angled triangle having one of its acute angles double the other: shew that the hypotenuse is double the shorter side. Ex. 87. Draw a straight line parallel to the base of a given triangle, so that the portion intercepted between the sides may be equal to a given straight line. PROB. 11. To construct a triangle, when two angles and a side opposite to one of them are given. Let A, B be the given angles, CD the given side opposite to the angle B : it is required to construct a triangle having two of its angles equal to A and B, and the side opposite to the angle equal to B equal to CD. At the point C, in the straight line CD, make the angle DCE equal to the angle A; Prob. 5. Prob. 5. at the point C, in the straight line CE, make the angle ECF equal to the angle B; through the point D draw DG parallel to CF meeting CE at G : Prob. 6. GCD shall be the triangle required. For, because DG is parallel to CF, therefore the angle DGC is equal to the alternate angle GCF, I. 23. 90. Through a given point O within given angle BAC draw a straight line BOC, such that BC may be bisected at O. 91. Through a given point O without a given angle BAC draw a straight line OBC, such that OB may be equal to BC. 92. Through a given point O within a given angle BAC draw a straight line BOC, such that BO may be double OC. 93. Through a given point O without a given angle BAC draw a straight line OBC, such that OB may be double BC. 94. Given a point, a closed curve which incloses the point, and a straight line outside the curve, draw a straight line terminated by the point and the straight line, and bisected by the curve. 95. Find a point in a given straight line, such that straight lines drawn from it to two given points on the same side of the given straight line may be equally inclined to the given straight line. 96. Construct a triangle having given the base, an angle djacent to the base, and the sum of the remaining sides. 97. Construct a triangle having given the base, an adjacent angle, and the difference of the remaining sides. 98. Construct a triangle having given the base, the opposite angle, and the sum of the remaining sides. 99. Construct a triangle having given the base, the opposite angle, and the difference of the remaining sides. 100. Construct a triangle having given the base, the difference of the base angles, and the difference of the remaining sides. 101. Construct a triangle having given the perimeter and the angles. 102. Construct a right-angled triangle, having given the hypotenuse and the perpendicular upon it from the right angle. SECTION V. PLANE LOCI. It may happen that the conditions given for the determination of a point do not suffice to fix its position absolutely, but are sufficient to limit it to some line, part of a line, or group lines. In such cases the point is said to have a locus. DEF. 44. If any and every point on a line, part of a line, or group of lines (straight or curved), satisfies an assigned condition, and no other point does so, then that line, part of a line, or group of lines, is called the locus of the point satisfying that condition. Hence, in order that a line or group of lines X may be properly termed the locus of a point satisfying an assigned condition A, it is necessary and sufficient to demonstrate the following pair of associated theorems : If a point satisfies A, it is upon X, (1) (2) If a point is not upon X, it does not satisfy A, and instead of (2) that If a point does not satisfy A, it is not upon X. LOCUS i. To find the locus of a point at a given distance from a given point. Let A be the given point, B the given distance: it is required to find the locus of a point at distance B from A. B A P With centre A, and with radius equal to B, draw the circle CDE: the circumference CDE shall be the locus required. Take any point P on the circumference CDE, join AP, then AP is a radius of the circle CDE, and is therefore equal to B. Take any point Q not in the circumference CDE, Def. 41. join AQ, and let AQ, produced if necessary, meet the circumference in R, since Q and R do not coincide, AQ is not equal to AR, Ax. a. but AR is equal to B, Def. 41. therefore AQ is not equal to B. Hence the circumference CDE is the locus of a point at a distance from A equal to B. LOCUS ii. To find the locus of a point at a given distance from a given straight line. Let D be the given distance, and AB the given straight line : it is required to find the locus of a point at the distance D from AB. Let Q be a point at the distance D from AB. Draw AP perpendicular to AB on the same side of AB as Q, and equal to D. Draw a straight line through P and Q, and draw QM perpendicular to AB. Then QM is equal to D, and therefore to PA, and it is perpendicular to AB, and therefore parallel to PA. Hence PQ is parallel to AM, I. 22 Cor. I. 31. therefore Q lies upon the straight line drawn through P parallel to AB. AB. Also every point on this line is at the same distance from |