Therefore bisect the given line in C, and cut its half CB so that the of the intermediate part shall be = to half the difference between two rs of half and the given quantity. Then it is evident from prop. 9, b. 2, that the 'rs of AD,DB are together to the given quantity. 9. By being given the hypothenuse and difference of the squares of the sides of a right angled triangle, we can find the sides. For the difference between the sum of the 'rs and difference of the 'rs is to two rs of the lesser side. 10. It can be proved by the sixth of the second book, that the rectangle under the sum and difference of two lines, is equal to the difference of their squares : or, that the rectangle under two lines, plus the square of their difference, is equal to the square of half their sum. 11. The common enunciation of the 5th and 6th is, that the rectangle under any two lines is equal to the difference between the squares of half their sum and half their difference. 12. The squares of the diagonals of any parallelogram are equal to the sum of the squares of the sides. For by the last scholium of the four sides are together half each diagonal, i. e. to the of prop. 13, b. 2, the *rs of 13: The squares of the difference, the sum of the squares, and the square of the sum of any two lines, are in arithmetical progression. 2 rs For they have a common difference, i. e. the difference between the of the difference and sum of the is two rectangles under the lines and the difference between the sum of the 'rs and of the sun is the same, 14. Given the hypothenuse and difference of the sides of a right angled triangle to find it, and its area. 2rs is to two The hypothenuse being given, the sum of the given, ... the difference between the sum of the 'rs and of the difference is given; which is rectangles under the sides, add this to the sum of the Dirs and their product is the of the sum, .. you have the sum of the sides, and their difference being given, you can find the sides and .. the triangle. 15. Given the sum of the sides and hypothenuse to find the triangle. The sum of the sides being given, the of the sum is given, and the hypothenuse being given, the difference between the of the sum and sum of the rs is given, .. the difference of the sides is given, and .. the sides themselves. 16. Given of any triangle the base, the line bisecting the base and sum of the sides to find the triangle, Take from the of the sum two rs of half the base and two'rs of the bisecting line, which are together to the 'rs of the sides, (cor. to 13, b. 2,) the remainder is to two rectangles under the sides; then you can cut the given sum so that the rect. under the parts shall be to half this difference, (by obsr、1. b. 2;) and.. you have the sides. 17. Given the sum of the sides and sum of the perpendicular and hypothenuse to find the triangle. 2 2 The of the sum of hypothenuse and perpendicular exceeds the 2 of the other sum by a ☐ of the perpendicular, .. you can find the perpendicular, .. the hypothenuse and... the triangle. 1 18. The square of the sum of the hypothenuse and perpendicular, exceeds the square of the sum of the sides, by a square of the perpendicular. 2 For the of the sum of the hypothenuse and perpendicular is to the sum of their 'rs and four times the area of the triangle, and the of the other sum is to the sum of their rs and four times the area of the triangle ;. take from both quantities the areas, and there remains on one side the sum of the 'rs of the sides and perpendicular, and on the other the sum of the 'rs of the sides, .. &c.. 19. Given the perimiter and sum of the squares of the sides of a right angled triangle to find it. The sum of the 'rs being given, the hypothenuse is given and ... the sum of the sides, .. you can find the triangle, (obser. 13, b. 3.) 20. In a right angled triangle, given one side [AB] and a segment [CD] of the hypothenuse made by a perpendicular, to find the triangle. Suppose it done, and that ABD is the given triangle. Then it is evident that the rectangle DCA with the of CA is to the 2 of AC the given side... the rectangle DAC is to the 2 of the given side. 2 Therefore produce the given segment until the rect, under the whole produced line and produced part is to the of the given side, (obser. 3. b. 2,) this whole produced line shall be the hypothenuse of the required triangle: .. you can find the triangle. 21. By having the area and hypothenuse of a right angled triangle, it can be found. For divide the area by half the hypothenuse and you will have the perpendicular, and then you can find the triangle. 22. Given the area and difference [AB] of the sides of a right angled triangle to find it. Suppose that ACD is the required triangle, and AB to the given difference, BC is to CD, but the rect angle AC,CD is to twice the area of ACD, .. the rect. ACB is to twice the given area. If therefore you produce the given difference until the rect. ACB is to twice the area, it is evident that AC and CB are the sides of the required triangle.. 23. Given the area and sum of the sides of a right angled triangle to find it. Cut the given sum so that the rect. under the parts may be to double the given area (by ob.I. b. 2.) 24. Given the segments of the hypothenuse to find the sides. Find a 2 to the rect. under their sum and lesser segment and you have the greater side, and also a2= to the rect, under their sum and greater seg. and your have the greater side. 25. Given the difference of the sides and difference of their squares to find the sides. Produce the given difference so that the rectangle under it and produced part may be to the difference of the rs, (ob. 3. b. 2,) this produced part shall be = to the sum of the sides, (prop. 5, b. 2,).. having the sum and difference you can find the sides. 26. To divide a given right line [AD] so that the sum of the squares of the whole line and one part, may be equal to three squares of the other part. ANALYSIS. = Suppose it is cut in the point B as required; then the 'rs of AD,DB are to three rs of AB, but the rs of AD,DB are together to a of AB, two rs of BD and two rect. ABD, .. two rectangles ABD with two rs of BD (i. e. two rect. ADB) are to two.rs of AB,.. the of AB is to a rect. ADB. 2 If then you cut the given line by prop. 11, b. 2, the 'rs of the whole line and lesser segment, will be together to three rs of the greater segment. 27. In a scelene triangle, the square of the greater side, the sum of the squares of the bisecting line and half the base and the square of the lesser, are in arithmetical progression. For they have a common difference, viz. double the rectangle under half the base and intercept between the perpendicular and bisector. 28. Given two sides of a triangle and a tine bisecting the third side, to find the area. Take double the of the bisecting line from the sum of the 'rs of the sides, and there will remain two 'rs of half the third side, .. this side can be found, and .. the triangle. 29. The difference of the segments made by a perpen◄ dicular from one angle of a triangle, on the opposite side, must be greater than the difference of the sides, For the reetangle under the sum and difference of the sides, is to a rectangle under the sum and difference of the segments, since the difference of their 'rs are , (oor. prop. 47, b. ).. it is evident, &c. &c. 30. The square of the hypothenuse of a right angled triangle, plus four times the area of the triangle, is equal to the square of the sum of the sides. For the of the sum is to the of the hypothe nuse and two rectangles under the sides, but two rect, under the sides are to four times the area of the tri angle, . &c. &c. 31. Given one side of a right angled triangle, and the difference between the hypothenuse and the other side, to find it. 2 The of the given side is to a rectangle under the sum and difference of the hypothenuse and third side, (prop. 5,, b. 2,) for it is to the difference of their ars, (cor. prop. 47, b. 1,) then produce the given difference so that the rectangle nnder it and produced part may |