PROP. XXII. PROB.-To describe a triangle of which the sides shall be equal to three given straight lines; but any two of these must be greater (I. 20) than the third. Let A, B, C be three given straight lines, of which any two are greater than the third: it is required to make a triangle of which the sides shall be equal to A, B, C, each to each. From E D G H K Take an unlimited straight line DE, and let F be a point in it, and make (I. 3) FG equal to A, FH to B, and HK to C. the centre F, at the distance FG, describe (I. post. 3) the circle GLM, and from the centre H, at the distance HK, describe the circle KLM. Now, because (hyp.) FK is greater than FG, the circumference of the circle GLM, will cut FE between F and K, and therefore the circle KLM cannot lie wholly within the circle GLM. In like manner, because (hyp.) GH is greater than C M HK, the circle GLM cannot lie wholly within the circle KLM. Neither can the circles be wholly without each other, since (hyp.) GF and HK are together greater than FH. The circles must therefore intersect each other: let them intersect in the point L, and join LF, LH; the triangle LFH has its sides equal respectively to the three lines A, B, C. Because F is the centre of the circle GLM, FL is equal (I. def. 30) to FG; but (const.) FG is equal to A: therefore (I. ax. 1) FL is equal to A. In like manner it may be shown that HL is equal to C; and (const.) FH is equal to B: therefore the three straight lines LF, FH, HL are respectively equal to the three A, B, C: and therefore the triangle LFH has been constructed, having its three sides equal to the three given lines, A, B, C: which was to be done.* PROP. XXIII. PROB.-At a given point in a given straight line, to make a rectilineal angle equal to a given one. Let AB be the given straight line, A the given point in it, and C the given angle: it is required to make an angle at A, in the straight line AB, that shall be equal to C. *It it evident that if MF, MH were joined, another triangle would be formed, having its sides equal to A, B, C. It is also obvious that in the practical construction of this problem, it is only necessary to take with the compasses FH equal to B, and then, the compasses being opened successively to the lengths of A and C, to describe circles or arcs from F and H as centres, intersecting in L; and lastly to join LF, LH. The construction in the text is made somewhat different from that given in Simson's Euclid, with a view to obviate objections arising from the application of this proposition in the one that follows it. In the lines containing the angle C, take any points D, E, and join them, and make (I. 22) the triangle AFG, the sides of which AF, AG, FG, shall be equal to the three straight lines CD, CE, DE, each to each. Then, because FA, AG are equal to DC, CE, each to each, and the base FG to the base DE, the angle A is equal (I. 8) to the angle C. Therefore, at the given point A in the given straight line AB, the angle A is made, equal to the given angle C: which was to be done.* АА D PROP. XXIV. THEOR.-If two triangles have two sides of the one equal to two sides of the other, each to each, but the angles contained by those sides unequal: the base of that which has the greater angle is greater than the base of the other. Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two DE, DF, each to each; but the angle BAC greater than EDF: the base BC is also greater than the base EF. Of the two sides DE, DF, let DE be that which is not greater than the other, and at the point D, in the straight line DE, make (I. 23) on the same side with EDF_the_angle EDG equal to BAC; make also DG equal (I. 3) to AC or DF, and join EG, GF.t Because AB is equal to DE, and AC to DG, the two sides BA, AC are equal to the two ED, DG, each to each; and the angle BAC is equal to the angle EDG: therefore the base BC is equal (I. 4) to the base EG. Again, because DG is equal to DF, the angles (I. 5) DFG, DGF are equal; but the angle DGF is greater D H G E *The construction of this proposition is most easily effected in practice, by making the triangles isosceles. In doing this, circular arcs are described with equal radii from C and A as centres, and their chords are made equal. It is evident that another angle might be made at A, on the other side of AB, equal to C. + Hence the angle DGE is not greater (I. 5, and 18) than DEG; but DHG is greater (I. 16) than DEG: therefore DHG is greater than DGH, and (I. 19) DG or DF is greater than DH; and consequently the point H lies between D and F, and the line EG above EF. If, on the contrary, a line equal to DE the less side, were drawn through D, making with DF, on the same side of it with DE, an angle equal to A, the extremity of that line might have fallen on FE produced, or above it or below it; and