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intersect, the line PQ will be equal to the sum of the lines B P and co.
14. The difference of the angles at the base of a triangle is double the angle contained by a line drawn from the vertex perpendicular to the base, and another line bisecting the angle at the vertex.
15. If in two opposite sides of a parallelogram two points be taken, one in each of those sides equidistant from two opposite angles of the figure, and if two other points be likewise taken, in the two other opposite sides, equidistant from the same two angles, the figure contained by the straight lines joining the four points so taken shall be a parallelogram.
16. The area of a trapezium is half that of a rectangle, whose base is one of the diagonals of the trapezium, and altitude the sum of the perpendiculars let fall upon the diagonal.
17. The lines which bisect the angles of a parallelogram, form a rectangle.
18. Two sides of a triangle being given, its area will be greatest when the given sides contain a right angle.
19. If the two exterior angles at the base of a triangle be bisected, and the bisecting lines produced until they intersect, the line drawn from this point to the vertical angle will bisect it.
20. In the figure E. 1. 47. prove that, if bg and ch be joined, these lines will be parallel.
EXERCISES ON BOOK II. 1. THEOREM. In any triangle, if a line be drawn from the vertex bisecting the base, the sum of the squares of the two sides of the triangle is double the sum of the squares of the bisecting line and of half the base.
Let ABC be a triangle; and from the vertex c let cd be drawn to the middle, D, of the base;, then the squares of AC, BC are together double the squares of DC, AD.
From c draw CE perpendicular to AB.
DE greater than the squares of DC, AD, by twice the rectangle AD, DE; and (E. 11. 13.) the square of bc is less than the
squares of DC, BD by twice the rectangle BD, DE; therefore . the squares of AC, BC are equal to twice the square of DC, and twice the square of AD, since AD is equal to BD.
= DC? + AD2 – 2 AD.DE, i AC + BC2 = 2 DC2 + 2 A D2. 2. The square upon the whole line is equal to four times the square upon half the line.
3. The rectangle under any two lines is equal to three times the rectangle under either of them and a third of the other.
4. If two lines be divided, each into any number of parts, the rectangle contained by the two lines, is equal to the sum of the rectangles contained by the several parts of the one and the several parts of the other.
5. In any triangle if a line be drawn from the vertex at right angles to the base, the difference of the squares of the sides is equal to the difference of the squares of the segments of the base.
6. If from one of the acute angles of a right-angled triangle, a line be drawn to the opposite side, the squares of that side and the line so drawn are together equal to the squares of the segment adjacent to the right angle and of the hypotenuse.
7. In any isosceles triangle, if a line be drawn from the vertex to any point in the base, the square upon this line, together with the rectangle contained by the segments of the base, is equal to the square upon either of the equal sides.
8. The rectangle contained by the sum and the difference of two lines is equal to the difference of their squares.
9. If a line be divided into five equal parts, the square of the whole line is equal to the square of the line which is made up of four of those parts, together with the square of the line which is made up of three of those parts.
10. The squares of the sides of a parallelogram are together equal to the squares of its diameters taken together.
11. If from the three angles of a triangle lines be drawn to the points of bisection of the opposite sides, four times the sum of the squares of these lines is equal to three times the sum of the squares of the sides of the triangle.
12. If from any point within a rectangle lines be drawn to
the angular points, the sums of the squares of those which are drawn to the opposite angles are equal.
13. The square of the base of an isosceles triangle is double the rectangle contained by either side and by the line intercepted between the perpendicular let fall upon it from the opposite angle, and the extremity of the base.
14. If two sides of a trapezium be parallel to each other, the squares of its diagonals are together equal to the squares of its two sides, which are not parallel, and twice the rectangle contained by its parallel sides.
15. The sum of the perpendiculars let fall from any point within an equilateral triangle, upon the three sides, is equal to the perpendicular let fall from one of the angles upon the opposite side.
16. The squares of the diagonals of a trapezium are toge. ther double the squares of the two lines joining the bisections of the opposite sides.
17. The squares of the diagonals of a trapezium are together less than the squares of the four sides, by four times the square of the line joining the points of bisection of the diagonals.
18. If from any point within (or without) a trapezium perpendiculars be let fall on every side, the sum of the squares of the alternate segments made by them will be equal.
