will come on to the road again by the other gate. To do this with the least amount of walking possible, where must he strike the stream? Let A and B be the E E and then to B, the amount of walking will be the same as if B had been situated as far from the stream as at present, but on the other side of it, as at B'. Now the shortest path from A to B' is along the straight line joining them. The point E, where this line cuts CD, is the point required. Synthesis. Draw BF CD, and produce it, making BF. Join AB' and EB. B'F = :: BF = .. BE B'F, FE is common, and BFE = .. AE + EB EB′ (I. 4). To prove AE + EB < lines drawn from A and B to any other point in CD, as G. Join GB'. As above, GB′ = GB, :. AG + GB = AG + GB'. .. E is the point required. COR. BED = AEC, for each is equal to ▲ FEB’. .. ▲ EFC = ▲ADE (I. 8). This suggests the following construction: Synthesis.-Draw EF || AB (I. 31), then DF is a ; = = .. EF BD (I. 34) AD (given), and AFEC (I. 29), and DEA = 2C, 3. Given the middle points of the sides of a triangle, to construct it. Let A, B, and C be the middle points of the sides. Analysis. Suppose DEF to be the triangle required. It is a well-known proposition (Ex. 100) that the line joining the middle points of two sides of a triangle is to the third E A F B side; AC is || DF, BC || DE, and AB || EF. Synthesis. The construction suggested is to draw through A a line || BC, through B a line || AC, through C a line || AB (I. 31); .. DC, AF, and BE are parallelograms; = .. AD CB AE (I. 34); . A is the middle point of DE, and similarly for B and C. .. EDF is the ▲ required. 4. A level field is bounded on one side by a stream flowing in a straight course, and on the opposite side by a straight road, at each end of which there is a gate. A person walking along the road comes to one gate; he wants to get a drink from the stream, and will come on to the road again by the other gate. To do this with the least amount of walking possible, where must he strike the stream? Let A and B be the two gates, and CD the stream, It is required to find a point in CD, as E, so that AE + EB shall be < lines drawn from A and B to any other point in CD. In going from A to E E and then to B, the amount of walking will be the same as if B had been situated as far from the stream as at present, but on the other side of it, as at B'. Now the shortest path from A to B' is along the straight line joining them. The point E, where this line cuts CD, is the point required. Synthesis. Draw BFL CD, and produce it, making B'FBF. Join AB' and EB. B'F, FE is common, and BFE = 4 B'FE, = .. AE + EB AE + EB' AB'. = = To prove AE + EB < lines drawn from A and B to any other point in CD, as G. Join GB'. As above, GB′ = GB, .. AG + GB AG + GB'. But AG + GB'> AB' (I. 20), i.e., >AE + EB, .. AG + GB > AE + EB. . E is the point required. COR.-BED = 4 AEC, for each is equal to FEB’. II. On Loci. 1. In a given line to find a point equidistant from two given points. Let A and B be the given points, and CD the given line. To find a point in CD equidistant from A and B. Analysis. If F is the pt. required, then FAB is an isosceles ▲; at E and EF joined, FE is and if AB is bisected AB (Ex. 16). Synthesis. Join AB, bisect it at E (I. 10), and draw EFAB (I. 11). Join FA, FB. EB, and FE is each being a rt. angle, com., and L AEF = < BEF, FB (I. 4). Q.E.F. It is also plain that any point in FE, or FE produced, is equidistant from A and B, so that if in the enunciation of the problem the condition "in a given line" had been omitted, any number of points could have been found to satisfy the remaining condition. Such a problem is said to be indeterminate, and arises in this way, that the given conditions are too few to fix definitely, or determine, the size or the position of the thing required. An example of this would be to find a point when only one condition is given; as, 2. To find a point at a given distance from a given point. Describe around the given point with the given distance as radius, and any point in the circumference of the circle satisfies the given condition, and no other point in the same plane does so. 3. To find a point at a given distance from a given line. If two lines are drawn || the given line at the given distance from it, one on each side of it, all points in these lines will satisfy the given condition, and no other point in the same plane will. 4. To find a point at a given distance from a given circle. Note. The distance of a point from a circle is measured along a radius of the circle, produced if necessary. Points satisfying the required condition will be found on circles concentric with the given circle, and having radii equal to the radius of the given circle increased or diminished by the given distance. 5. To find a point equally distant from two given intersecting lines. DGA are rt. angles, ▲ DGA can be proved equal to = DAG; AD E |