the parallels AB, CD. But the lines AE, EF, determine this plane; therefore, so do the parallels, AB, CD.

PROPOSITION III. THEOREM.

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

Let the two planes AB, CD, cut one another, and let E and F be two points of their common section. Draw the straight line EF. This line lies wholly in the plane AB, and also, wholly in the plane CD (B. I., D. 9): therefore, it is in both planes at once. But

B

A

E

D

since a straight line and a point out of it cannot lie in two planes at the same time (P. II., C. 1), EF contains all the points common to both planes, and consequently, is their common intersection.

PROPOSITION IV. THEOREM.

If a straight line be perpendicular to two straight lines at their point of intersection, it will be perpendicular to the plane of those lines.

Let MN be the plane of the two lines BB, CC, and let AP be perpendicular to each of them at their point of intersection P; then will AP be perpendicular to every line of the plane passing through P, and consequently to the plane itself (D. 1).

The triangle BAC in like manner gives,

AC2+AB2=2AQ2+2QC2.

Taking the first of these equals from the second, and observing that the triangles APC, APB, being right-angled at P, give

AC-PCAP, and AB'-PBAP2,

we shall have,

2

AP2+AP2

2AQ'—2PQ®.

Therefore, by taking the halves of both, we have

AP2AQ-PQ, or AQ2== AP2+PQ2;

hence, the triangle APQ is right-angled at P; hence, AP is perpendicular to PQ.

Scholium. Thus, it is evident, not only that a straight line may be perpendicular to all the straight lines which pass through its foot, in a plane, but that it always must be so, whenever it is perpendicular to two straight lines drawn in the plane: hence, a line and plane may fulfil the conditions of the first definition.

Cor. 1. The perpendicular AP is shorter than any oblique line AQ; therefore, it measures the shortest distance from the point A to the plane MN.

Cor. 2. At a given point P, on a plane, it is impossi ble to erect more than one perpendicular to the plane; for, if there could be two perpendiculars at the same point P, draw through these two perpendiculars a plane, whose section with the plane MN is PQ; then these two perpen

diculars would be both perpendicular to the line PQ, at the same point, which is impossible (B. I., P. 14, c.)

It is also impossible to let fall from a given point, out of a plane, two perpendiculars to that plane; for, if AP, AQ, be two such perpendiculars, the triangle APQ will have two right angles APQ, AQP, which is impossible (B. I., P. 25, c. 3).

PROPOSITION V. THEOREM.

If, from a point without a plane, a perpendicular be drawn to the plane, and oblique lines be drawn to its different points: 1st. The oblique lines which meet the plane at points equally distant from the foot of the perpendicular, are equal:

2d. Of two oblique lines which meet the plane at unequal dis tances, the one passing through the remote point is the longer.

Let AP be perpendicular to the plane MN; AB, AC, AD, oblique lines intercepting the equal distances PB, PC, PD, and AE a line intercepting the larger distance PE: then will AB=AC=AD; and AE will be greater than AD.

For, the angles APB, APC, APD, being right angles, and the distances PB, PC, PD, equal to each other, N the triangles APB, APC, APD, have in each an equal angle contained by two equal sides: therefore they are equal (B. I., P, 5); hence, the hypothenuses, or the oblique lines AB, AC, AD, are equal to each other.

P

BE

M

Again, since the distance PE is greater than PD, or its equal PB, the oblique line AE is greater than AB, or its equal AD (B. 1, p. 15).

Cor. All the equal oblique lines, AB, AC, AD, &c., terminate in the circumference BCD, described from P, the foot of the perpendicular, as a centre; therefore, a point A being given out of a plane, the point P at which the per

pendicular let fall from A would meet that plane, may be found by marking upon that plane three points, B, C, D, equally distant from the point A, and then finding the centre of the circle which passes through these points; this centre will be P, the point sought.

Scholium. The angle ABP is called the inclination of the oblique line AB to the plane MN; which inclination is evidently equal with respect to all such lines AB, AC, AD, as make equal angles with the perpendicular; for, all the triangles ABP, ACP, ADP, &c., are equal to each other.

PROPOSITION VI. THEOREM.

If from the foot of a perpendicular a line be drawn at right angles to any line of a plane, and the point of intersection be joined with any point of the perpendicular, this last line will be perpendicular to the line of the plane.

Let AP be perpendicular to the plane NM, and PD perpendicular to BC; join D with any point of the perpendicular, as A; then will AD also be perpendicular to BC.

Take DB= DC, and draw PB, PC, AB, AC. Now, since DB is equal to DC, the oblique line PB is equal to PC (B. 1, p. 5): and since PB is equal to PC, the oblique line AB is equal to AC (P. 5); therefore, the line AD has two of its points A and D equally distant from the extremi

M

A

C

P

D

N

Ε

ties B and C; therefore, AD is a perpendicular to BC, at its middle point D (B. I., P. 16, c.)

Cor. It is evident, likewise, that BC is perpendicular to the plane of the triangle APD, since it is perpendicu lar to the two straight lines AD, PD of that plane (P. 4).

Scholium 1. The two lines AE, BC, afford an instance of two lines which are not parallel, and yet do not meet, because they are not situated in the same plane. The short

est distance between these lines is the straight line PD, which is at once perpendicular to the line AP and to the line BC. The distance PD is the shortest distance between them because, if we join any other two points, such as A and B, we shall have AB>AD, AD>PD; therefore, still more, AB>PD.

M

P

E

Scholium 2. The two lines AE, CB, though not situated in the same plane, are conceived as forming a right angle with each other; because AE and the line drawn through any one of its points parallel to BC, would make with each other a right angle. In the same manner, AB, PD, which represent any two straight lines not situated in the same plane, are supposed to form with each other the same angle, as would be formed by AB and a straight line drawn through any point of AB, parallel to PD.

PROPOSITION VII. THEOREM.

If one of two parallel lines be perpendicular to a plane, the other will also be perpendicular to the same plane.

Let ED, AP, be two parallel lines; if AP is perpendicular to the plane NM, then will ED be also perpendic ular to it.

For, through the parallels AP, DE, pass a plane; its intersection with the plane MN will M be PD; in the plane MN draw BD perpendicular to PD, and then draw AD.

Now, BD is perpendicular to the plane APDE (P. 6, c.) there

A

Ε

P

B

C

N

fore, the angle BDE is a right angle; but the angle EDP is also a right angle, since AP is perpendicular to PD, and DE parallel to AP (B. I., P. 20, c. 1); therefore, the line DE is perpendicular to the two straight lines DP, DB; consequently it is perpendicular to their plane MN (P. 4).