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For bisect AB in C (I. 7.), draw (I. 5. cor.) the perpendicular BD equal to BC, join AD and continue it until DE be equal to DB F.
or BC, and on AB produced take AF T) equal to AE : The line AF is the re- G. quired extension of AB. For make DG r
A C B IS’ equal to DB or BC ; and because (Il. 17. cor. 2.) the rectangle EA, AG, together with the square of DG or DB, is equivalent to the square of DA, or to the squares of AB and DB ; the rectangle EA, AG, or FA, FB, is equivalent to the square of AB.
13. Proposition twenty-third. This proposition is of great use in practical geometry, since it enables us to divide a triangle, of which all the sides are given, into two right-angled triangles, by determining the position, and consequently the length, of the perpendicular.
Thus, suppose the base of the triangle to be 15, and the two sides 13 and 14: Then 15* + 13* – 14* = 225 + 169 – 196 = 198, which shows that the perpendicular falls
within the triangle; and o = 6.6, the segment adjacent to
the short side, whence the perpendicular= y((13)”—(6.6/*) = w/(169–43.56)= 11.2. The area is therefore 15 × 5.6= 84.
Again, let the base be 9, and the two sides 17 and 10: Then 17*—9°–10* = 289–81 — 100 = 108, indicating that the perpendicular falls without the base. Wherefore, }=
the area of the triangle. Lastly, if the base were 10, and the sides 21 and 17 : Then 21*—17*—10* = 441 – 289 – 100 = 52, which shows that the perpendicular falls somewhat beyond the base. Whence 52
26 = 2.6, the external segment; and V (17* — 2.6*) = w/ (289–676) = y282.24 = 16.8, which gives 84 for the
area, as in the first example. The same results are obtained by applying the Twenty-First
Proposition. Thus, in the first example, the distance of the
14* — 13” "
*—so-- 9. and therefore the segments of the base are 8.4, and 6.6. In the second example, the distance of the perpendicular from
the middle of the base id:Ho = 10.5, and consequently r i
the segments of the base are 15 and 6. In the last example, the distance of the perpendicular from the middle part of the 21*–17* –35– 12.6 and 2.6.-The length of the perpendicular and the area of the triangle are, in each case, therefore, easily deduced from these data.
perpendicular from the middle of the base is
base is =76, and the segments of that base are hence
14. From the corollary to the last proposition is derived a very simple construction of the problem, “to find a square equivalent to a given rectangle.”
Let ABCD be the given rectangle, F. C of which the side AD is greater than L^2. AB. In AB or its production, take AE equal to the half of AD and place
it from E to F; then AF being joined, * HF-4D is the side of the equivalent square. To : A For (II. 23. cor. El..) since the sides *S., ; A' AE and EF of the triangle AEF are o
equal, the square of AF is equivalent to the rectangle under twice AE and AB, that is, from the construction, the rectangle under AD and AB.
The same construction might likewise be deduced from the demonstration of the celebrated property of the rightangled triangle. For, in the figure of page 52, suppose BO were drawn to the hypotenuse AC, making an angle ABO
equal to BAO or BAC'; since the two acute angles are together equal to a right angle, the angle BCA is equal to the remaining portion CBO of the right angle at B, and consequently the triangles AOB and COB are isosceles, and the sides OA, OB and OC all equal. Wherefore AB, the side of a square equivalent to the rectangle ADMN or that under AK and AN, is determined by making AO equal to the half of AK or AC and inserting it from O to B.—The inspection of the same figure also points out the mode of dissecting the rectangle, and thence compounding the square; for a perpendicular let fall from K on AB is evidently equal to GB or AB. Hence, on AF, in the original construction, let fall the perpendicular DG, transpose the triangle FBA in the situation DHI, and slide the quadrilateral portion into the place of KAHI ; the rectangle ABCD is now transformed into the square KGDI-A slight modification will be required when AB is less than the half of AD. In this construction of the problem, the application of the circle which (III. 37. El.) is indispensably required, is only not brought into view.—When the side AD is double of AB, the point G coincides with F, and the rectangle is resolved into three triangles, which combine to form a square.
15. To this Book some neat propositions may be subjoined. PROP. I. THEOR.
If, from the hypotenuse of a right-angled triangle, portions be cut off equal to the adjacent sides ; the square of the middle segment thus formed, is equivalent to twice the rectangle contained by the extreme segments.
Let ABC be a triangle which is right-angled at B ; from the hypotenuse AC, cut off AE equal to AB, and CD equal to CB: Twice the rectangle under AD and CE is equivalent to the square of DE. For the straight line AC being divided into three portions, the squares of AE and CD, together with twice the rectangle AD, CE are equivalent to the squares of AC K-5-HTC and DE (art. 10.). But the squares of AB and BC, or those of AE and CD, are equivalent to the square of AC (II. 10.). There consequently remains twice the rectangle AD, CE equivalent to the square of DE. By an inverse process of reasoning it will appear, that if twice the rectangle AD, CE be equal to the square of DE, the straight line AC, so composed, is the hypotenuse of a right-angled triangle, of which AB and BC are the sides. This proposition will furnish another convenient method of discovering the numbers which represent the sides of a rightangled triangle : For since DE*=2AD.CE, it is evident that #DE*=AD.CE; and consequently, expressing DE by an even whole number, and resolving #DE” into the factors AD and CE, AD+DE and CE--DE will represent the two sides, and AD+CE+DE the hypotenuse. Thus, if 2 be taken, the factors of half its square are 1 and 2, which produce the numbers 3, 4, and 5. Again, if 4 be assumed, the factors are 2 and 4, or 1 and 8; whence result these numbers, 6, 8, and 10, or 5, 12, and 18. In this way, a very great variety of numbers can be found, to express the sides of a right-angled triangle.
PROP. II. THEOR.
The squares of lines drawn from any point to the opposite corners of a rectangle are together equivalent.
If from a point E, either within or without the rectangle ABCD, straight lines be drawn to the 'four corners, the squares of AE, EC are together equivalent to the squares of BE, ED,
For join E with F, the intersection of the diagonals AC, BD. Because it follows readily from Prop.27. Book I, that these diagonals are equal, and bisect each other, the lines AF, BF, CF, and B C DF are all equal. Wherefore the squares of AE, EC are equivalent to twice the square of AF, and twice the square of EF (II. 22.); and the squares of BE, ED are likewise equivalent to twice the square of BF and twice the same square of EF; consequently, the squares of AF and BF being equal, the squares of AE, EC, are together equivalent to the squares of BE, ED.
PROP. III. THEOR.
If straight lines be drawn from the angular points of a triangle to bisect the opposite sides, thrice the squares of these sides are together equivalent to four times the squares of the bisecting lines.
Let the sides of the triangle ABC be bisected in D, E, and F, and straight lines drawn from these points to the opposite vertices; thrice the squares of the sides AB, BC, and AC are together equivalent to four times the squares of BD, CE and AF.
For, by Proposition II. 22. the squares of AB, BC are equivalent to twice the square of BD and twice the square of AD, that is, half the square of AC ; the squares of BC, AC are equivalent to twice the squares of CE and half the square of AB; and the squares of AC, AB are equivalent to twice the square of AF and half the square of BC. Whence the squares of the sides of the tri