thus the proof would require three cases, while the method here given, which is that of Simson, is universally applicable. We should also have an easy proof by bisecting the angle FDG by a straight line cutting EG in a point, which call K, and joining DK and KF. For KG would be equal (I. 4) to KF; and by adding EK, we should have EG equal to EK, KF, and, therefore (I. 20) EG greater than EF. Let the student compare this proposition and the following with the fourth and eighth of this book. (I. ax. 9) than EGF: therefore the angle DFG is greater than EGF; and much more is the angle EFG greater than the angle EGF. Then, because the angle EFG of the triangle EFG is greater than its angle EGF; the side EG is greater (I. 19) than the side EF: but EG is equal to BC; and therefore also BC is greater than EF. Therefore, if two triangles, &c. PROP. XXV. THEOR.-If two triangles have two sides of the one equal to two sides of the other, each to each, but their bases unequal; the angle contained by the sides of that which has the greater base, is greater than the angle contained by the sides equal to them, of the other. Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two DE, DF, each to each; but the base BC greater than the base EF: the angle A is likewise greater than the angle D. For, if it be not greater, it must either be equal to it or less; but the angle A is not equal to D, because then (I. 4) the base BC would be equal to EF; but it is not. Neither is it less; because then (I. 24) the base BC would be less than EF; but it is not: therefore the angle A is not less than the angle D; and it has been shown, that it D Δ B C E is not equal to it: therefore, the angle A is greater than D. Wherefore, if two triangles, &c. PROP. XXVI. THEOR.-If two angles of one triangle be equal to two angles of another, each to each, and if a side of the one be equal to a side of the other similarly situated with respect to those angles; (1) the remaining sides are equal, each to each; (2) the remaining angles are equal; and (3) the triangles are equal to one another. Let ABC, DEF be two triangles which have the angles ABC, BCA equal to the angles DEF, EFD, each to each, viz. ABC to DEF, and BCA to EFD; also one side equal to one side: and, First, let those sides be equal to which the angles are adjacent, that are equal in the two triangles; viz. BC to EF: the other sides are equal, each to each, viz. AB to DE, and AC to DF; and the remaining angles BAC, EDF are equal; also the areas of the triangles are equal. G B D E F For, if AB be not equal to DE, one of them must be the greater. Let AB be the greater, and make (1.3) BG equal to DE, and join GC; then because GB is equal to DE, and (hyp.) BC to EF, the two sides GB, BC are equal to the two DE, EF, each to each; and (hyp.) the angle GBC is equal to the angle DEF: therefore the angle GCB is equal (I. 4, part. 3) to DFE. But DFE is (hyp.) equal to ACB: wherefore, also the angle GCB is equal to ACB, a part to the whole, which (I. ax. 9) is impossible: therefore AB is not unequal to DE, that it, it is equal to it. Then, since BC is equal to EF, the two AB, BC are equal to the two DE, EF, each to each; and the angle ABC to DEF: therefore (I. 4) the base AC is equal to the base DF, the triangle ABC to the triangle DEF, and the third angle BAC to the third angle EDF. A A B H C F Next, let the sides which are opposite to equal angles in each triangle be equal to one another, viz. AB to DE; likewise in this case the other sides are equal, viz. BC to EF, and AC to DF; also the triangle ABC to DEF, and the third angle BAC to the third angle EDF. For, if BC be not equal to EF, let BC be the greater, and make (I. 3) BH equal to EF, and join AH. Then, because BH is equal to EF, and (hyp.) AB to DE; the two AB, BH are equal to the two DE, EF, each to each; and they contain equal angles: therefore (I. 4, part 3) the angle BHA is equal to EFD; but EFD is equal (hyp.) to BCA: therefore also the angle BHA is equal to BCA, that is, the exterior angle BHA of the triangle AHC is equal to its interior and remote angle BCA, which (I. 16) is impossible: wherefore BC is not unequal to EF, that is, it is equal to it. Then, since AB is equal (hyp.) to DE, and the angle B to the angle E: therefore (I. 4) the base AC is equal to the base DF, the third angle BAC to the third angle EDF, and the triangle ABC to the triangle DEF.