III. (1) [P. P.] (3) . . . m — o = n —p, [§ 14, 2.] or C = B. [S 14, 7.] EXERCISES. Q. E. D. 7. Can you prove that an equilateral▲ is equiangular? = 8. Given the square A B C D. Draw B D, and make DN BK. Draw arc E F with center A and radius A E. Prove (1) A BD ZA DB; (2) ZA KN=ZANK; (3) AEF=/ AFE. ZA = = 9. In the second figure, A B is diameter of O; AOC F is a square erected on radius A O; B D is a chord, and E its middle point. Prove, (1)A C=CB; (2) ZO AC=/OCA; (3) A C B sum of the /s C A B and C BA; (4) ZOB D =ZODB; (5) The sum of the SOC B and ODB=/ CBD. 10. Construct A B C an equilateral, and A B D an isosceles A, on the same base, A B. Prove the CAD = ZC B D, whether the As are on the same side or on opposite sides of A B. (1) Join D and C. Produce D C, if necessary, until A B is cut. Is A B bisected? Prove. Make further deductions. 11. If x and y are the middle points of the equal sides A B and A C, respectively, of the isosceles AA BC, prove in two ways that C x By. Make deductions. E and F are points in the base B C, aud BEC F. Prove A E = A F. 42. PROPOSITION IV. Draw two acute-angled As, A B C and D E F, so that A C =DF, A B=DE, B C = E F. Place DEF on ▲ A B C so that E F falls on B C and vertex D falls opposite to vertex A. Join A and D. How many isosceles As are formed? Can you prove D equal to / A? What follows? Draw again, making s at E and B obtuses. follows? Write a general statement and call it Prop. IV. What 43. Definition. A right bisector of a line meets it at right angles and bisects it. 44. PROPOSITION V. Draw any straight line, A B, and fix two points, P and Q, equidistant from the ends of A. B. Do these two points determine the right bisector of A B? Draw P Q and if necessary produce it to meet A B at C. Can you prove the angles at C are right angles? What is a straight angle? a right angle? Can you prove A B is divided into two equal parts? Can you state this proposition? If you fail to state Prop. V., or to prove it, do not get discouraged (the proof is difficult for the beginner), but carefully study the "hints" given below. Master each step before reading the next, and just so soon as you discover the proof, finish it without reading further until after you have carefully prepared your own proof in full. Ability to originate a single step in a demonstration will give you power with which to attack future demonstrations. Struggle, repeatedly and with determination, for this power to originate. It is the highest order of mental achievement. [Hint.-Let A B be the given line and let P and Q be two points equidistant from the ends.] Το prove that P Q, or PQ produced, is the rt. bisector of A B. B Let P Q intersect A B at C. Think of the definition of a rt. bisector. We must prove what lines are equal to prove that A B is bisected? We have learned to prove lines equal in what ways? If B C is superposed on C A, can we prove that the lines must coincide and are consequently equal? If not, can we prove B C and CA similar sides of equal figures, and are consequently equal? Draw two lines which will form two triangles of which B C and CA are sides, respectively. Do you know any sides and angles of these triangles equal? Are you able to prove the triangles equal? (Think of the previous propositions you have proved, and just what is necessary to prove the triangles equal by using Prop. I., or Prop. II., or Prop. IV.) If you do not know sufficient sides and angles of these triangles equal to prove the triangles equa by either of the previously proved propositions, you must form other triangles and try to prove them equal. (Note well just what was lacking to prove the above triangles equal-was it not the angles at P, or at Q?) If you need the angles at P equal in order to prove AACP ABC P, by joining other points you will have two new triangles which contain these angles at P. Now if you can prove the new triangles, A P Q and B PQ, equal (redraw figure on new page if necessary), you can then prove angles at P equal, and then the triangles ACP and B C P equal, and finally the sides B C and C A equal. [The pupil must understand each step in the order presented. Review until you clearly see just why each step is necessary.] Then B C C A, [P. P.] and A B = B C + CA; [Ax. ?] ... A B is bisected by P Q, and PQ is the bisector of A B. But it is required to prove that points P and Q determine the right bisector; hence it is necessary to prove angles A C P and B C P right angles. What are right angles? (See the definition.) What is the sum of two adjacent angles at C? Can you prove them equal? Carefully prove Prop. V. Compare with proof given to Prop. III. if necessary. EXERCISES. 12. Letter the intersection of A D and B C, in § 42, O. What pairs of equals in the figure? Give the equal homologous or corresponding parts resulting. What angles at O are equal? 13. If 2 oblique lines are drawn from the same point in a perpendicular cutting off equal distances from the foot of the 1, how are these two lines related? Problem. 45. PROPOSITION VI. To bisect a given straight line. [What proposition treats of the bisection of a line? What is necessary to bisect it? How can you find the necessary points?] Given: A B a straight line. Required: To bisect A B. Construction: [Let the pupil construct the figure.] (1) With centers A and B and any radius greater than one-half of A B, describe arcs which intersect at C and D. (2) Draw CD, cutting A B at E. (2) Also A D = BD; [?] (3) ... A B is bisected by C D at E. [?] Q. E. F. Problem. 46. PROPOSITION VII. To bisect a given angle. [Draw an angle and bisect it. The construetion has been learned in Constructional Geometry, but it is now required to prove that the angle has been bisected. You are required to prove angles equal; review the propositions in which this is done. By joining fixed points, construct two triangles which contain the angles. Prove these triangles equal. There are two pairs of triangles which may be drawn.] Given: The angle A B C. Required: To bisect A B C. N. B.— Construction: [Let the pupil construct the figure.] (1) With center B and any radius less than B A or B C, |