6. Construct angles of 60°, 120°, and 30°. 7. Construct angles of 45° and 135°. SUGGESTION. Bisect a right angle and extend one side. 122. PROBLEM. Given two sides of a triangle and an angle opposite one of them, to construct the triangle. α B SOLUTION. Let A and the segments a and b be the given parts. On one side of A lay off AB = b, and let the other side be extended to some point K. With B as a center and a radius equal to a, construct arcs of circles as shown in the figure. The following cases are possible: (1) If a equals the perpendicular distance from B to AK, the arc will meet AK in but one point, and a right triangle is the solution. (2) If a <b and greater than the perpendicular, the arc cuts AK in two points, and there are then two triangles containing the given parts, as shown by the dotted lines in the figure. (3) If a > b, the arc will cut AK only once on the right of 4, and hence only one triangle will be found. Repeat this construction, making a separate figure for each case. Make the construction when A is a right angle. Are all three cases possible then? Make the construction when A is obtuse. What cases are possible then? 1. A carpenter bisects an angle A as follows: Lay off AB AC. Place a steel square so that BD CD as shown in the figure. F 2 C D = = Draw the line AD. Is this method A 2. In the triangle ABC, AC BC. The points B D, E, F are so placed that AD = BD and AF = BE. Compare DE and DF. Prove your conclusion. = 3. If in the figure 2 21 15°, and 24 = 120, find each angle of the triangle. 4. If in ▲ ABC, AC = BC, and if AC is extended to D so that AC CD, prove that = 5. In ▲ ABC, CA DB is perpendicular to AB. = CB, AD = BE. Prove ▲ ADB ≈ ▲ ABE. 6. In the triangle KLN, NM is perpendicular to KL, and KM = MN = ML. Prove that KLN is an isosceles right triangle. 7. If in the isosceles ▲ ABC a point D lies in the base, and 21: = Z2, determine whether there is any position for D such that DE = DF. 8. If the bisectors AD and BE of the base angles of an isosceles triangle ABC meet in O, what pairs of equal angles are formed? What pairs of equal segments? Of congruent triangles? 9. If the middle points of the sides of an equilateral triangle are connected as shown in the figure, compare the resulting four triangles. 10. Triangle ABC is equilateral. AD = BE = CF. Compare the triangles DBE, ECF, FAD. 11. Two railway tracks cross as indicated in the figure. What angles are equal and what pairs of angles are supplementary? State a theorem involved in each case. 12. In a ▲ ABC does the bisector of A also bisect the side BC (1) if AC = BC but AC <AB, (2) if AC = AB? A E C B = AC and if E 13. If in the triangle ABC, AB AECBA CDB. 14. If in the figure ZA = 60°; if ZA = 40°. = AB AC, find 21 if Show that whatever the B value of ZA, 21 = { 2 A + rt. 2. B 17. To cut two converging timbers by a line AB which shall make equal angles with A them, a carpenter proceeds as follows: Place two squares against the timbers, as shown in the figure, so that AO = BO. Show that AB is the required line. DETERMINATION OF LOCI. 124. THEOREM. Every point on the bisector of an angle is equidistant from the sides of the angle. Given P, any point on the bisector of the angle A, and PC and PB perpendicular to the sides. To prove that PC = PB. Proof: By the hypothesis 21= 22. Show that ≤3 = ≤4 and complete the proof. 3 P 125. THEOREM. If a point is equally distant from the sides of an angle, it lies on the bisector of the angle. Given an angle A, any point P and the perpendiculars PB and PC equal. To prove that PA bisects the angle A. Proof: Give the argument in full to show that ΔΑΒΡΩ ΔΑΡC and thus show that 1= ≤2. Hence AP is the bisector of the angle 4. Ᏼ 3 P 126. The two preceding theorems enable us to assert the following: (1) Every point in the bisector of an angle is equidistant from its sides. (2) Every point equidistant from the sides of an angle lies in its bisector. For these reasons the bisector of an angle is called the locus of all points equidistant from its sides. The word locus means place or position. It gives the location of all points having a given property. 127. All points in a plane which satisfy some specified condition, as in the case preceding, will in general be restricted to a certain geometric figure. This figure is called the locus of the points satisfying the required condition, provided: (1) Every point in the figure possesses the required property. (2) Every point in the plane which possesses the required property lies in the figure. 128. THEOREM. The locus of all points equidistant from the extremities of a given line-segment is the perpendicular bisector of the segment. Given the perpendicular bisector / meeting the segment AB at D. To prove that (a) every point in 7 is equidistant from A and B; (b) every point which is equidistant from A and B lies in 7. Proof: (a) Let P be any point in l. Draw PA and PB. Α (§ 112) (b) Let P be any point in the plane such that PA = PB. Bisect APB with the line PD. ADP = ▲ BDP. Now show that A ADP≈ ▲ BPD. And hence that AD = BD and Hence PD is the perpendicular bisector of AB, and since there is only one such, this is the line 7 (§ 67). Thus the perpendicular bisector of the segment fulfills the two requirements for the locus in question. Steps (a) and (b) together show that a point not on the line is unequally distant from A and B. |