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; which proves the first Definition to be accurate. Cor. 1. The perpendicular AP is shorter than any oblique line AQ; therefore it measures the true distance from the point A to the plane MN. Cor. 2. At a given point P on a plane, it is impossible to erect more than one perpendicular to that plane; for if there could be two perpendiculars at the same point P, draw through these two perpendiculars a plane, whose intersection with the plane MN is PQ; then these two perpendiculars would be perpendicular to the line PQ, at the same point, and in the same plane, which is impossible (Book I. Prop. XIV. Sch.). It is also impossible to let fall 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. PROPOSITION V. THEOREM. If from a point without a plane, a perpendicular be drawn to the plane, and oblique lines be drawn to different points, 1st. Any two oblique lines equally distant from the perpendicular will be equal. 2d. Of any two oblique lines unequally distant from the perpendicular, the more distant will be the longer. Let AP be perpendicular to the plane MN; AB, AC, AD, oblique lines equally distant from the perpendicular, and AE a line more remote: then will AB=AC=AD; and AE will be greater than AD. For, the angles APB, APC, APD, being right angles, if we suppose the distances PB, PC, a M E N PD, to be equal to each other, the triangles APB, APC, APD, will have in each an equal angle contained by two equal sides; therefore they will be equal; hence the hypothenuses, or the oblique lines AB, AC, AD, will be equal to each other. In like manner, if the distance PE is greater than PD or its equal PB, the oblique line AE will evidently be greater than AB, or its equal AD. Cor. All the equal oblique lines, AB, AC, AD, &c. terminate in the circumference BCD, described from P the foot of the M perpendicular as a centre; therefore a point A being given out of a plane, the point P at E P B which the perpendicular let fall from A would meet that plane, may be found by marking upon N 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 are equally distant from the perpendicular; for all the triangles ABP, ACP, ADP, &c. are equal to each other. PROPOSITION VI. THEOREM. If from a point without a plane, a perpendicular be let fall on the plane, and from the foot of the perpendicular a perpendicular be drawn to any line of the plane, and from the point of intersection a line be drawn to the first point, this latter line will be perpendicular to the line of the plane. Let AP be perpendicular to the plane NM, and PD perpendicular to BC; then will AD be also perpen- M dicular to BC. Take DB DC, and draw PB, PC, AB, AC. Since DB=DC, the oblique line PB PC: and with regard to the perpendicular AP, since PB= PC, the oblique line AB=AC (Prop. V. Cor.); therefore the line AD has two of its points A and D equally distant from the extremities B and C ; therefore AD is a perpendicular to BC, at its middle point D (Book I. Prop. XVI. Cor.). 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. Scholium. The two lines AE, BC, afford an instance of two lines 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 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 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 one of the points of AB. 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 the lines ED, AP, be parallel; if AP is perpendicular to the plane NM, then will ED be also perpendicular to it. Through the parallels AP, DE, pass a plane; its intersection with the plane MN will be PD; in the plane MN draw BC perpendicular to PD, and draw AÐ. By the Corollary of the preceding Theorem, BC is perpendicular to the plane APDE; therefore 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 (Book I. Prop. XX. Cor. 1.); therefore the line DE is perpendicular to the two straight lines DP, DB; consequently it is perpendicular to their plane MN (Prop. IV.). Cor. 1. 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 erected to the same plane, which is impossible (Prop. IV. Cor. 2.). Cor. 2. Two lines A and B, parallel to a third 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. The three lines are supposed not to be in the same plane; otherwise the proposition would be already known (Book I. Prop. XXII.). PROPOSITION VIII. THEOREM. If a straight line is parallel to a straight line drawn in a plane, it will be parallel to that plane. For, if the line AB, which lies in the plane ABDC, could meet the plane MN, this could only be in some D B point of the line CD, the common intersection of the two planes but AB cannot meet CD, since they are parallel; hence it will not meet the plane MN; hence it is parallel to that plane (Def. 2.). PROPOSITION IX. THEOREM. Two planes which are perpendicular to the same straight line, are parallel to each other. Let the planes NM, QP, be perpendicular to the line AB, then will they be parallel. P M A D N B For, if they can meet any where, let O be one of their common points, and draw OA, OB; the line. AB which is perpendicular to the plane MN, is perpendicular to the straight line OA drawn through its foot in that plane; for the same reason AB is perpendicular to BO; therefore OA and OB are two perpendiculars let fall from the same point O, upon the same straight line; which is impossible (Book I. Prop. XIV.); therefore the planes MN, PQ, cannot meet each other; consequently they are parallel. PROPOSITION X. THEOREM. If a plane cut two parallel planes, the lines of intersection will be parallel. Let the parallel planes NM, seq out of 09 QP, be intersected by the plane w EH; then will the lines of inter- H1-07 no section EF, GH, be parallel.oll on N For, if the lines EF, GH, lying only w in the same plane, were not par- lg lofiets allel, they would meet each other when produced; therefore, the planes MN, PQ, in which those lines lie, would also meet; and hence the planes would not be parallel. If two planes are parallel, a straight line which is perpendicular to one, is also perpendicular to the other. M |