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and Fge, and the angles .at G are right angles, therefore (E. I. 4.) the triangles CoE and FgE are equal, and the angle ceg is equal to the angle FEG; but (E. 1. 15.) the angle deh is equal to the angle FEG, therefore the angle CEG is equal to the angle DEH.
5. On a given line to describe an isosceles triangle, whose perpendicular height is equal to the base.
6. To make an angle equal to half a right angle. 7. Given the diagonal of a square to construct it.
8. Given the base, the perpendicular height, and one of the angles at the base of a triangle, to construct it.
9. Trisect two right angles.
12. Given the hypotenuse of a right-angled triangle, and the difference of the two acute angles, to construct the triangle.
13. To construct a right-angled triangle, having given the hypotenuse, and one of the acute angles equal to one third a right angle. Show that the side opposite to the given angle is equal to one half the hypotenuse.
14. Given the two sides, and an angle opposite to one of the given sides of a triangle, to construct it. Show that, in general, there are two triangles answering the conditions of the problem.
15. Given the base, one of the other sides, and the perpendicular on the base of a triangle, to construct it.
16. On a given base, to construct an isosceles triangle having the angle at the vertex equal to one third a right angle.
17. From the vertex of a given triangle, to draw to the base a straight line which shall be less than the greater side by a given line.
18. Given the perpendicular and the equal side of an isosceles triangle, to construct it.
19. Through a given point P, to draw a straight line, which shall cut off equal parts from two straight lines AB and AC, cutting one another in A.
20. Given the vertical angle and the perpendicular height of an isosceles triangle, to construct it.
21. Given the base, the less angle at the base, and the difference of the sides of a triangle, to construct it.
22. From a given point, to draw the shortest line possible to a given straight line.
23. In a straight line given in position, but indefinite in length, to find a point, which shall be equidistant from each of two given points, on the same side of the given line, and in the same plane with it.
24. To draw a straight line from a given point to meet another straight line, which shall make with it an angle equal to a given rectilineal angle.
25. To find a point, in either of the equal sides of a given isosceles triangle, from which, if a straight line be drawn, perpendicular to that side, so as to meet the other side produced, it shall be equal to the base of the triangle.
26. From a given point to draw a straight line cutting two parallel straight lines, so that the part of it intercepted between them shall be equal to a given finite straight line, not less than the perpendicular distance of the two parallels.
27. To trisect a triangle by lines drawn from the vertex to the base.
THEOREMS. No. 1. 1. The straight line which joins the points of bisection of the two sides of a triangle is parallel to the base.
Let ABC be a triangle, D and E the points of bisection of the sides AC and BC respectively; then the straight line DE, which joins the points D and E, is parallel to the base A B.
Join the points A and E, B and D. Because AD is equal to DC, therefore (E. I. 38.) the triangle ABD is half the triangle ABC; and for the same reason the triangle A BE is also half the triangle ABC; therefore the triangles ABD and A B E are equal to one another; hence (E. 1. 39.) DE is parallel to A B.
2. If the sides of a trapezium be bisected, and the adjacent points be joined, the figure thus formed will be a parallelogram.
• Should the student fail in solving any of these problems by the ordinary synthetic method, let him employ the method of analysis given at page 107.
Let. ACBH be a trapezium, whose sides are bisected in the points D, E, G and F; let the adjacent points be joined ; then the figure DEGF is a parallelogram.
Draw the diagonals AB, CH. Since AC, BC are bisected in D and E (Theorem 1. No. 1.), therefore DE is parallel to AB; and for the same reason Fg is parallel to AB; therefore (E. I. 30.) DE is parallel to FG.
In the same way it may be shown that Fd is parallel to GE; therefore the figure DEGF is a parallelogram.
3. Each of the angles of an equilateral triangle is equal to two-thirds of a right angle.
4. The diagonals of a parallelogram bisect one another. 5. The diagonals of a square cut each other at right angles.
6. The three sides of a triangle taken together exceed the double of any one side, and are less than the double of any two sides.
7. If two opposite sides of a quadrilateral figure be equal to one another, and the two remaining sides be also equal to one another, the figure is a parallelogram.
