.: :: AEFC = A ADE (1.8). This suggests the follow ing construction : Synthesis.-Draw EF || AB (1. 31), then DF is a 0; :: EF = BD (1.3.4) AD (given), and LA = _FEC (I. 29), and _DEA = LC, :: A ADE AEFC (1. 26); : AE Q.E.D. = EC. a 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 proposi tion (Ex. 100) that the A 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; CB AE (I. 34); :. A is the middle point of DE, and similarly for B and C. :: EDF is the required. E :. AD 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, a point in CD, as E D other point in CD. In going from A to 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 1 CD, and produce it, making B'F = BF. Join AB' and EB. :: BF BʼF, FE is common, and <BFE = LB'FE, :: BE EB' (1. 4). :. AE + EB = AE + EB , 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' (1. 20), i.e., >AE + EB, :. AG + GB > AE + EB. .: E is the point required. CoR.- - BED = AEC, for each is equal to Z FEB'. AB'. .. .. :: AEFC = A ADE (1.8). This suggests the follow ing construction : Synthesis.-Draw EF | AB (I. 31), then DF is a 0); :: EF = BD (1. 34) AD (given), and LA FEC (1. 29), and _DEA = LC, .: A ADE = A EFC (1. 26); :: AE EC. Q.E.D. 3. Given the middle points of the sides of a triangle, to construct it. Let A, B, and C be the middle F points of the sides. Analysis.-Suppose DEF to be the triangle required. It is a well-known proposi tion (Ex. 100) that the А 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 A 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 'ate; 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, F It is required to find a point in CD, as E other point in CD. In going from A to B 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 I CD, and produce it, making B'F BF. Join AB' and EB. :: BF = B'F, FE is common, and <BFE = _ B'FE, EB' (1.4). :. AE + EB AE + EB 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 - AEC, for each is equal to 2 FEB'. .: BE AB'. :: AEFC = AADE (1.8). This suggests the follow ing construction: Synthesis.-Draw EF || AB (I. 31), then DF is a 0); :: EF = BD (1.34) = AD (given), and LA = _FEC (I. 29), and _DEA = LC, .: A ADE = AEFC (1. 26); :: AE EC. Q.E.D. 3. Given the middle points of the sides of a triangle, to construct it. Let A,B, and C be the middle F points of the sides. Analysis.-Suppose DEF to be the triangle required. It is a well-known proposi tion (Ex. 100) that the А 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); CB AE (I. 34); :. A is the middle point of DE, and similarly for B and C. :: EDF is the A required. E 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 te; he wants to get a drink from the stream, and |