and the angle A is equal to the angle D, the side AC will fall on the side DF; and, these sides being equal, the point C will fall on the point F; and, consequently, the line BC will coincide with the line EF. Hence, the two triangles are identical, having all the parts in the one equal to the corresponding parts in the other, viz. the side BC to the side EF, the angle B to the angle E, the angle C to the angle F, and the whole triangle ABC to the triangle DEF. THEOREM II.—If two triangles, as ABC, DEF (fig. 30, p. I.), have two angles as B and C in the one, respectively equal to two angles, as E and F in the other, and the sides BC, and EF, adjacent to these angles also equal, the triangles are equal in all respects, and have those sides equal which are opposite the equal angles. For, conceive the point B to be laid on the point E, and the line BC on the line EF, then, because these lines are equal, the point C will fall on the point F. And, as BC coincides with EF, and the angle B is equal to the angle E, the side BA will fall on the side CD; and, for a like reason, AC will fall on DF, and the points A and D will consequently coincide; the triangles are, therefore, identical, having the parts in one respectively equal to the corresponding parts in the other. THEOREM III.-In any isosceles triangle, as ABC (fig. 31, p. I), the angles B and C, opposite the equal sides A B and A C, are equal. For, conceive the angle BAC to be bisected by the line AD, then, as the two triangles ABD, ACD, have AC equal to AB, AD, common, and the angles BAD and DAC equal, they are identical (Theorem I.), and have therefore the angles B and C equal. Cor. 1. The line which bisects the angle included between the equal sides of an isosceles triangle, bisects the third side also, and is perpendicular to it. Cor. 2. An equilateral triangle is also an equiangular one. THEOREM IV.-If a triangle, as ABC (fig. 32, p. I.), have two equal angles, as A and B, the sides AC and BC, opposite those angles, are also equal. For, if AC and BC are not equal, suppose one of them, as AC, to be longer than the other, and conceive AD to be the part of AC which is equal to BC, and join BD. Then, as AD is equal to BC, AB common to both, and the angles CAB and C B A equal, the triangles ABD and ABC are themselves equal (Theorem I.), the less equal to the greater which is impossible. AC and BC are, therefore, not unequal, that is, they are equal. Cor. An equiangular triangle is also an equilateral one. THEOREM V-If two triangles, as ABC, DEF (fig. 33 p. I.), have the sides AC and DF, AB and DE, and BC and EF, respectively equal, the triangles are identical, and have the angles equal, which are opposite the equal sides. For let the point A be laid on D, and the line AB on the line DE; then, as these lines are equal, the points B and E will coincide. Let the point C fall at G, and join FG. Then, as DF and DG are equal, the angles DFG and DGF are equal (Theorem III.); and in the same way may the angles EFG and EGF be shown to be equal. Hence the angles DF E, and DG E, or AC B, are equal; and, consequently (Theorem I.), the triangles A B C, and D E F, are identical, and have the angles BAC, and EDF, equal, and the angles BAC, and EDF, also equal. THEOREM VI.-The angles DBA, DBC (fig. 34, p. I.), which one straight line, DB, makes with another, A C, on the same side of it, are either two right angles, or, together, they are equal to two right angles. For, if BD is perpendicular to AC, each of the angles A B D and DBC is a right angle. If BD is not perpendicular to AC, let B E be perpendicular to A; then ABI exceeds the right angle ABE by the right angle EBD, and DBC is less than the right angle EBC by the same angle EBD; the angles A B D and DBC are, together, equal to two right angles. Cor. All the angles that can be made at a point, as B, by any number of lines drawn on the same side of a line, as AC, are, together, equal to two right angles; and, as all the angles that can be made at the same point by lines drawn on the other side of AC are also together equal to two right angles, we may infer that all the angles that can be made in the same plane about any point, are, together, equal to two right angles. THEOREM VII.-If two straight lines, as A B, DE (fig. 35 p. I.), intersect each other, any two vertical or opposite angles, as A CD, BC È, are equal. For, as A C meets D E, the angles ACD and AC E are, together, equal to two right angles (Theorem VI.); and, as EC meets AD, the angles ACE and ECB are also, together, equal to two right angles. Hence the sum of the angles ACD and ACE is equal to the sum of the angles ACE and EC B, and, if the common angle ACE be omitted from each sum, the remaining angles, ACD and ECB, are equal. THEOREM VIII.