fore (f) the bases EF, GF are equal; the angles E, G are equal; also DFE equals DFG. But G is made equal to B, and DFG to C; there fore B is equal to E, and C to DFE (g). Wherefore, if two triangles have, &c. Q. E. D. Recite (a) p. 23, 1; (b) p. 32, 1; (d) p. 11, 5; (e) 9, 5; (c) p. 4, 6: (g) ax. 1, 1. 7 Th. If two triangles have two angles equal, and the sides about other two angles proportionals; then, whether the third two be oblique or right angles, the triangles shall be equiangular, and have those angles equal which are contained by the proportional sides. Let two triangles ABC, DEF, have equal angles at A and D, and have AB to BC as DE to EF; their angles at B and E are equal, also those at C and F. If then it be said, that the angles at B and E are unequal, make the angle ABG equal to E (d); and it follows, that AB : BG :: DE: EF (b); that AB has to BC and BG the same ratio; that BG equals BC (c), and that the angles BGC and BCG are equal (d Then since the angles at C and F may be right or oblique: 1. If Cand F be acute, BGC is acute (d), and the adjacent angle BGA is obtuse (e); and so, F is obtuse, because the triangles ABG, DEF assume to be equiangular. 2. If Cand F be obtuse, BGC is obtuse (d), and the adjacent angle BGA is acute (e); and so, F is acute, because the triangles ABG, DEF assume to be equiangular. ity of the angles 3. If Cand F be right angles, BGC is a right angle (d) and so two angles of a triangle are not less than two right angles (f). Therefore to deny the equal- B at and E. B ဂ.၈ makes the same angle F acute and obtuse, and two angles of a trian gle equal to two right angles, which are both absurd. Wherefore, if two triangles have two angles equal, &c. Q. E. D. 8 Th. In a right angled triangle, if a perpendicular be drawn from the right angle to the base, the triangles on each side of it are similar to the whole and to each other. Given the triangle ABC, right angled at C: the perpendicular CD divides ABC into two triangles similar to the whole and to each other. Because the angle BCA equals BDC (a), and that the angle B is common to the two triangles BCA, BDC, the third angles BAC, BCD, are equal to each other (6); therefore the triangles BCA and BDC are equiangular, and the sides about the equal A angles are proportionals (c): therefore the trian gles are similar (d). (d) D B In like manner, it may be shown, that the triangles ADC, BDC are equiangular and similar: and the triangles BDC, ADC, being each equiangular and similar to BCA, are equiangular and similar to each other. Therefore, in a right angled triangle, &c. Recite (a) ax. 10, 1; (d) def. 1, 6. (b) p. 32, 1; (c) p. 4, 6; E. D. Cor. From the equiangular triangles, as above, come the following proportions namely, BD: BC:: BC: BA, AD: AC:: AC: AB, and BD: DC::CD: DA. Therefore cach side of the triangle ABC is a mean proportional between its adjacent segment of the base and the base complete; and the perpendicular is a mean proportional between the segments of the base 9 P. From a given straight line (AB), to cut off any part required. From the point A draw AC, at any angle with AB. In AC take a point D, and make AC the same multiple of AD that AB is of the part to be cut off from it: join BC, and draw DE parallel to it; AE is the part required. Because ED is parallel to BC, a side of the triangle ABC (a), CD: DA:: BE: EA; and by composition, CA: AD::BA: AE (6): but CA is a multiple of AD, and BA is the same multiple B of AE (c): therefore, whatever part AD is of AC, the same part is AE of AB. A E D C Wherefore, from the straight line AB the required part is cut off; which was to be done. Recite (a) p. 2, 6; (b) p. 18, 5; (c) p. D, 5, 10 P. To divide a given straight line into parts, having the same ratios as the parts of a divided straight line given. Given the straight line AC, divided in D, E; and AB to be divided into similar parts, or having the same ratios. A Place AB, AC, so as to contain any angle; join BC; and through the points D, E draw DF, EG parallel to BC (a); and through D draw DHK parallel to AB. Now, in the parallelograms FH, HB, the side DH is equal to FG, and HK to GB (b). And because, in the triangle DKC, HE is parallel to KC, CE is to ED as KH is to HD, or as BG is to GF: and because, in the triangle AGE, FD is parallel to GE, ED is to DA as GF is to FA (c). And since it proves that CE:ED::BG:GF, and ED: DA::GF:FA; Therefore, the given straight line AB is divided similarly to AC; which was to be done. Recite (a) p. 31, 1; (b) p. 34, 1; (c) p. 2, 6. 