Hint.-Prove from Corollary I. that one perpendicular OP to the plane MN can be drawn. No other line, as OP' through O can be perpendicular to MN. 536. The minimum line from a point to a plane is perpendicular to that plane. GIVEN the plane MN and the point P without it, and PO, the minimum line from P to MN. TO PROVE that PO is perpendicular to MN. In the plane MN through the point O draw any straight line AB. Since PO is the shortest line from P to the plane MN, it is the shortest line from P to the line AB in that plane. Therefore PO is perpendicular to AB. $96 That is, PO is perpendicular to any or every straight line in MN through its foot O. Therefore PO is perpendicular to MN. 537. COR. From a point without a plane one and only one perpendicular to the plane can be drawn. Hint.-Apply the Proposition and § 34. 538. Def.-By the distance from a point to a plane is meant the shortest distance, and therefore the perpendicular. distance. PROPOSITION V. THEOREM 539. If oblique lines are drawn from a point to a plane: I. Those meeting the plane at equal distances from the foot of the perpendicular are equal. II. Of two unequally distant, the more remote is the greater. I. GIVEN the oblique lines AC and AD meeting the plane MN at the equal distances BC and BD from the perpendicular AB. In the triangles ABC and ABD, AB is common; BC=BD by hypothesis; and the angles ABC and ABD are equal, being right angles. Therefore the triangles are equal, and AC=AD. $79 Q. E D. II. GIVEN the oblique lines AF and AD meeting MN so that I. Equal oblique lines from a point to a plane meet the plane at equal distances from the foot of the perpendicular. II. Of two unequal oblique lines the greater meets the plane at the greater distance from the foot of the perpendicular. Hint.-Prove as in § 100. 541. Remark.-Article 540 supplies practical methods of drawing a straight line perpendicular to a plane, as a floor or a blackboard. I. To erect a perpendicular to a plane at a given point in it. With the given point as centre, describe a circumference in the given plane. Take three strings of equal length somewhat longer than the radius of the circumference. To each of three points on the circumference attach an end of one string. Unite the three remaining ends in a knot and pull the strings taut. A line through the given point and the knot is the perpendicular required. Prove the method correct by supposing if possible that the foot of the perpendicular from the knot is not in the given point, and apply § 103. II. To draw a perpendicular to a given plane from a given point without it. From the point with a string of convenient length measure three equal distances to the plane. The centre of the circumference which passes through the three points thus found is the foot of the required perpendicular. (Why?) PARALLEL LINES AND PLANES 542. Def.-A straight line and a plane are parallel to each other if they cannot meet, however far produced. 543. Def.-Two planes are parallel to each other if they cannot meet, however far produced. 544. If two parallel planes are cut by a third plane, their intersections with this plane are parallel. GIVEN the parallel planes MN and PQ cut by the plane AD in the lines AC and BD. Since the planes MN and PQ cannot meet, the lines AC and BD lying in them cannot meet. Moreover these lines lie in the same plane AD. Therefore they are parallel. $31 Q. E. D. 545. COR. Parallel lines AB and CD intercepted between parallel planes are equal. PROPOSITION VII. THEOREM 546. If a straight line is parallel to a plane, the intersection of the plane with a plane passed through the line is parallel to the line. GIVEN the line BA parallel to the plane MN and a plane BF passing through BA and intersecting MN in EF. These lines lie in the same plane. They cannot meet, for BA cannot meet the plane MN in which EF lies. Therefore they are parallel. 31 Q. E. D. |