-If any side, as A B (fig. 36, p. I.), be produced, as to D, the outward angle CBD is greater than either of the inward and opposite angles A and C. For bisect BC in E, join A E, and produce it till EF is equal to A E, and join BF. Then as AE is equal to EF, BE to EC, and the angles A EC and BEF are also equal (Theorem VII.), the triangles AEC and BEF are identical (Theorem I.), and the angle EBF is equal to the angle ACE; consequently the whole angle CBD is greater than the angle Č. By producing CB as to C, it may be shown in the same way that the angle ABG, which is equal to CBD is greater than BAC. THEOREM IX. In any triangle, as ABC (fig. 37, p. I.), if a side AB be greater than another side A C, the angle A C B, opposite the greater side A B, is greater than the angle A B C, opposite the less side A C. For let AD be the part of AB which is equal to AC, and join CD. Then the angles ACD and ADC are equal (Theorem III.); but the angle ADC is greater than the angle B (Theorem VIII.); therefore the angle ACD is also greater than B, and consequently the angle ACB is much greater than B. THEOREM X.-In any triangle, as ABC (fig. 37, p. I.), the greater angle, as C, has the side A B opposite to it, greater than the side AC, opposite the less angle B. For, if AB were equal to A C, the angles B and C would be equal (Theorem III.); and, if AB were less than AC, the angle C would be less than the angle B, both of which conclusions are inconsistent with the hypothesis that the angle C is greater than the angle B; hence the side A B must be greater than the side A C. THEOREM XI.-If two triangles, as ABC, DEF (fig. 38 p. I.), have two sides AB, BC, of the one, respectively equal to two sides DE, EF, of the other, but the angle ABC greater than the angle DEF; then the side AC will be greater than the side DF. Let ABG be the part of the angle ABC which is equal to DEF, and let BG be equal to E For EC. Then (Theorem I.) the triangles ABG and DEF are identical, and have the sides AG and DF equal. And, as BG and BC are equal, the angles BGC and BCG are equal; hence BGC is greater than ACG, and much more is AGC greater than ACG; hence (Theorem X.) AC is greater than AG, or than its equal D F. THEOREM XII.-If two triangles, as ABC, DEF (fig. 38, p. I.), have two sides A B, BC, of the one respectively equal to DE, EF, two sides of the other; but the third side AC of the one greater than the third side DF of the other, the angle ABC will be greater than the angle DEF. For, if ABC and DEF were equal, AC and DF would be equal (Theorem I.), which, by hypothesis, they are not; and, if ABC were less than DEF, AC would be less than DF (Theorem XI.), which it is not. Hence, as ABC is neither equal to nor less than DE F, it must be greater. THEOREM XIII. In any triangle, as ABC, (fig. 39, p. I.), the sum of any two of its sides, as AC and CB, is greater than the third side AB. Produce A C till CD is equal to C B, and join DB; then (Theorem III.) the angles D and DBC are equal, therefore the angle ABD is greater than the angle D, and hence (Theorem X.) AD, or the sum of AC and CB, is greater han AB. THEOREM XIV.-In any triangle, as ABC (fig. 39, p. I.), the difference of any two of its sides, as AC and A B, is less than the third side BC. For let A E be the part of AB the greater, which is equal to AC the less; then, as AC and CB (Theorem XIII.) are together greater than AE and EB; if the equal parts, AC and A E, be taken from each, EB, the difference of AC and A B, will remain less than C B. THEOREM XV.-If a line, E F (fig. 40, p. I.), intersect two parallel lines, A B and CD, the alternate angles BEF, CFE will be equal to each other. If they are unequal, suppose BEF to be the greater, and that DEF is the part of it which is equal to CF E. Then, as A B is parallel to CD, ED is not parallel to C D, it will, therefore, meet it in some point, as D; and in the triangle, EFD, the outward angle E F C, is greater than the angle DEF (Theorem VIII.), which was supposed equal to it, which is impossible. The angles CF E, and BE F, are therefore equal. THEOREM XVI.