562. Two parallel lines determine a plane.

For, by the definition of parallel lines, the two lines are in the same plane; and, as only one plane can be drawn to contain one of the lines and any point in the other line, it follows that only one plane can be drawn to contain both lines.

THEOREM II.

563. If two planes cut one another, their common section must be a straight line.

HYPOTHESIS. Let AB and CD be two intersecting planes.
CONCLUSION. Their common section is a straight line.

PROOF. Let M and N be two points common to both planes. Draw the straight line MN. Therefore, by 554, since M and N are in both planes, the straight line MN lies in both planes.

And no point out of this line can be in both planes; because then two planes would each contain the same line and the same point without it, which, by 560, is impossible.

Hence every point in the common section of the planes lies in the straight line MN.

564. CONTRANOMINAL OF 554. A line which does not lie altogether in a plane may have no point, and cannot have more than one point, in common with the plane.

Therefore three planes which do not pass through the same line cannot have more than one point in common; for, by 563, the points common to two planes lie on a line, and this line can have only one point in common with the third plane.

565. All planes are congruent; hence properties proved for one plane hold for all. A plane will slide upon its trace.

PRINCIPLE OF DUALITY.

566. When any figure is given, we may construct a reciprocal figure by taking planes instead of points, and points instead of planes, but lines where we had lines.

The figure reciprocal to four points which do not lie in a plane will consist of four planes which do not meet in a point. From any theorem we may infer a reciprocal theorem.

Two points determine a line.
Three points which are not in

a line determine a plane.

A line and a point without it determine a plane.

Two lines in a plane determine a point.

Two planes determine a line.

Three planes which do not pass through a line determine a point.

A line and a plane not through it determine a point.

Two lines through a point determine a plane.

There is also a more special principle of duality, which, in the plane, takes points and lines as reciprocal elements; for they have this fundamental property in common, that two elements of one kind determine one of the other. Thus, from a proposition relating to lines or angles in axial symmetry, we get a proposition relating to points or sects in central symmetry.

The angle between two corresponding lines is bisected by the

axis.

The sect between two corresponding points is bisected by the

center.

THEOREM III.

567. If a line be perpendicular to two lines lying in a plane, it will be perpendicular to every other line lying in the plane and passing through its foot.

HYPOTHESIS. Let the line EF be perpendicular to each of the lines AB, CD, at E, the point of their intersection.

CONCLUSION. EF 1 GH, any other line lying in the plane ABCD, and passing through E.

PROOF. Take AE = BE, and CE = DE.

Join AD and BC, and let G and H be the points in which the joining lines intersect the third line GH.

Take any point F in EF. Join FA, FB, FC, FD, FG, FH.
Then

Δ AED = Δ BEC,

(124. Triangles having two sides and the included angle equal are congruent.)

(128. Triangles having two angles and the included side equal are congruent.)

Then

▲ AEF≈ ▲ BEF,

(124. Triangles having two sides and the included angle equal are congruent.)

(129. Triangles having three sides in each respectively equal are congruent.)

Then

¥ DAF = × CBF.

▲ AFG ≈ ▲ BFH,

(124. Triangles having two sides and the included angle equal are congruent.)

Then

FG = FH.

Δ FEG = Δ FEH,

(129. Triangles having three sides in each respectively equal are congruent.)

As GH is any line whatever lying in the plane ABCD, and passing through E,

:. EF 1 every such line.

568. A line meeting a plane so as to be perpendicular to every line lying in the plane and passing through the point of intersection, is said to be perpendicular to the plane. Then also the plane is said to be perpendicular to the line.

569. COROLLARY. At a given point in a plane, only one perpendicular to the plane can be erected; and, from a point without a plane, only one perpendicular can be drawn to the plane.

570. INVERSE OF 567. All lines perpendicular to another line at the same point lie in the same plane.

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B

F

HYPOTHESIS. AB any line 1 BD and BE. BC any other line AB.

CONCLUSION. BC is in the plane BDE.

PROOF. For if not, let the plane passing through AB, BC, cut the plane BDE in the line BF. Then AB, BC, and BF are all in one plane; and because AB 1 BD and BE, therefore, by 567,

AB 1 BF.

But, by hypothesis, AB 1 BC;

therefore, in the plane ABF we have two lines BC, BF, both LAB at B, which, by 105, is impossible;

.. BC lies in the plane DBE.

571. COROLLARY. If a right angle be turned round one of its arms as an axis, the other arm will generate a plane; and when this second arm has described a perigon, it will have passed through every point of this plane.

572. Through any point D without a given line AB to pass a plane perpendicular to AB. In the plane determined by the line AB and the point D, draw DB 1 AB; then revolve the rt. X ABD about AB.