7. Then, point B will fall on point F, and line BE will coincide with line EF; that is, BEEF. 8. Hence, since points A and E are each equally distant from B and F, line AE is 1 BF. [Two points, each equally distant from the extremities of a str. line, determine a at its middle point.] (§ 56) 9. Since AE is any str. line, AB is to every str. line drawn through A in MN, and is therefore I to plane MN. [A str. line is said to be to a plane when it is to every str. line drawn in the plane through its foot.] (§ 349) PROP. V. THEOREM 353. All the perpendiculars to a straight line at a given point lie in a plane perpendicular to the line. B Given AC, AD, and AE any three Is to line AB at A. To Prove that they lie in a plane L to AB. Proof. 1. Let MN be the plane determined by AC and AD; then, plane MN is LAB. [A str. line to each of two str. lines at their point of intersection is I to their plane.] (§ 352) 2. Let the plane determined by AB and AE intersect MN in line AE'; then, AB L AE'. [AL to a plane is 1 to every str. line drawn in the plane through its 3. In plane ABE, only one L can be drawn to AB at A. (§ 15) 4. Then, AE' and AE coincide, and AE lies in plane MN. 5. Then, all the is to AB at A lie in a plane 1 AB. 354. It follows from § 353 that Through a given point in a straight line, a plane can be drawn perpendicular to the line, and but one. 355. Through a given point without a straight line, a plane can be drawn perpendicular to the line, and but one. If Cis the given point without line AB, draw line CB LAB, and BD any other to AB at B. Then, the plane drawn through BC and BD will be L AB. -C B D [A str. line to each of two str. lines at their point of intersection is I to their plane.] (§ 352) Again, every plane through CL AB must intersect the plane determined by AB and BC in a line from CL AB. [A to a plane is to every str. line drawn in the plane through its foot.] But only one I can be drawn from C to AB. (§ 350) [From a given point without a str. line, but one I can be drawn to the line.] ($ 59) Then, every plane through CLAB must contain BC, and be to AB at B. But only one plane can be drawn through BLAB. [Through a given point in a str. line, but one plane can be drawn to the line.] Hence, but one plane can be drawn through CL AB. (§ 354) Ex. 4. If we have five points, four of which are in same plane, how many planes do they determine? If no four of the five were in the saine plane, would the number of planes change? PROP. VI. THEOREM 356. If from a point in a perpendicular to a plane, oblique lines be drawn to the plane, I. Two oblique lines cutting off equal distances from the foot of the perpendicular are equal. II. Of two oblique lines cutting off unequal distances from the foot of the perpendicular, the more remote is the greater. Given line AB plane MN at B; AC and AD oblique lines meeting MN at equal distances from B, and AE an oblique To Prove ACAD, and AE > AC. Proof. 1. Draw lines BC, BD, and BE. 2. In ▲ ABC, ABD, AB= AB; and by hyp., BC= BD. [AL to a plane is to every str. line drawn in the plane through its foot.] ($ 350) [Two A are equal when two sides and the included ▲ of one are equal respectively to two sides and the included of the other.] [In equal figures, the homologous parts are equal.] (§ 46) (§ 48) 6. Again, on BE take_BF= BC, and draw line AF; then, since BF BC, AF AC. are [If from a point in a 1 to a plane, oblique lines be drawn to the plane, (§ 356, I) [AL to a plane is to every str. line drawn in the plane through its foot.] 8. Then, AE > AF, or AC. (§ 350) [If oblique lines be drawn from a point to a str. line, of two oblique lines cutting off unequal distances from the foot of the L from the point to the line, the more remote is the greater.] ($ 64, II) |