-If the alternate angles BEF and CFE (fig. 40, p. I.) are equal, A B and C D are parallel. For, if AB is not parallel to CD, let ED drawn from E be parallel to it. Then the angle CFE equal to the alternate angle FED (Theorem XV.), and also by hypothesis equal to the FEB, the angles FED and FEB must be equal, which is impossible. Hence no line drawn through, except EB, can be parallel to CD. THEOREM XVII.—If AB, C D (fig. 40, p. I.), two parallel lines, be cut by another line EF, any outward angle, as A EG, is equal to the inward and opposite one CFE on the same side of EH, and any two inward angles, as A EF and EFC, are together equal to two right angles. For the angle BEF being equal to the angle AEG (Theorem VII.), and also equal to the angle CFE (Theorem XV.), the angles A EG and CFE are equal. And as the angles A E F and B E F, together, are equal to two right angles (Theorem VI.), and the angle BEF is equal to the angle CFE, the angles A EF and CFE are together equal to two right angles. THEOREM XVIII.-If two lines, as AB, EF (fig. 41 p. I), be each parallel to another line as CD, AB and E F are parallel to each other. Let the lines be cut by the line GHIKL; then as the angle HIC is equal to BHI (Theorem XV.), and also equal to IKE (Theorem XVII.), the angles BHI and IKE are equal, and consequently (Theorem XVI.) the lines A B and EF are parallel. THEOREM XIX.-If any side, as AB (fig. 42, p. I.) of a triangle ABC be produced, the outward angle CBD, is equal to both the inward and opposite angles A and C. Let BE be a line parallel to AC: then (Theorem XV.) the angles ACB and CBE are equal; and (Theorem XVII.) the angles CAB and EBD are equal; hence the angle CBD is equal to the sum of the angles A and C. Cor. 1. As the angles CBD and CBA are together equal to two right angles (Theorem VI.) and the angle CBD is equal to the sum of the angles A and C, the three interior angles of any triangle are equal to two right angles. Cor. 2. Either of the angles C or A is the difference between the other and the outward angle C BE. THEOREM XX.-All the interior angles of any rectilineal figure, ABCDE (fig. 43, p. I.), are together equal to twice as many right angles as the figure has sides wanting four right angles. For from any point, as F, within the figure, let lines be drawn to its angular points, dividing the figure into as many triangles as it has sides; then the interior angles of each triangle being equal to two right angles, the interior angles of all the triangles (which are the interior angles of the figure, and the angles about the point F, or the interior angles of the figure and four right angles), are equal to twice as many right angles as the figure has sides. Hence the interior angles of the figure are equal to twice as many right angles as the figure has sides wanting four right angles. Cor. 1. All the interior angles of a quadrilateral figure are together equal to four right angles. Cor. 2. If the sides of the figure be produced the sum of the outward angles will be four right angles. For, each interior with its corresponding exterior angle being equal to two right angles (Theorem VI.), all the interior and exterior angles will be equal to twice as many right angles as the figure has sides, or equal to the interior angles and four right angles; therefore the exterior angles are equal to four right angles. THEOREM XXI.-In any parallelogram, as A BCD (fig. 44, p. I.), the opposite sides are equal to each other, and so are also the opposite angles, and the diagonal BD divides it into two equal triangles. For AB and C D being parallel, and AD and CB also parallel, the angle ABD is equal to the angle CDB and the angle ADB to the angle CBD (Theorem XV.); and, as D Bis common to both triangles, the triangles are (Theorem II.) identical, having A B equal to C D, AD to BC, and the angle A to the angle C, and as the two parts of the angle ADC are equal to the two parts of the angle ABC, the angles ADC and A BC are also equal. Cor. If a parallelogram have one right angle, all its angles are right angles, and consequently all the angles of a rectangle are right angles. THEOREM XXII.—Any quadrilateral, as A B C D(fig. 44, p. I.), whose opposite sides are respectively equal is a parallelogram. For, DB being joined, the triangles ABD and CDB are (Theorem V.) identical, having the angles CDB and A B D equal, and the angles AĎ B and C B D equal. Hence (Theorem XVI.) A B and C D are parallel, and AD and BC are parallel; and consequently (Def. XXI.) the figure ABCD is a parallelogram. Cor. Hence a square is a parallelogram. THEOREM XXIII.