PROPOSITION VIII. THEOREM. In a right-angled triangle (ABC), if a perpendicular (AD) be drawn from the right angle to the base; the triangles on each side of it are similar to the whole triangle, and to one another. DEMONSTRATION. Because the angles BAC and ADB are equal, being both right angles, and that the angle at B is common to the two triangles ABC and ABD; the remaining angle at C is equal to the remaining angle BAD (a): therefore the triangles ABC and ABD are equiangular, and the sides about their equal angles are proportionals (6); wherefore the tri B D (a) I. 32. angles are similar (c): in the like manner it may be demonstrated, that the triangles ADC and ABC are equiangular and similar: and the triangles ABD and ADC, being both equiangular and similar to ABC, are equiangular and similar to each other. COROLLARY. From this it is manifest, that the perpendicular drawn from the right angle of a right-angled triangle to the base, is a mean proportional between the segments of the base: and also that each of the sides is a mean proportional between the base, and its segment adjacent to that side because in the triangles BDA, ADC, BD is to DA, as DA is to DC (b); and in the triangles ABC, DBA, BC is to BA, as BA is to BD (6); and in the triangles ABC, ACD, BC is to CA, as CA is to CD (6). PROPOSITION IX. PROBLEM. From a given finite straight line (AB) to cut off any required part, or submultiple. SOLUTION. From the point A draw a straight line AC, making any angle with AB; and in AC take any point D, and take AC the same multiple of AD, that AB is of the part which is to be cut off from it; join BC, and draw DE parallel to it: then AE is the part required to be cut off. DEMONSTRATION. Because ED is parallel to BC, one of the sides of the triangle ABC; as CD is to DA, so is BE to EA (a); and by composition, CA is to AD, as BA is to AE (6): but CA is a multiple of AD; therefore BA is the same multiple of AE (c): whatever part, therefore, AD is of AC, AE is the same part of AB: wherefore, from the straight line AB the part required is cut off. PROPOSITION X. PROBLEM. To divide a given straight line (AB) similarly to a given divided straight line (AC); that is, into parts proportional to the parts of the given divided straight line. SOLUTION. Let AC be divided in the points D, E; and let AB, AC be placed so as to contain any angle, and join BC, and through the points D, E, draw DF, EG, parallels to it (a); and through D draw DHK parallel to AB. B G H DEMONSTRATION. Because each of the figures FH, HB, is a parallelogram, DH is equal to FG (b), and HK to GB (b); and because HE is parallel to KC, one of the sides of the triangle DKC, as CE is to ED, so is KH to HD (c) but KH is equal to BG, and HD to GF; therefore, as CE is to ED, so is BG to GF: again, because FD is parallel to EG, one of the sides of the triangle AGE, as ED is to DA, so is GF to FA: but it has been proved that CE is to ED, as BG is to GF; and as ED is to DA, so is GF to FA; therefore the given straight line AB is divided similarly to AC. PROPOSITION XI. PROBLEM. To find a third proportional to two given straight lines (AB and AC). SOLUTION. Let the two given straight lines AB and AC be so placed as to contain any angles, and produce them to the points D, E; make BD equal to AC; and having joined BC, through D draw DE parallel to it (a). K (a) I. 31. (b) I. 34. (c) VI. 2. B (a) I. 31. DEMONSTRATION. Because BC is parallel to DE, a side of the triangle ADE, AB is to BD, as AC is to CE (6): but BD is equal to AC; therefore, as AB is to AC, so is AC to CE. Wherefore to the two given straight lines AB, AC, a third proportional CE is found. PROPOSITION XII. PROBLEM. To find a fourth proportional to three given straight lines (A, B, and C). SOLUTION. Take two straight lines DE, DF, containing any angle D, and upon these make DG equal to A, GE equal to B, and DH equal to C; and having joined GH, draw EF parallel to it through the point E (a); then HF is the fourth proportional required. DEMONSTRATION. Because GH is parallel to EF, one of the sides of the triangle DEF, DG is to GE, as DH is to HF (b); but DG is equal to A, H (a) I. 31. B C GE to B, and DH to C; therefore as A is to B, so is C to HF. Wherefore, to the three given lines, A, B, C, a fourth proportional HF is found. PROPOSITION XIII. PROBLEM. To find a mean proportional between two given straight lines (AB and BC). SOLUTION. Place AB, BC in a straight line, and upon AC describe the semicircle ADC; from the point B draw BD at right angles to AC (a), and join AD, DC. Then DB is the mean proportional required. Because DEMONSTRATION. the angle ADC in a semicircle is a right angle (6), and be B (a) I. 11. cause in the right-angled triangle ADC, DB is drawn from the right angle perpendicular to the base, DB is a mean proportional between AB BC the segments of the base (c). PROPOSITION XIV. THEOREM [1.]-If equal parallelograms (AB and BC) have an angle of the one equal to an angle of the other, their sides about the equal angles are reciprocally proportional (DB is to BE, as GB is to BF). [2.] And if parallelograms (AB and BC) have an angle of the one equal to an angle of the other, and their sides about the equal angles reciprocally proportional, they are equal to one another. CONSTRUCTION. Let the sides DB, BE, be placed contiguous, in the same straight line, with the parallelograms on opposite sides of DE; then FB, BG are in one straight line (a). Complete the parallelogram FE. DEMONSTRATION [1] Because the parallelogram AB is equal to BC, and that FE is another parallelogram, AB is to FE, as BC is to FE (6): but as AB is to FE, so is the base DB to BE (c); and, as BC is to FE, so is the base GB to BF; therefore, as DB is to BE, so is GB to BF (d). Where fore the sides of the parallelograms AB, BC, about their equal angles are reciprocally proportional. [2.] Let the same construction remain, then, because DB is to BE, as GB is to BF; and DB is to BE, as the parallelogram AB is to the parallelogram FE, and GB is to BF, as the parallelogram BC is to the parallelogram FE; therefore AB is to FE, as BC is to FE (d) wherefore, the parallelogram AB is equal to the parallelogram BC (e). PROPOSITION XV. THEOREM [1.]-If equal triangles (ABC and ADE) have an angle of the one equal to an angle of the other, their sides about the equal angles are reciprocally proportional (CA is to AD, as EA is to AB). [2.] And if triangles (ABC and ADE) have an angle in the one equal to an angle in the other, and their sides about the equal angles reciprocally proportional, they are equal to one another. CONSTRUCTION. Let the sides CA, AD, be placed contiguous, in the same straight line, with the triangles on opposite sides of CD; then EA, AB are in one straight line (a). Join BD. DEMONSTRATION [1.] Because the triangle ABC is equal to ADE, and that ABD is another triangle, the triangle CAB is to the triangle BAD, as the triangle EAD is to the triangle DAB (6): but as the triangle CAB is to the triangle BAD, so is the base CA to AD (c); and, as the triangle EAD is to the triangle DAB, so is the base EA to AB (c); therefore, as CA is to AD, so is EA to AB (d). Wherefore, the sides of the triangles ABC, ADE, about the equal angles are reciprocally proportional. [2.] Let the same construction remain, then, because CA is to AD, as EA is to AB; and CA is to AD, as the triangle ABC is to the triangle BAD (c); and EA is to AB, as the triangle EAD is to the triangle BAD (c); therefore the triangle BAC is to the triangle BAD, as the triangle EAD is to the triangle BAD (d); wherefore the triangle ABC is equal to the triangle ADE (e). |