beam from the vertical motion of the piston. Let O A be half of the balance-beam oscillating about the point O, and TT' the vertical line

along which the top of the

piston-rod moves. It is evident that tne rod cannot be attached to a fixed point in the beam; for while the former describes a straight line, the latter describes the arc of a circle. Watt fixed to the balance-beam OA a jointed parallelogram, ABCM, a point, C, of which he attached by a rod, CI, to a fixed point, I. Let us suppose the beam to oscillate about its centre, O, and let OA, O A', OA” represent different positions of the beam. The point A is rigidly attached to a point O, so that it describes the arc A A' A", of which

FIG. 78.

O is the centre, C describes the arc C C' C", which has the point I as centre; and then M describes the straight line M M'M". Consequently if M be jointed to the end of the piston-rod, by moving along the line M M'M" the rod will cause the beam to swing about O. This arrangement answers in practice if the angle A OH does not exceed 20°.

THEOREMS ON CHAPTER VI.

DEFINITIONS.

36. A quadrilateral is a four-sided figure.

37. A parallelogram is a quadrilateral whose opposite sides are parallel.

38. A rhombus is a quadrilateral all of whose sides are equal.

39. A rectangle is a right-angled parallelogram.

40. A square is a rectangle all of whose sides are equal.

THEOREM XIX.

The opposite sides and angles of a parallelogram are equal, and the diagonal bisects it.

Let A B CD be a parallelogram and D B a diagonal, then shall DC = AB, DA = CB,

B

C

=

4 DAB ▲ BCD, ▲ ADC = 4 CBA, and ▲ ADB = ▲ CBD. Since A B and DC are parallel, the alternate angles CD B, A B D are

equal. Since AD and BC are parallel, the alternate angles A D B, C B D are equal. By addition, therefore, ▲ ADC = 4CBA.

Again, the two triangles ADB, CBD have the two angles A D B, A B D equal respectively to the two angles CBD, CDB and the side D B common; consequently ▲ ADB = ▲ CB D in all respects, and DC = AB, DA CB, and ▲ DAB = 4 BCD.

THEOREM XX.

The diagonals of a parallelogram bisect one another; And conversely, if the diagonals of a quadrilateral bisect one another the quadrilateral is a parallelogram.

Let ABCD be a quadrilateral and AC, DB its diagonals intersecting in O; and first

let the figure be a parallelogram, then shall the diagonals AC and DB bisect one another.

The triangles OCD, OBA have CD =

A

B

C

A B, because ABCD is a parallelogram; ▲ CDO=2AB O, because they are alternate angles; <DCO = < BAO, because they are alternate angles. Consequently the triangles are equal in all respects, and DO=BO,CO=A O.

Secondly, let A B C D be a quadrilateral, and let the diagonals bisect one another in the point O, then shall ABCD be a parallelogram.

Triangles DOC, BOA have

CO=AO, DOBO and COD 4 AOB;

Therefore the triangles are equal in all respects, and 4 CDO = 4ABO.

But these are alternate angles, consequently CD and A B are parallel.

Similarly it may be shown that AD and BC are parallel.

THEOREMS FOR EXERCISE.

30. If a parallelogram have one angle a right angle, all its angles are right angles.

31. A quadrilateral which has its opposite angles equal is a parallelogram.

F

32. A quadrilateral which has its opposite sides equal is a parallelogram.

33. When two sides of a quadrilateral are equal and parallel the quadrilateral is a parallelogram.

34. The diagonals of a rectangle are equal.

35. When the diagonals of a parallelogram are equal the parallelogram is a rectangle.

36. The diagonals of a square are at right angles.

37. When the diagonals of a rectangle are at right angles the rectangle is a square.

38. The sum of the four sides of a quadrilateral is greater than the sum and less than twice the sum of the diagonals.

39. Any straight line, drawn through the point of intersection of the diagonals of a parallelogram and terminating in the sides, is bisected at the point.

40. When two straight lines bisect one another the lines joining their extremities form a parallelogram.

41. If any two lines be drawn through the centre of a parallelogram, and the points in which they meet the sides be joined, the figure formed will be a parallelogram.

42. If a straight line be drawn through the middle point of a side of a triangle parallel to the base it will bisect the other side.

43. The line joining the middle points of two sides of a triangle is equal to half the third side and is parallel to it.

44. If from points in the base of an isosceles triangle perpendiculars be drawn to the sides, their sum is the same for each of the points.

45. The straight lines joining the middle points of the sides of a quadrilateral form a parallelogram.

46. Two equal straight lines, AB, CD, which are drawn between two parallels, AC, DB, intersect in the point O. Prove that A O CO and BO=DO.

47. Two straight lines are drawn at right angles through the centre of a square and terminating in the sides. Prove that the lines joining their extremities form a square.

48. The diagonal of a rectangle is larger than any other straight line intercepted by the sides of the rectangle.

49. Straight lines are drawn from a point O within a parallelogram, ABCD, to the angular points. Prove that the triangles A B O, DCO are together equal to half the parallelogram.

50. Through the angular points of a triangle straight lines are drawn, parallel to the opposite sides. Show that a triangle is formed which is four times the original triangle.