* Therefore, if two triangles, &c. *By the fifth corollary to the 32d proposition of this book, if two angles of one triangle be equal to two angles of another, their remaining angles are also equal. This proposition, therefore, if it be postponed till the student has learned the 32d, may be demonstrated in a single case, the same as either the first or second in Euclid's method, given above; and it may be so postponed without impropriety, as the 32d does not depend on it, either directly or indirectly. If the proposition be taken thus, the enunciation may be as follows: If two triangles be equiangular to one another, and if a side of the one and a side of the other, which are opposite to equal angles, be equal; then (1) the remaining sides are equal, each to each, viz. those which are opposite to equal angles; and (2) the triangles are equal. This is the third of the propositions in which triangles are demonstrated to be in every respect equal. They are proved to be such in the fourth proposition, when two sides and the contained angle of the one are respectively equal to two sides and the contained angle of the other: in the eighth, when the three sides of the one are equal to the three sides of the other, each to each: and in the twenty-sixth, when two angles and a side of the one are respectively equal to two angles and a side, similarly situated, of the other. There is only one other case in which two triangles are in all respects equal; that is, when a side and the opposite angle of one of them are respectively equal to a side and the opposite angle of the other, and a second side in the one equal to a second side in the other, the angles opposite to these latter sides being of the same kind; that is, both right angles, both acute, or both obtuse. This proposition, which is a case of the 7th of the sixth book, is of little use, except that it completes the theory; but it may be a proper exercise for the student to prove it. It will be seen from these remarks, that two triangles are in every respect equal, when of the sides and angles any three, except the three angles in one of them, are B PROP. XXVII. THEOR.-If a straight line falling upon two other straight lines, in the same plane, make the alternate angles equal to one another, these two straight lines are parallel. Let the straight line EF, which falls upon the two straight lines AB, CD, in the same plane, make the alternate angles, AEF, EFD, equal to one another; AB is parallel to CD. A C H Б B G F D For, if it be not parallel, AB and CD, being produced, will meet either towards B, D, or towards A, C: let them be produced and meet towards B, D, in the point G. Therefore GEF is a triangle, and its exterior angle AEF is greater (I. 16) than the interior and remote angle EFG; but (hyp.) it is also equal to it, which is impossible: therefore, AB and CD, being produced, do not meet towards B, D. In like manner, it may be demonstrated, that they do not meet towards A, C. But those straight lines which are in the same plane, and which meet neither way, though produced ever so far, are parallel (I. def. 11) to one another: AB therefore is parallel to CD.* Wherefore, if a straight line, &c. K PROP. XXVIII. THEOR.-Two straight lines are parallel to one another, (1) if a straight line falling upon them make the exterior angle equal to the interior and remote, upon the same side of that line; and (2) if it make the interior angles upon the same side together equal to two right angles. Let the straight line EF, which falls upon the two straight lines, AB, CD, make the exterior angle EGB equal to the interior and remote angle GHD upon the same side of EF, or make the interior angles on the same side, BGH, GHD, together equal to two right angles; AB is in each case parallel to CD. respectively equal to the corresponding sides or angles, in the other, with only the one limitation in the proposition just enunciated, that in it the angles opposite to two equal sides must be of the same kind: and from this it appears, that of the sides and angles of a triangle, three must be given to determine the triangle, and these three cannot be the three angles. Were only the three angles given, the sides, as will appear from the 32d proposition of this book, might be of any magnitudes whatever. The student may exercise himself in proving this proposition by superposition; and also by producing BA and BC, instead of cutting off parts of them. * In this proof ABG and CDG must be regarded as straight lines. It is not necessary, indeed, to make the actual construction here given; for the student will see that AB and CD cannot meet on either side, since, if they did, a triangle would thus be formed, and the exterior angle would be (I. 16) greater than the interior and remote angle, which is contrary to the hypothesis. The alternate angles as they are understood by Euclid, or the interior alternate angles, as they might be called with perhaps more propriety, are the interior remote angles on opposite sides of the line which falls on the other two. AEH and KFD, and also BEH and CFK, may be called exterior alternate angles; and it is easy to prove that if these be equal, the lines are parallel. |