PROPOSITION XXXVI. PROBLEM. 315. To inscribe a circle in a given triangle. From E as centre, with radius EH, describe the O KHM. Proof. Since E is in the bisector of the A, it is equidistant from the sides AB and AC; and since E is in the bisector of the C, it is equidistant from the sides AC and BC. $ 162 ..a O described from E as centre, with a radius equal to EH, will touch the sides of the ▲ and be inscribed in it. NOTE. The point E is called the in-centre of the triangle. 316. The intersections of the bisectors of the exterior angles of a triangle are the centres of three circles, each of which will touch one side of the triangle, and the two other sides produced. These three circles are called escribed circles; and their centres are called the ex-centres of the triangle. X Q. E. F. PROPOSITION XXXVII. PROBLEM. 317. Through a given point, to draw a tangent to a given circle. A E H CASE 1. When the given point is on the circumference. Through C draw AM to OC. Then AM is the tangent required. CASE 2. When the given point is without the circle. Let O be the centre of the given circle, E the given point. Draw OE § 301 § 253 On OE as a diameter, describe a circumference intersecting the given circumference at the points M and H. Draw OM and EM. Then EM is the tangent required. ZOME is a right angle. Proof. .. EM is tangent to the circle at M. § 290 § 253 In like manner, we may prove EH tangent to the given O. Q. E. F. Ex. 146. To draw a tangent to a given circle, so that it shall be parallel to a given straight line. 318. Upon a given straight line, to describe a segment of a circle in which a given angle may be Let AB be the given line, and M the given angle. Construct the ▲ ABE equal to the ▲ M. § 305 Bisect the line AB by the OF. $ 302 § 301 From the point B draw BO 1 to EB. From O, the point of intersection of FO and BO, as a cen tre, with a radius equal to OB, describe a circumference. The segment AKB is the segment required. Proof. The point O is equidistant from A and B. § 160 .. the circumference will pass through A. But BE is 1 to OB. .. BE is tangent to the O, (a straight line 1 to a radius at its extremity is tangent to the O). .. ZABE is measured by arc AB, Const. § 253 $ 295 (being an formed by a tangent and a chord). But any as K inscribed in the segment AKB is measured by arc AB. § 289 .. the M may be inscribed in the segment AKB. Q. E. F. SOLUTION OF PROBLEMS. 319. If a problem is so simple that the solution is obvious from a known theorem, we have only to make the construction according to the theorem, and then give a synthetic proof, if a proof is necessary, that the construction is correct, as in the examples of the fundamental problems already given. 320. But problems are usually of a more difficult type. The application of known theorems to their solution is not immediate, and often far from obvious. To discover the mode of application is the first and most difficult part of the solution. The best way to attack such problems is by a method resembling the analytic proof of a theorem, called the analysis of the problem. 1. Suppose the construction made, and let the figure represent all parts concerned, both given and required. 2. Study the relations among the parts with the aid of known theorems, and try to find some relation that will suggest the construction. 3. If this attempt fails, introduce new relations by drawing auxiliary lines, and study the new relations. If this attempt fails, make a new trial, and so on till a clue to the right construction is found. 321. A problem is determinate if it has a definite number of solutions, indeterminate if it has an indefinite number of solutions, and impossible if it has no solution. A problem is sometimes determinate for certain relative positions or magnitudes of the given parts, and indeterminate for other positions or magnitudes of the given parts. 322. The discussion of a problem consists in examining the problem with reference to all possible conditions, and in deter mining the conditions necessary for its solution. Ex. 147. PROBLEM. To construct a circle that shall pass through a given point and cut chords of a given length from two parallels. Ꭰ Analysis. Suppose the problem solved. Let A be the given point, BC and DE the given parallels, MN the given length, and O the centre of the required circle. Since the circle cuts equal chords from two parallels its centre must be equidistant from them. Therefore, one locus for O is FG | to BC and equidistant from BC and DE. F B M E A G N Draw the bisector of MN, cutting FG in P. PM is the radius of the circle required. With A as centre and radius PM describe an arc cutting FG at O. Then O is the centre of the required circle. Discussion. The problem is impossible if the distance from A to FG is greater than PM. Ex. 148. PROBLEM. To construct a triangle, having given the perimeter, one angle, and the altitude from the vertex of the given angle. Analysis. Suppose the problem solved, and let ABC be the ▲ required, ACB the given 2, and CD the given altitude. Produce AB both ways, and take AE = AC, and BF=BC, then EF the given perimeter. Join CE and CF, form ing the isosceles A CAE and CBF. In the AECF, ZE+ZF+ E D/B LECF 180° (why ?), but ECF = 2 ECA +2 FCB+2ACB. = = Since ZE ZECA and F = LFCB, we have ECF = LE+LF + LACB. ..2ZE+2ZF+ZACB: = 180°. :. LE + ZF+ZACB = 90°, and ZE + ≤ F = 90° – ‡ ZACB. By substitution, ECF 90° + ≤ ACB. .. LECF is known. = Construction. To find the point C, construct on EF a segment that will contain the ECF (§ 318), and draw a parallel to EF at the distance CD, the given altitude. To find the points A and B, draw the bisectors of the lines CE and CF, and the points A and B will be vertices of the required A. Why ? |