PROPOSITION I. THEOREM.

446. If two planes cut one another their intersection is

Let MN and PQ be two planes which cut one another.

We
Te are to prove their intersection a straight line.

Let A and B be two points common to the two planes.

Draw the straight line A B.

Since the points A and B are common to the two planes, the straight line A B lies in both planes.

Now, no point out of this line can be in both planes ;

for, if it be possible, let C be such a point.

§ 428

But there can be but one plane embracing the point C and

the line A B.

.. C does not lie in both planes.

8 442

.. every point in the intersection of the two planes lies in the straight line A B.

Q. E. D

PROPOSITION II. THEOREM.

447. From a point without a plane only one perpendicular can be drawn to the plane; and at a given point in a plane only one perpendicular can be erected to the plane.

Let CD (Fig. 1) be a perpendicular let fall from the point C to the plane MN.

We are to prove that no other can be drawn from the point

C to the plane MN.

If it be possible, let C B be another to the plane M N, and let a plane PQ pass through the lines C B and C D. The intersection of P Q with the plane MN is a straight line B D.

§ 446

Now if C D and CB be both to the plane, the ▲ CBD would have two rt. 4, CBD and CDB, which is impossible.

§ 102

Let DC (Fig. 2) be a perpendicular to the plane M N at the point D.

If it be possible, let DA be another to the plane from the point D,

and let a plane PQ pass through the lines DC and D A. The intersection of P Q with the plane M N is a straight line. Now if DC and DA could both be to the plane M N at D, we should have in the plane PQ two straight lines to the line DQ at the point D, which is impossible.

$61

Q. E. D.

448. COROLLARY. A perpendicular is the shortest distance from a point to a plane.

PROPOSITION III. THEOREM.

449. If a straight line be perpendicular to each of two straight lines drawn through its foot in a plane it is perpendicular to the plane.

D

Let DC be perpendicular to each of the two lines ACA' and BC B' drawn through its foot in the plane MN.

We are to prove DC to the plane M N.

Then A B and A' B' are symmetrical with respect to C, § 426 (their extremities being symmetrical).

Through C draw any line HCH' in the plane MN.

H and H' are symmetrical,

$ 422

Then (being corresponding points in the symmetrical lines A B and A' B'). About C, the centre of symmetry, revolve A B, keeping A C and BCL to CD, until it comes into coincidence with A' B'. Then the point I will coincide with its symmetrical point H',

and DC H will coincide with, and be equal to, ≤ DCH'. .. DCH and D CH' are rt. .

.. DC is to HCH'.

$ 25

Now since DC is to any line, H CH', drawn through its

450. Oblique lines drawn from a point to a plane at equal distances from the foot of the perpendicular are equal; and of two oblique lines unequally distant from the foot of the perpendicular the more remote is the greater.

B

Let the oblique lines BC, BD, and BE, be drawn at equal distances, A C, AD, and A E, from the foot of the perpendicular BA; and let BC' be drawn more remote from the foot of the perpendicular than BC.

Ᏼ Ꭰ,

.. A BAC=▲ BAD,

(being homologous sides of equal ▲).

II. Since A C" is > A C,

451. COROLLARY 1. Equal oblique lines from a point to a plane meet the plane at equal distances from the perpendicular; and of two unequal oblique lines, the greater meets the plane at the greater distance from the perpendicular.

452. COR. 2. All equal oblique lines BC, BD, etc., drawn from a point to a plane terminate in the circumference CDE described from A as a centre with a radius equal to A C. Hence, to draw a perpendicular from a point to a plane, draw any oblique line from the given point to the plane; revolve this line. about the point, tracing the circumference of a circle in the plane, and draw a line from the point to the centre of the circle.

PROPOSITION V. THEOREM.

453. If three straight lines meet at one point, and a straight line be perpendicular to each of them at that point, the three straight lines lie in the same plane.

Let the straight line A B be perpendicular to each of the straight lines BC, B D, and B E, at B.

We are to prove BC, BD, and BE in the same plane M N.

If not, let B D and BE be in the plane MN, and BC without it; and let PH, passing through A B and B C, cut the plane MN in the straight line B H.

Now A B, BC, and B H are all in the plane P H,

and since AB is 1 to BD and BE, it is to the

plane MN,

§ 449

(if a straight line be to each of two straight lines drawn through its foot in a plane, it is to the plane).

.. A B is (a to a plane is

That is

But

to BH, a straight line in the plane MN, § 430

to every straight line in that plane drawn through

its foot).

454. COROLLARY. The locus of all perpendiculars to a given straight line at a given point is a plane perpendicular to this given straight line at the given point.

455. SCHOLIUM. In the geometry of space the term locus has the same signification as in plane geometry, only it is not limited to lines, but is extended to include surfaces.