The triangle BAC will, in like manner, give AC2+AB2 = 2AQ2+2QC2. Taking the first equation from the second, and observing that the triangles APC, APB, which are both right-angled at P, give AC2 — PC2 = AP2, and AB2 – PB2 = AP2, we shall have 2AP22AQ2 - 2PQ2, or, that is, AP2 = AQ2 — PQ2; AQ2 AP2+PQ2. Hence the triangle APQ is right-angled at P, and therefore 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. Cor. At a given point P on a plane, it is impossible to draw more than one perpendicular to that plane. For, if there could be two perpendiculars at the same point P, draw along these two perpendiculars a plane, whose intersection with the plane MN is PQ; then those two perpendiculars would be perpendicular to the line PQ at the same point, and in the same plane, which is impossible. It is also impossible to draw, from a given point out of a plane, two perpendiculars to that plane. For, let AP, AQ be these two perpendiculars, then the triangle APQ would have two right angles APQ, AQP, which is impossible. THEOREM V. If from a point without a plane, a perpendicular and several oblique lines be drawn to this plane: I. The perpendicular will be shorter than any oblique line. II. Any two oblique lines which terminate at equal distances from the foot of the perpendicular, will be equal. III. Of two oblique lines, terminating at unequal distances from the foot of the perpendicular, the one at the greater distance will be the longer. Let O be the given point, MN the given plane, OP perpendicular, and OA, OA', OA", and OB oblique lines. First. Since OA is an oblique line, the right-angled triangle OPA gives OP < OA (B. I., T. XXVII.). Secondly. If PA = PA', the two right-angled triangles OPA, OPA' will be equal (B. I., T. XX., C.), and give OA' OA. Thirdly. If PB >PA, take on PB a distance PA' PA, and draw OA', and we shall have OB> OA' (B. I., T. XII.); but we have OA'=OA, consequently OB>OA. M A P A' N Cor. I. The perpendicular drawn from any point to a plane, measures the true distance of the point from the plane. Cor. II. Any point whatever in a line perpendicular to a plane, is equally distant from all the points of the circumference described about the foot of the perpendicular as a centre, with any radius whatever. Scholium I. This perpendicular is called the axis of the circle. All the points of the axis may be regarded as centres of the circle, and the oblique lines as corresponding radii, any one of which may be used for describing the circle. Scholium II. By means of an extended thread, one extremity being fixed at the point O, and the other extremity attached to a pencil, we may determine any three points, as A, A', A", equally distant from the point O. If we then describe the circumference of a circle passing through these three points, its centre will be the foot of the perpendicular drawn from O to the plane. Scholium III. When a plane passes perpendicularly through the middle of a straight line: 1. All points of this plane are equally distant from the extremities of this line. 2. All points situated without this plane are unequally distant from these extremities. THEOREM VI. If, from a point without a plane, a perpendicular be drawn to that plane, and from the foot of the perpendicular a line be drawn perpendicular to a line in the plane, and from the point of intersection a line be drawn to the first point, this last line will be perpendicular to the line in the plane. Let AP be a line perpendicular to the plane MN, and PD perpendicular to BC; then will AD be perpendicular to BC. N A M C D P B E Take DB = DC, and join PB, PC, AB, AC. Since PD is perpendicular at D, the middle point of BC, PB is equal to PC (B. I., T. XII.); consequently the oblique lines AB, AC are equal, and the two triangles ADB, ADC have the three sides of the one equal to the three sides of the other; therefore they are equal (B. I., T. XXV.), and the angle ADB = ADC; hence each is a right angle, and AD is perpendicular to BC (B. I., D. XII.). Cor. It is evident, likewise, that BC is perpendicular to the plane APD, since BC is at once perpendicular to the two straight lines AD, PD (T. IV.). Scholium. The two straight lines AE, BC afford an instance of two lines not parallel which do not meet, because they are not situated in the same plane. The shortest distance between these lines is the straight line PD, which is perpendicular both to the line AP and to the line BC. For if we join any other two points, such as A and B, we shall have AB > AD, AD > PD, therefore AB > PD. 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 one of its points parallel to BC would make with each other a right angle. In the same manner, the line AB and the line PD, which represent any two straight lines not situated in the same plane, are supposed to form with each other the same angle which would be formed by AB and a straight line parallel to PD drawn through any point of AB. THEOREM VII. If one of two parallel lines is perpendicular to a plane, the other will also be perpendicular to this plane. Let AP and ED be parallel lines, of which AP is perpendicular to the plane. MN; then will ED be also perpendicular to this plane. N D E A M P B Along the parallels AP, DE extend a plane; its intersection with the plane MN will be PD. In the plane MN draw BC perpendicular to PD, and join AD. Then BC is perpendicular to the plane APDE (T. VI., Cor.), and therefore the angle BDE is right. But the angle EDP is right also, since AP is perpendicular to PD, and DE parallel to AP (B. I., T. XV., Cor.); therefore the line DE is perpendicular to the two straight lines DP, DB, hence it is perpendicular to their plane MN (T. IV.). Cor. I. Conversely, if the straight lines AP, DE are perpendicular to the same plane MN, they will be parallel. For, if they be not so, draw through the point D a line parallel to AP; this parallel will be perpendicular to the plane MN; therefore, through the same point D more than one perpendicular might be drawn to the same plane, which (T. IV., C.) is impossible. Cor. II. Two lines, A and B, parallel to a third line C, are parallel to each other. For, conceive a plane perpendicular to the line C; the lines A and B, being parallel to C, will be perpendicular to the same plane; therefore, by the preceding Corollary, they will be parallel to each other. THEOREM VIII. If a straight line without a plane is parallel to a line in the plane, it is parallel to the plane itself. Let the straight line AB, without the plane MN, be parallel to the line CD of this plane; then will AB be parallel to the plane MN. B A M D N For, if the line AB, which lies in the plane ABDC, could meet the plane MN, this could only be in some point of the line CD, the intersection of the two planes; but AB cannot meet CD, since they are parallel; hence it will not meet the plane MN, therefore (D. IV.) it is parallel to that plane. THEOREM IX. If two planes are perpendicular to the same line, they are parallel. is perpendicular to the straight line AO drawn through its foot in that plane. For the same reason, AB is perpendicular to BO. Therefore OA and OB are two perpendiculars drawn from the same point O, upon the same straight line, which is impossible; hence the planes MN, PQ cannot meet each other, and consequently they are parallel. THEOREM X. The intersections of two parallel planes with a third plane are parallel. Let the two parallel planes MN and PQ intersect the plane EH; then will EF be parallel to HG. For, if the lines EF, GH, lying in the same plane, were not parallel, they would meet each other when produced; there E M N F G P H fore the planes MN, PQ, in which those lines are situated, would also meet, and the planes would not be parallel. |