8. If the sides of a triangle are 3, 4, and 5, taken from a scale of equal parts, it is a right-angled triangle.
9. Any side of a triangle is greater than the difference between the other two sides.
10. Any one side of a quadrilateral figure is less than the sum of the remaining sides.
11. If either of the equal sides of an isosceles triangle be produced towards the vertex, the straight line which bisects the exterior angle is parallel to the base.
12. If the opposite angles of a quadrilateral figure be equal to each other, the figure is a parallelogram.
13. The diameters of an 'equilateral four-sided figure bisect one another at right angles.
14. The diameters of a rectangle are equal to one another.
15. The line which joins the points of bisection of any two sides of a triangle is equal to half the remaining side.
16. The diameters of any parallelogram divide it into four. equal triangles.
17. The two triangles formed by drawing straight lines, from any point within a parallelogram, to the extremities of
either pair of opposite sides, are together half of the parallelogram.
18. The angle at the base of an isosceles triangle is equal to, or is less, or greater, than the half of the vertical angle, accordingly as the triangle is a right-angled, an obtuseangled, or an acute-angled triangle.
19. If a perpendicular be drawn bisecting a given straight line, any point in this perpendicular is at equal distances from the extremities of the line.
20. Of all straight lines which can be drawn from a given point to an indefinite straight line, that which is nearer to the perpendicular is less than the more remote. And from the same point there cannot be drawn more than two straight lines equal to each other, viz. one on each side of the perpendicular.
21. The perimeter of an isosceles triangle is greater than the perimeter of a rectangular parallelogram, which is of the same altitude with and equal to the given triangle.
22. If from any point in the diameter (or diameter produced) of a parallelogram straight lines be drawn to the opposite angles, they will cut off equal triangles.
23. The two perpendiculars let fall from the extremities of the base of an isosceles triangle upon the sides, are equal to each other.
24. If a line be drawn bisecting a given angle, any point in that line will be equally distant from the two lines forming the angle.
25. Any straight line which bisects the diagonal of a parallelogram also bisects the parallelogram.
26. The perpendiculars drawn from the opposite angles of a parallelogram upon the diagonal are equal to one another.
27. If two straight lines bisect each other perpendicularly, the lines joining the extremities will form an equilateral parallelogram, or rhombus.
28. The area of a rhombus is equal to half the rectangle constructed upon the two diagonals of the rhombus.
29. Parallelograms whose sides and angles are equal are themselves equal.
30. The lines which join the middle points of the three sides of a triangle, divide it into four equal triangles.
31. The inscribed parallelogram DEGF, in Exercise 2., is equal to half the trapezium ACBH.
32. If perpendiculars be let fall from the angles at the
base of a triangle, upon the line drawn from the vertex to the middle of the base, these perpendiculars will be equal.
33. If two opposite sides of a trapezium be parallel to one another, the straight line joining their bisections, bisects the trapezium.
34. If from the extremities of the base of an isosceles triangle lines be drawn perpendicular to the sides, the angles made by these perpendiculars with the base are each equal to half the vertical angle.
35. In Theorem 1. No. 1., prove that the triangle DCE is one-half the triangle ABD.
PROBLEMS. No. 2. 1. To draw a perpendicular to a straight line through its extremity without producing it.
Let A be the extremity of the line at: It is required to draw Ac perpendicular to AE.
Take any portion AB, and on it construct the equilateral triangle ABD; produce BD to C, making Dc equal to DB; join Ac; then AC will be the perpendicular required.
Since AD is equal to DC (E. 1. 5.) the angle dac is equal to the angle DCA; and because ABD is an equilateral triangle the angle DAB is equal to the angle ABC; therefore the angles DAC and DAB taken together are equal to the sum of the angles DCA and ABC; but the angles DAC and DAB are together equal to the angle BAC; therefore the angle Bac is equal to the sum of the angles DCA and ABC; hence (E. 1. 32.) the angle BAC is equal to a right angle.
2. Given the sum of the three sides, and the angles at the base of a triangle, to construct it.
Let AB be the sum of the three sides, and c and D the two given angles : It is required to describe a triangle, which shall have the sum of its sides equal to A B, and the angles at the base equal to c and D respectively.