11 P. To find a third proportional to two given straight lines, AB, AC. Place the given lines AB, AC so as to contain any angle; join BC; produce AB, so that BD shall equal AC; through D, draw DE parallel to BC (a), and produce AC to meet DE (6). Because, in the triangle ADE, BC is parallel to DE, we have AB to BD, or to its equal AC, as AC to CE (c). Therefore, CE is the third proportional to AB, AC; which was sought. Recite (a) p. 31, 1; B A C 12 P. To find a fourth proportional to three given straight lines A, B, C. equal to A, B, C: E F Therefore, A: B::C: HF; and so, HF is the fourth proportional sought. Recite (a) p. 31, 1; (b) p. 2, 6. 13 P. To find a mean proportional between two given straight lines AB, BC. Place AB, BC in a straight line AC; bisect it (a); and, upon the bisectional point, describe a semicircle to pass through A, C; through B draw BD at right angles to AC (b); join AD, CD. A B C Because the angle ADC, in a semicircle, is a right angle (c); and that, in the right angled triangle ADC, a perpendicular BD is drawn from the right angle to the base; DB is a mean proportional between the segments of the base, which are the given straight lines AB, BC (d): Wherefore, BD is the mean proportional sought. Recite (a) p. 10, 1; (b) p. 11, 1; (c) p. 31, 3; (d) Cor. p. 8, 6. 14 Th. Equal parallelograms, which have equal angles, two and two, have their sides about equal angles reciprocally proportional: and parallelograms are equal, which have equal angles, two and two, and their sides about equal angles reciprocally proportional. Given two equal parallelograms AB, BC, hav- H F ing equal angles at B. Place the parallelograms 1. Because the parallelograms AB, BC are ᄇ D 2. Because DB: BE::GB: BF; and that DB: BE:: AB: FE, and GB:BF:: BC: FE (c); therefore AB: FE:: BC: FE (d); and so, the parallelograms AB and BC having the same ratio to the parallelogram FE, are equal to each other (e). Therefore, equal parallelograms, &c. Recite (a) p. 15, 1; (b) p. 7, 5; (d) p. 11, 5; (e) p. 9, 5. Q. E. D. (c) p. 1, 6; NOTE. The opposite sides and angles of a parallelogram are equal: p. 34 of b. 1. 15 Th. Equal triangles which have an angle of one equal to an angle of the other, have their sides about the equal angles reciprocally proportional: and triangles are equal, which have an angle of one equal to an angle of the other, and the sides about the equal angles reciprocally proportional. Given two equal triangles ABC, ADE, having equal angles at A. Place the triangles so that the angles at A may be vertical (a) to each other, and reciprocal sides CA, AD and EA, AB in straight lines: join BD. Then 1, CA: AD::EA:AB; 2, ABC=ADE. A D E 1. Because the triangles ABC, ADE are equal; and that ABD is another triangle, c ABC: ABD::ADE: ABD (b): but ABC:ABD::CA: AD; and ADE: ABD:: EA: AB (c); therefore CA: AD :: EA: AB (d); and, taken in this order, these proportionals are reciprocal sides. 2. Because CA: AD::EA: AB; and that CA:AD::ABC: ABD, and EA: AB:: ADE: ABD (c); therefore ABC:ABD::ADE: ABD (b); and so, the triangles ABC, ADE, having the same ratio to ABD, are equal to each other (d). Therefore, equal triangles, &c. Recite (a) p. 14, 15, 1; def. 2, 6; (c) p. 1, 6; NOTE.-Reciprocal sides, def. 2, b. 6. Q. E. D. (b) p. 7, 5; 16 Th. If four straight lines be proportionals, the rectangles of the extremes and means are equal: and if the rectangles of the extremes and means be equal the four straight lines are proportionals. 1. Given AB to CD as E to F; sought ABXF=CDXE. 2. Given FXAB=EXCD; sought AB: CD:: E: F. From the points A, C, and at right angles to AB, CD, draw AG=F, and CH=E (a). Complete the rectangles BG, DH. 1. Because AB: CD: E: F, and that E, F are equal to CH, AG; therefore E H AB:CD::CH: AG (6). Therefore the F sides of the parallelograms BG, DH, are G tangle of CD and EH, or E: therefore ABXF=CDXE. D 2. Because F times AB equals E times CD; and that E, F are equal to CH, AG: therefore ABXAG=CDXCH. But parallelograms that are equal and equiangular, have their sides about equal angles reciprocally proportional (c): therefore AB: CD::CH: ÁG; that is, AB:CD::E:F. Therefore, if four straight lines, &c. Recite (a) p. 11, 1; (b) p. 7, 5; (c) p. 14, 6. Q. E. D. |