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SOLID GEOMETRY.

BOOK VI.

LINES AND PLANES IN SPACE.

DIEDRAL

ANGLES. POLYEDRAL ANGLES.

394. Def. A plane is said to be determined by certain lines or points when one plane, and only one, can be drawn through these lines or points.

PROP. I. THEOREM.

395. A plane is determined

I. By a straight line and a point without the line. II. By three points not in the same straight line. III. By two intersecting straight lines.

IV. By two parallel straight lines.

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I. Given point C without str. line AB.

To Prove that a plane is determined by AB and C.

Proof. If any plane MN be drawn through AB, it may be revolved about AB as an axis until it contains point C.

Hence, a plane can be drawn through line AB and point C; and it is evident that but one such plane can be drawn.

•C

B

II. Given A, B, and C three points not in the same str. line.

To Prove that a plane is determined by A, B, and C.

Proof. Draw line AB; then a plane, and only one, can be drawn through line AB and point C.

[A plane is determined by a str. line and a point without the line.] (§ 395, I) Then, a plane, and only one, can be drawn through A, B, and C.

B

III. Given AB and BC intersecting str. lines.

To Prove that a plane is determined by AB and BC. Proof. A plane, and only one, can be drawn through line AB and point C.

[A plane is determined by a str. line and a point without the line.] (§ 395, I) And since this plane contains points B and C, it must contain line BC.

[A plane is a surface such that the str. line joining any two of its points lies entirely in the surface.] (§ 9) Then, a plane, and only one, can be drawn through AB and BC.

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To Prove that a plane is determined by AB and CD.

Proof. The Is AB and CD lie in the same plane.

[Two str. lines are said to be when they lie in the same plane, and cannot meet however far they may be produced.]

($ 52)

And only one plane can be drawn through AB and point C. [A plane is determined by a str. line and a point without the line.] (§ 395, I) Then, a plane, and only one, can be drawn through AB and CD.

PROP. II. THEOREM.

396. The intersection of two planes is a straight line.

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Given line AB the intersection of planes MN and PQ.

To Prove AB a str. line.

Proof. Draw a str. line between points A and B.

This str. line lies in plane MN, and also in plane PQ.

[A plane is a surface such that the str. line joining any two of its points lies entirely in the surface.]

(§ 9) Then it must be the intersection of planes MN and PQ. Hence, the line of intersection AB is a str. line.

397. Defs. If a straight line meets a plane, the point of intersection is called the foot of the line.

A straight line is said to be perpendicular to a plane when it is perpendicular to every straight line drawn in the plane through its foot.

A straight line is said to be parallel to a plane when it cannot meet the plane however far they may be produced. A straight line which is neither perpendicular nor parallel to a plane, is said to be oblique to it.

Two planes are said to be parallel to each other when they cannot meet however far they may be produced.

398. Sch. The following is given for convenience of reference:

A perpendicular to a plane is perpendicular to every straight line drawn in the plane through its foot.

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399. At a given point in a plane, one perpendicular to the plane can be drawn, and but one.

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To Prove that a can be drawn to MN at P, and but one. Proof. At any point A of indefinite str. line AB, draw lines AC and AD L to AB.

Let RS be the plane determined by AC and AD.

Let AE be any other str. line drawn through point A in plane RS; and draw line CD intersecting AC, AE, and AD at C, E, and D, respectively.

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Produce BA to B', making AB' = AB, and draw lines BC, BE, BD, B'C, B'E, and B'D.

In ABCD and B'CD,

CD = CD.

And since AC and AD are 1 to BB' at its middle point,

BC = B'C and BD = B'D.

[If a be erected at the middle point of a str. line, any point in the is equally distant from the extremities of the line.] (§ 41, I)

.. ABCD = ▲ B'CD.

[Two ▲ are equal when the three sides of one are equal respectively to the three sides of the other.] (§ 69)

Now revolve ▲ BCD about CD as an axis until it coincides with ▲ B'CD.

Then, point B will fall at point B', and line BE will coincide with line B'E; that is, BE = B'E.

Hence, since points A and E are each equally distant from B and B', line AE is 1 BB'.

[Two points, each equally distant from the extremities of a str. line, determine a at its middle point.]

(§ 43)

But AE is any str. line drawn through A in plane RS. Then, AB is to every str. line drawn through A in plane RS. Whence, AB is

to plane RS.

[A str. line is said to be to a plane when it is to every str. line drawn in the plane through its foot.]

(§ 397) Now apply plane RS to plane MN so that point A shall fall at point P; and let AB take the position PQ.

Then, PQ will be 1 MN.

Hence, a can be drawn to MN at P.

If possible, let PT be another to plane MN at P; and let the plane determined by PQ and PT intersect MN in line HK.

Then, both PQ and PT are 1 HK.

[A to a plane is to every str. line drawn in the plane through its foot.] (§ 398) But this is impossible; for, in plane HKT, only one can be drawn to HK at P.

[At a given point in a str. line, but one to the line can be drawn.] Then only one I can be drawn to MN at P.

(§ 25)

400. Cor. I. A straight line perpendicular to each of two straight lines at their point of intersection is perpendicular to their plane.

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