but when ABC <rt. , then thoses are internal. when ABC becomes a rt. L, BD and BE coincide; In the general proposition, this isosc. A DBE, is what the perpendicular is, when the given A is rt. angled. .. the sides of this A DBE, and the triangles under them, and the sides of the given ▲ ABC, possess many of the properties already proved in the case of a rt. dA; for instance, as in Cor. 1, Pr. 8, VI.; 1. AC AB = AB : AD, where AB is a mean proportional. 2. AC CB = CB : CE, where CB is a mean proportional. AB = BC : BE, 3. AC 4. AD BD BD : CE, where BD is a mean proportional. 5. 2. : or since BD = BE; AD: BE = BE: CE, where BE is a mean propl. .' AD : BD = AB : BC, .. the segments AD and EC are in the duplicate ratio of the sides AB and BC. "Hence in a right-angled triangle the segments of the hypotenuse by the perpendicular, are in the duplicate ratio of the sides." LARDNER'S Euclid, pp. 187, 188. USE & APP. 1. The first Corollary of Pr. 8, bk. VI. supplies the principle on which a very clear and brief demonstration may be given of Prop. 47, bk. I. 2. Also according to this Proposition, and by aid of a square, inaccessible distances, as DB, may be measured. At D raise the perp. DA, and measure it; And at A place a square, so that by looking along one of its sides Ab, the point B may be seen in the same st. line with Ab; And along the other side Ac, the point C may be seen; and measure DC; Then, . CD : DA = DA : DB; .. DB = Ex. Suppose AD = 3, and CD = 2.25; then DB = DA2 CD 9 2.25 3. In a circle any chord, as BA, is a mean proportional between the diameter BC, and that segment of the diameter BD, which is drawn from one extremity of the chord, B, and cut off by a perpendicular, AD, let fall from A the other extremity of the chord. '.' As BAC and ADB have each a rt., and B common; ..the ABAC is eq. ang. to A ADB, and.. BC: BA = BA: BD; i. e. the chord BA is a mean propl. to BC & BD. And. As BAC, BDA and ADC are similar, the segments of the hypotenuse are in the duplicate ratio of the sides. PROP. 9.-PROB. From a given st. line to cut off any part required, i. e., any measure or submultiple. CON.-3, I. 31, I. DEM.-2, VI. 18, V. Componendo. If Ms, taken separately, be propls, : 3rd: D, V-If 1st : 2nd 3rd 4th, if 1st a m or pt of the 2nd, the 3rd is the same m or pt of the 4th. Def. 1, V.-A less M is a pt of a greater when the less measures the 31, I. 4 Sol. D. 1 C. 3. 22, VI. 3 18, V. 4 C. D, V. Let AB be a given st. line; to cut off from it any pt. From. A draw AC making any in AC take any. D, and make AC the same m of AD join BC, and draw ED || BC; then AE is the submultiple required; .. CD: DA = BE: EA; and compon., CA : AD = BA : AE; but CA is a m of AD; .. BA the same m of AE; 5 Def. 1,V... AE the same pt of AB, that AD is of AC. 6 Recap. . From a given st. line, &c. Q. E. F. SCH. Prop. 10, Bk. I. by which a rt. line may be bisected, and its bisections also bisected, is a particular case of this Problem. USE & APP. A simple extension of the Problem enables us; 1st, to divide a given line, AL, into any number of equal parts. 2. Join EL, and through D, C, B draw ||s to EL and cutting AL; then AL is divided into four eq. pts. in b, c, d, L. ...Ab is the same pt. of AL that AB is of A E; and ⚫. Ab, bc cd dL; AL is divided into four eq. parts. To divide a triangle, ABC, into any number of eq. parts, say four, by lines from a given point, P, in one of the sides, as BC. C. 1 Sch. 1. 9, VI. 2 31, I. 3 Pst. 1, I. 4 Sol. Divide BC into four eq. pts. in D, E, F, and join AP; = AABC; i. e. by lines from P, ▲ ABC is divided into 4 eq. pts D. 1 37, I. A DPG: =A GAD, both being on GD and between the same s GD, AP; 2 Add. Ax. 2, I. to each add ▲ BGD, .. ▲ BPG = ▲ ADB. 3 C 1. 5 D. 3 6 Sim. 7 Sub. Next, . BD = DE = EF = FC and the altitude com. .. As ABD, ADE, AEF, AFC, are equal; and each is ▲ ABC; but ▲ BPG ▲ ADB, ... BPG : = ▲ ABC. So A GPA ▲ DPA, and ▲ HPA = A APE, 8 Ax. 3, I. D. 4... rem. A GPH rem. ▲ AED = ▲ ABC. Given the nth part of a line AB, to find the (n + 1) th part. C. 1 46, I. Pst. 1, I. On AB desc. a sq. ABEF, and join AF, EC cutting in G; through G draw HGK || AE, cutting AB, EF, in H & K; AB then AH, or EK =—; or (n+ 1) AH n+1 = AB. As AHG, FKG, are eq. ang.; and As AGC & EGF; AH: FK, or BH, AG: GF AC: EF, or AB. and .. BH = n. AH; and AB = (n+1) AH;' To divide a given st. line similarly, i. e., proportionally, to a given divided st. line; or to divide a given st. line into parts that shall have the same ratios to one another which the parts of the divided given st. line have. Or, “To divide a given undivided line similarly to a given divided line.” CON. Pst. 1, I. 31, I. 1, I. |