EXERCISES. 7. If, in the figure of Prop. IV., ▲ AOD = 137°, how many degrees are there in BOC? in AOC? in BOD? 8. Two angles are supplementary, and the greater is seven times the less. How many degrees are there in each angle? PROP. V. THEOREM. 41. If a perpendicular be erected at the middle point of a straight line, I. Any point in the perpendicular is equally distant from the extremities of the line. II. Any point without the perpendicular is unequally distant from the extremities of the line. I. Given line CD 1 to line AB at its middle point D, E any point in CD, and lines AE and BE. Proof. Superpose figure BDE upon figure ADE by folding it over about line DE as an axis. [But one str. line can be drawn between two points.] .. AE = BE. (Ax. 3) II. Given line CDL to line AB at its middle point D, Fany point without CD, and lines AF and BF. Proof. Let AF intersect CD at E, and draw line BE. BE+EF> BF. [A str. line is the shortest line between two points.] (Ax. 4) [If a be erected at the middle point of a str. line, any point in the is equally distant from the extremities of the line.] (§ 41, I) Substituting for BE its equal AE, we have AE + EF > BF, or AF> BF. 42 Cor. I. Every point which is equally distant from the extremities of a straight line, lies in the perpendicular erected at the middle point of the line. 43. Cor. II. Since a straight line is determined by any two of its points (§ 18), it follows from § 42 that Two points, each equally distant from the extremities of a straight line, determine a perpendicular at its middle point. 44. Cor. III. When figure BDE is superposed upon figure ADE, in the proof of § 41, I., EBD coincides with ▲ EAD, and BED with AED. That is, EAD = ZEBD, and AED = / BED. Then, if lines be drawn to the extremities of a straight line from any point in the perpendicular erected at its middle point, 1. They make equal angles with the line. 2. They make equal angles with the perpendicular. PROP. VI. THEOREM. 45. From a given point without a straight line, a perpendicular can be drawn to the line, and but one. Given point C without line AB. To Prove that a can be drawn from C to AB, and but one. Proof. Let line HK be to line FG at H. [At a given point in a str. line, a to the line can be drawn.] (§ 25) Apply line FG to line AB, and move it along until HK passes through C; let point H fall at D, and draw line CD. Then, CD is 1 AB. If possible, let CE be another 1 from C to AB. Produce CD to C', making C'D = CD, and draw line EC". By cons., ED is to CC' at its middle point D. ../CED = / C'ED. [If lines be drawn to the extremities of a str. line from any point in the erected at its middle point, they make equal ▲ with the 1.] CED is a rt. Z; :: ≤ CED + ≤ C'ED But by hyp., Then line CEC" is a str. line. (§ 44) then, ≤ C'ED is a rt. ≤. two rt. s. [If the sum of two adj. is equal to two rt. 4, their ext. sides lie in the same str. line.] (§ 37) (Ax. 3) that But this is impossible, for, by cons., CDC" is a str. line. [But one str. line can be drawn between two points.] Hence, CE cannot be AB, and CD is the only can be drawn. PROP. VII. THEOREM. 46. The perpendicular is the shortest line that can be drawn from a point to a straight line. Given CD the from point C to line AB, and CE any Proof. Produce CD to C', making CD CD, and draw line EC'. By cons., ED is L to CC at its middle point D. [If a be erected at the middle point of a str. line, any point in the is equally distant from the extremities of the line.] [A str. line is the shortest line between two points.] (§ 41) (Ax 4.) Substituting for DC' and EC" their equals CD and CE, respectively, we have 2 CD <2 CE. .. CD < CE. 47. Sch. The distance of a point from a line is understood to mean the length of the perpendicular from the point to the line. Ex. 9. Find the number of degrees in the angle the sum of whose supplement and complement is 196°. PROP. VIII. THEOREM. 48. If two lines be drawn from a point to the extremities of a straight line, their sum is greater than the sum of two other lines similarly drawn, but enveloped by them. Given lines AB and AC drawn from point A to the extremities of line BC; and DB and DC two other lines similarly drawn, but enveloped by AB and AC. [A str. line is the shortest line between two points.] Adding EC to both members of the inequality, (Ax. 4) BA+AC > BE + EC. Again, DE+EC> DC. Adding BD to both members of the inequality, BE+EC> BD+DC. Since BA + AC is greater than BE + EC, which is itself greater than BD + DC, it follows that AB+ AC > DB + DC. EXERCISES. 10. The straight line which bisects an angle bisects also its vertical angle. (If OE bisects ZAOC, and these are equal to respectively.) ZAOE = LCOE; E A Ο B X |