parallel lines not in the same plane: then will the triangles ACE and BDF be equal, and their planes parallel.

For, AB being equal and parallel to EF, the figure ABFE is a parallelogram, and consequently, AE is equal and parallel to BF For a like reason, AC is equal and parallel to BD: hence, the included angles CAE and DBF are equal and their planes parallel (P. XIII.). Now, the triangles CAE and

A

DBF have two sides and their

included angles equal, each to each: hence, they are equal in all their parts. The triangles are, therefore, equal and their planes parallel; which was to be proved.

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If two straight lines are cut by three parallel planes, they will be divided proportionally.

Let the lines AB and CD be cut by the parallel lanes MN, PQ, and RS, in the points A, E, B, and C, F, D; then Ç,

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The plane ACD intersects the parallel planes MN and PQ, in the parallel lines AC and GF: hence,

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Combining these proportions (B. II., P. IV.), we have,
AE : EB :: CF: FD;

which was to be proved.

Cor. 1. If two straight lines are cut by any number of parallel planes, they will be divided proportionally.

Cor. 2. If any number of straight lines are cut by three parallel planes, they will be divided proportionally.

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