-The lines AC and BD (fig. 44, p. I.).which join the corresponding extremities of equal and parallel lines, as A B and CD, are themselves equal and parallel. For, BC being joined, the angles A B C and DCB are equal (Theorem XV.) and consequently (Theorem I.) the triangles ABC and DCB are identical, and have AC equal to BD, and the angles ACB and DBC equal, when AC and BD are parallel (Theorem XVI.). THEOREM XXIV.-If two parallelograms, as ABCD, ABEF (fig. 45, p. I.), be on the same base A B, and between the same parallels A B, and FC, those parallelograms are equal to each other. For, (Theorem XXI.) FE and DC are each equal to A B, they are therefore equal to each other, and if ED be added to each, FD and EC will be equal; and as A F is equal to B E, and AD to BC, the triangles A D F, and BCE, are equal (Theorem V). If therefore each of these equal triangles be taken from the whole figure ABCF, the remainders, or the parallelograms ABEF and ABCD are equal. Cor. If the diagonals A E and DB be drawn, the triangles A E B and AD B, halves of equal parallelograms, are equal. THEOREM XXV.-Parallelograms, as ABCD, EFGH (fig. 46, p. I.), on equal bases A B and EF, and between the same parallels AF and DG, are equal to each other. For HG being equal and parallel to EF (Theorem XXI.) is equal and parallel to AB, hence (Theorem XXIII.) AH is equal and parallel to BG, therefore AHGB is a parallelogram; and each of the parallelograms AC and EG being equal to the parallelogram AG, they are equal to each other. Cor. 1. If BD and EG be joined, the triangles ABD and EFG, the halves of the equal parallelograms AC and E G are equal. Cor. 2. If a parallelogram and a triangle be on the same or on equal bases, and between the same parallels, the parallelogram will be double the triangle. THEOREM XXVI.-Equal triangles, as A B C, ADB (fig. 47, p. I.) on the same base AB, are between the same parallels. For if DC is not parallel to AB, let DE meeting AC in E be parallel to A B, and join BE. Then the triangles A B D and AB E are equal (Cor. Theorem XXIV.); and the triangles A BD and A BC are equal, by hypothesis; hence the triangles A BC and ABE are equal, the less to the greater, which is impossible. No line therefore, drawn through D, except DC, can be parallel to A B, and DC and AB are therefore parallel. THEOREM XXVII.-If A B (fig. 48, p. I.) be a line, and BC another divided into any number of parts, as BD, DE, EC, the rectangle contained by A B and BC is equal to the rectangles contained by A B and B D, A B and DE, and AB and EĆ. For let AC be the rectangle contained by AB and BC, and let DF and EG be drawn parallel to AB; then the rectangle AC is the sum of the rectangles AD, FE, and EC. Now AD is contained by AB and BD, FE by FD and DE, and GC by GE and EC; and, as DE and EG are each equal to A B, their rectangles are equal to rectangles contained by AB and BD, AB and D E, and AB and EC. Cor. 1. If AB be equal to BD, AD will be. a square on BD, and DH will be a rectangle under BD and DC; therefore the rectangle under a line, and a part of itself, is equal to the square of that part, and the rectangle of the two parts. Cor. 2. If BA be equal to BC, BH will be a square on BC; hence the rectangles under a line, and the several parts of itself, are equal to the square of that line. THEOREM XXVIII.-The square of the sum of two lines, as AF and F B (fig. 49, p. I.), is equal to the square of each line, and twice the rectangle contained by these lines. Let AC be the square on AB, and FI the square on F B, and produce FG and IG till they meet DC and A'D in E and H; then HI and EF are equal, being equal to A B and B C, sides of the same square; if from these equals, the equal parts GI and FG be taken, the re maining parts GE and GH will be equal. But HIG is equal to D E, and GE to HD, and the angle D being a right angle, H E is a square on H G, or it is equal to the square of AB. But the parallelograms AC and G C are rectangles, each equal to the rectangle contained by AF and FB; for FG and GI are each equal to F B, and I C is equal to A F. Hence the square of A B, or the square of the sum of AF and F B, is equal to the squares of AF and F B, and twice the rectangle of AF and F B. Cor. The square of any line is four times the Suare of half that line. THEOREM XXIX.