19. To divide a given line into two parts, such that the squares described upon them shall be equal to a given square. Show when the problem becomes impossible.
20. Divide a given straight line into two parts, such that twice the rectangle contained by them may be equal to the square of one of the parts.
21. Divide a given straight line into two parts, such that nine times the square described upon one part may be equal to the square described upon the other.
22. Divide a given straight line into two parts, such that the square of the whole line may be equal to the square of one of the parts, together with four times the rectangle contained by the two parts.
23. Divide a given straight line into three parts, such that the square upon the sum of the greatest and least parts may be four times the square upon the remaining part; and the sum of the squares upon the two least parts may be equal to the square upon the greatest.
24. Upon a given line, as an hypotenuse, to describe a
right-angled triangle, such that the hypotenuse, together with the less of the two sides, shall be double the greater. 25. Show that the algebraical proposition
(a+b) + (a-b) = 2a + 20%, is equivalent to Props. 9 and 10 of the Second Book of Euclid.
26. Find algebraical propositions equivalent to Propositions 1, 2, 3, 4, 5, 6, 7, and 8 of the Second Book of Euclid.
EXERCISES ON BOOK III.
1. To describe a circle which shall touch a given straight line in a given point, and also touch a given circle.
Let FK be the given circle, and c the given point in the straight line xy: It is required to describe a circle which shall touch xy in the point c, and also touch the circle FK.
Through c draw (E. 1. 11.) the line DCB perpendicular to XY; find (E. III. 1.) the centre A of the circle FK, and draw any radius AF; make cd equal to AF, and join A, D; from a draw (E. 1. 23.) AB, making the angle DAB equal to the angle AD B, and let DCB meet AB in the point B. Then B is the centre of the circle required.
Since the angle DAB is equal to the angle ADB (constr.), therefore (E. 1. 6.) bd is equal to BA; and cp is equal to AF or AK; therefore the remainder Bc is equal to the remainder BK, and consequently a circle described from B as a centre, with the radius BC, will pass through K, and (E. III. 16.) it will touch xy in c, and moreover it will touch the circle Fk in K, for a line drawn perpendicular at K will form a common tangent to the circles FK and cK.
2. Given the base, the vertical angle, and the perpendicular height of a triangle, to construct it. Let AB be the given base: Upon AB
DF describe (E. III. 33.) a segment of a circle ADB, which shall contain an angle equal to the given vertical angle; from B draw BF perpendicular to A B, and make BF equal to
the given height of the triangle; through r draw FD parallel to AB, meeting the arc of the circle in D; join B, D and A, D: Then ABD is the triangle required.
Draw Ds parallel to FB, then BFDs is a parallelogram, and (E. I. 34.) Ds is equal to FB; and (constr.) because the angle sbf is a right angle, therefore (E. 1. 29.) the angle DSA is a right angle, that is, Dş is perpendicular to A B, and it has been shown to be equal to FB, which was taken equal to the given perpendicular; moreover (constr.) the angle ADB is equal to the given vertical angle; therefore ADB is the triangle which was to be constructed.
3. To determine the position of a point of observation, D, at which lines drawn from three objects (A, B, C), whose distances from each other are given, shall make with each other angles equal to given angles ; that is, let the angles ADB and ADC be given angles.
Let A, B, C be the three given points or objects; join A B, and on it describe (E. 111. 33.) a segment of a circle ADB, which shall contain an angle equal to the angle which the lines drawn from A and B are to include. Describe the whole circle, ADBE, and make (E. 1. 23.) the angle A BE equal to the angle which the lines drawn from A and c are to include. Join C, E; and produce it to D), the circumference of the circle ADBE; then d is the position of the point required.
Join Å, E; then (E. 111. 21.) the angle ADC is equal to the angle ABE, which was made equal to the angle formed by the lines drawn from a and c; and, moreover, the angle ADB is, by construction, equal to the angle formed by the lines drawn from A and B.
4. Through a given point within a circle, to draw a chord which shall be bisected in that point.
5. To draw a tangent to a circle, which shall be parallel to a given straight line.
6. Given the hypotenuse and the perpendicular let fall upon it from the opposite angle, to construct the right-angled triangle.