Proposition 2. Theorem. 496. If oblique lines are drawn from a point to a plane: (1) Two oblique lines meeting the plane at equal distances from the foot of the perpendicular are equal. (2) Of two oblique lines meeting the plane at unequal distances from the foot of the perpendicular, the more remote is the longer. Hyp. Let OP be to the plane (2) Proof. On PC take PB = PA, and join OB. 497. COR. 1. The perpendicular is the shortest distance from a point to a plane; therefore, by the distance of a point from a plane is meant the perpendicular distance from the point to the plane. (493) 498. COR. 2. Equal oblique lines from a point to a plane meet the plane at equal distances from the foot of the perpendicular; and of two unequal oblique lines, the greater meets the plane at the greater distance from the foot of the perpendicular. 499. Cor. 3. The locus of the point in a plane at a given distance from a fixed point without the plane, is a circle whose centre is the foot of the perpendicular. Proposition 3. Theorem. 500. If a straight line is perpendicular to each of two straight lines at their point of intersection, it is perpendicular to the plane of those lines. Hyp. Let OP be to PA, PB at the pt. P. To prove OP is to the plane MN of these lines. Proof. Join AB, and through P draw in MN any other st. line PC cutting AB in C. Produce OP to O' making PO' = PO, and join O, O' to each of the pts. A, B, C. M B Since PA, PB are 1 to 00' at its mid. pt., (Hyp.) (Cons.) .. OA = O'A, and OB = O'B. (66) Δ ... ▲ OAB = ▲ O'AB, and ... ZOAC = ZO'AC. (108) having two sides and the included equal, each to each (104). ... OC = O'C, and ... PC is to OO' at its mid. pt. P. (67) .. OP is ¦ to any st. line in MN passing through its foot P. ... OP is to the plane MN. (487) Q.E.D. 501. COR. 1. At a given point in a plane, only one perpendicular to the plane can be erected. For, if there could be two is at the same pt. P, pass a plane through them whose intersection with the plane MN is AP; then these two Is would be both to the line AP at the same pt. P, which is impossible. (51) 502. COR. 2. From a given point without a plane only one perpendicular can be drawn to the plane. For, if OP, OA be two such Ls, the ▲ OPA contains two rt. 8, which is impossible. (101) Proposition 4. Theorem. 503. Conversely, all the perpendiculars to a straight line at the same point lie in a plane perpendicular to the line. Hyp. Let PA, PB be two st. lines I to OP at P, and PC any other line to OP at P. To prove PC is in the same plane with PA, PB. Proof. Let the the plane passing through PO and PC cut the plane APB in the line PC'. B A Then OP is to PC'. (48%) But in the plane OPC only one can be drawn to OP at P, (51) ... PC and PC' coincide, and PC lies in the plane APB. Q. E. D. NOTE.-Hence a plane is determined by one point and the normal to the plane at that point. 504. COR. 1. At a given point in a straight line one plane, and only one, can be drawn perpendicular to the line. 505. COR. 2. If a right angle be turned round one of its arms as an axis, the other arm will generate a plane. 506. COR. 3. Through a given point without a straight line one plane, and only one, can be drawn perpendicular to the line. For, in the plane of OP and the pt. C, the CP can be drawn to OP; then the plane generated by turning PC round OP will be to OP; and it is clear that there is only one such plane. Proposition 5. Theorem. 507. If from the foot of a perpendicular to a plane a straight line is drawn at right angles to any line in the plane, and its intersection with that line is joined to any point of the perpendicular, this last line will be perpendicular to the line in the plane. Hyp. Let OP be a to the plane MN, PA a from P to any line BC in MN, and OA a line joining A with any pt. O in OP. M. C and ... PB = PC, ... OB = OC. (66) (496) Then, since O and A are each equally distant from B and C, .. OA is to BC. (67) Q. E.D. 508. COR. 1. The line BC is perpendicular to the plane of the triangle OPA. For, it is to the st. lines AP, AO at pt. A. (500) COR. 2. The line PA measures the shortest distance between OP and BC. EXERCISES. 1. Find the locus of points equally distant from two given points. 2. Given a straight line and any two points: find a point in the straight line equally distant from the two points. Proposition 6. Theorem. 509. Two straight lines perpendicular to the same plane are parallel. Hyp. Let AB, CD be to the plane MN at the pts. B, D. M ... CD, AD, BD, are all in the same plane, all the 1s to a st. line at the same pt. lie in the same plane (503). ... AB and CD lie in the same plane, 510. COR. 1. If one of two parallel lines is perpendicular to a plane, the other is also perpendicular to that plane. C For, if AB is I to CD, and to the plane MN, then a L to MN at D will be || to AB (509), and will coincide with CD (501). ... CD is to MN. 511. COR. 2. Two straight lines that are parallel to a third straight line are parallel to each other. For, each of the lines AB, CD, is 1 to a plane MN that is to EF (510); ... AB and CD are || . (509) CE M D |