-The square of AC, the 'difference of two lines A B and BC (fig. 50, p. I.), is equal to the sum of the squares of those lines, all but twice the rectangle contained by them. For let AD be the square on the difference AC, AF the square on the greater A B, and CI the square on the less BC; and produce ED to meet F C in II. Then GH and HK are the excess of the two squares A Fana CI above the square AD. Now as A B is equal to AG, and A C to AE, EG is equal to BC; and, A B being equal to GF, G H is equal to a rectangle under AB and BC. Again, as AC is equal to C D, and C B to C K, AB is equal to DK; and, CB being equal to KI, HK is equal to a rectangle under A B and BC; hence the square of the difference of A B and BC is equal to the square of each of those lines, diminished by twice their rectangle. THEOREM XXX.-The difference of the squares of any two unequal lines, as A B, AC (fig. 51, p. I.), is equal to a rectangle under their sum and difference. For let AD be the square on A B, and A F the square on AC; produce BD till BK is equal to AC; draw IK parallel to AB, and produce F C till it meet IK, and ED in I and H. Then the rectangles EF and CD together form the difference of the two squares AD and AF. But the rectangles EF and BI are equal; for GF and BK are each equal to A C, and GE is equal to BC, as they are each equal to the difference of A B and A C, or of their equals AE and AG. The rectangles EF and CD are therefore together equal to the rectangles BI and CD, or to the whole rectangle DI, which is therefore equal to the difference of the squares AD and A F But DI is a rectangle contained under D K the sum, and BC the difference of A B and A C; hence the difference of the squares of A B and AC is equal to the rectangle under their sum and difference. THEOREM XXXI.-If two parallelograms AG, CE (fig. 52, p. II.), be described on AB, BC, two sides of a triangle ABC, and the outward sides FG, DE; if the parallelograms be produced till they meet in H; and if HB be joined, and produced to meet the base AC in K ; and A L, C M be drawn parallel to H K, meeting FH and HD in L and M; then, if LM be joined, AM will be a parallelogram, and equal to the parallelograms AG and CE. For, as A H and H C are parallelograms, AL and CM are, each equal and parallel to BH (Theorem XXI.), they are equal and parallel to each other (Theorem XXIII.), therefore A C and L M are also equal and parallel, and hence AM is a parallelogram. Now (Theorem XXIV.) the parallelograms AG and AH are equal, for they are on the same base A B, and between the same parallels A B and C H. But the parallelogram A H is equal to the parallelogram A N, for they are on the same base A L, and between the same parallels AL and KH; hence the parallelograms AG and AN, being each equal to AH, are equal to each other. Similarly the parallelograms CE and CN may be shown to be equal; and consequently the two parallelograms AG and CE are equal to the whole parallelogram AM. Cor. If the parallelograms AG and CE be squares, and the angle ABC a right angle, then the parallelogram A M will also be a square, and equal to the two squares AG and CE. THEOREM XXXII.-In any triangle as ABC (fig. 53, p. II.), obtuse angled at A, if a perpendicular CD be drawn from C on the base produced, then B C2— B A2 + A C2 + 2 BA Ď. For (Theorem XXXI., Cor.) BC2 = B D2 +D C2; but (Theorem XXVIII.) BD2= BA2 + A D2 + 2 BA AD; therefore B C2 = BA' +AD' + DC2+2 BA AD. But (Theorem XXXI. Cor.) A D2 + D C2 = A C2, hence BC BA+AC2 + 2 BAAD. = THEOREM XXXIII.-In any triangle, as ABC (fig. 54, p. II.), if CD be a perpendicular, drawn from C on the opposite side A B, then BC BA2+ AC2-2 BA· AD. For (Theorem XXXI. Cor.), A C2 = A D2 + DC3 and (Theorem XXVIII.) A B2 = A D2+ D B2 + 2 BD · AD; hence A B+ A C2 = B D2 + DC2 + 2 A D2 + 2 B DA D. But (Theorem XXXI. Cor.) BD2 + DC2 = BC2, and (Theorem XXVII. Cor. 1,) 2 A D2 + 2 B D, AD=2BA AD; therefore A B2+ AC2 = BC2+2 BA AD, or B C2 — A B2 + A C2— 2 BAAD. THEOREM XXXIV.-In any triangle as ABC (figs. 53 & 54, p. II.) the rectangle under the sum and difference of two sides, as BC and AC is equal to the rectangle of the segments, AD and B D, of the third side, made by a perpendicular from the angle C. For BC2 BD2 + D C3, and A CADa + DC, (Theorem XXXI. Cor.); hence BC - A C2 (B D2 + DC2) — (A D2 + DC2)= BD2--A D2. But (Theorem XXX ), B C2 — A C2 = B C + AC BC-AC; and B D2— AD2=BD+AD·BD-AD; therefore BC + AC · B C — A C = BD + AD · BD-AD. THEOREM XXXV.-If the base A B of a triangle, as ABC (fig. 55, p. II.), be bisected in D, then A C2+BC2= 2A D2 + 2D C2. For let C E be a perpendicular from Con AB; then (Theorem XXXII.) A C2 — A D2 + DC 2 AD DE; and (Theorem XXXIII.) BC BD + D C2 - 2 BD DE AD2 +DC'. 2 AD DE; hence A C2+BC* = 2 A D2 + 2 D C2. |