PROP. XXI. THEOREM. If from the ends of the side of a triangle two straight lines be| drawn to a point within the triangle, these shall be together less than the other two sides of the triangle, but they shall contain a greater angle. Let ABC be a triangle, and from B and C, the ends of the side BC, let the two straight lines BD and CD be drawn to a point D within the triangle. Then it is to be proved that 1. The two lines BD and DC are together less than the two sides BA and AC; but 2. The angle BDC is greater than the angle BAC. CONSTRUCTION.-Produce BD to meet AC in E. PROOF.-Because in the triangle BAE, the sides BA and AE are together greater than BE (I. 20), add to each of these EC; then the sides BA, AE, and EC, i.e. the sides BA and AC, are together greater than the sides BE and EC (ax. 4). Again, because in the triangle CED the sides CE and ED are together greater than CD (I. 20), add to each of these DB, then the sides CE, ED, and DB, i.e. the sides CE and EB, are together greater than the sides CD and DB (ax. 4). But we have already proved that the sides BA and AC are together greater than the sides CE and EB; and therefore the sides BA and AC are together much greater than the sides CD and BD. Therefore, it is proved, as required, that 1. The two lines BD and DC are together less than the two sides BA and AC. Next, because the exterior angle BDC of the triangle DCE is greater than the interior and opposite angle CED (I. 16), i.e. the angle CEB (note 2 def. 15), and also the exterior angle CEB is greater than the interior and opposite angle BAE (I. 16), i.e. the angle BAC, Therefore, it is proved, as required, that 2. The angle BDC is greater than the angle BAC. Wherefore, If from the ends, &c. Q. E. D. Exercises. 1. If from a point A two straight lines, AB and AC, be let fall upon another straight line ED, the line AB being perpendicular to ED; prove that AB subtends, or is opposite to, an acute angle. 2. Prove, in Prop. XX., that, as stated, the sides AB and BC are together greater than the side AC. 3. If a point A and a straight line BC be given; prove that the shortest line that can be drawn from A to BC, say AD, is perpendicular to BC. PROP. XXII. PROBLEM. To make a triangle of which the sides shall be equal to three given straight lines, any two of which are together greaterthan the third. Let A, B, and C be three given lines any two of which are together greater than the third. It is required to make a triangle having its sides A, B, and C, each to each. CONSTRUCTION.--1. Take a straight line DE, terminated at D, but unlimited towards E. 2. In this line make DF: = A, make FG = B, and GH = C (I. 3). 3. From centre F with radius FD describe the circle DKL. 4. From centre G with radius GH describe the circle HKL. 5. Join KF and KG. Then it is to be proved that The triangle KFG is the triangle required. PROOF. Because F is the centre of the circle DKL, therefore FK FD (def. 15), and FD = A (cons.), therefore FK A (ax. 1). = Again, because G is the centre of the circle HKL, therefore GK GH (def. 15), and GHC (cons.), therefore GK = C (ax. 1); also FG B (cons.), therefore, the triangle KFG has its three sides KF, FG, and GK = the three given lines A, B, and C, each to each. = Therefore, it is proved, as required, that The triangle KFG is the triangle required. Q. E. F. At a given point in a given straight line, to make a Let AB be the given straight line, A the given point in it, and DCE the given rectilineal angle. It is required to make at A a rectilineal angle: tilineal angle DCE. C the rec CONSTRUCTION.-1. In CD and CE take any points D and E and join DE. 2. Make the triangle FAG = the triangle DCE, so that the three sides FA, AG, and GF = the sides DC, CE, and ED, each to each (I. 22). Then it is to be proved that The angle FAG is the angle required. the PROOF.-Because in the triangles FAG and DCE we have the three sides FA, AG, and GF in the one three sides DC, CE, and ED in the other, each to each (cons.), therefore the angle FAG = the angle DCE (I. 8). Therefore, it is proved that The angle FAG is the angle required. Q. E. F. Exercise. If ABC be a triangle with the side AB greater than AC; prove that the difference between AB and AC is less than the side BC. PROP. XXIV. THEOREM. If two triangles have two sides of the one equal to two sides of the other, each to each, but the angles contained by these sides unequal, then their bases or third sides shall be unequal, and the base of that triangle which has the greater angle shall be greater than the base of the other. Let ABC and DEF be two triangles, with the sides AB and AC in the former = the sides DE and DF, in the latter, each to each, but with the angle BAC greater than the angle EDF. Then it is to be proved that The base BC is greater than the base EF. CONSTRUCTION.-Of the two sides DE and DF, let DE be not greater than DF. 1. At the point D in the straight line ED make the angle EDG= the angle BAC (I. 23). 2. Make DG= AC or DF, and join EG and GF. PROOF.-Because in the triangles BAC and EDG we have the sides BA and AC and their angle BAC, in the former = the sides ED and DG and their angle EDG in the latter, each to each (hyp. and cons.), therefore the base BC= the base EG (I. 4). Again, because DGDF (cons.), therefore the angle DGF = the angle DFG (I. 5). But the angle DGF is greater than the angle EGF (ax. 9); therefore the angle DFG is also greater than the angle EGF. |