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THEOREM M.

84. The opposite sides of a parallelogram are equal, as well as the opposite angles.

Draw the diagonal BD. The triangles ADB, DBC, » c

have the common side DB ; also, because of the parallels, / ^n. /

AB, CD, the angle ABD is equal to CDB {theorem 22); L B

and, because of the parallels AD, BC, the angle ADB is equal to DBC; therefore the triangles (theorem 6) and the sides AB, DC, which are opposite the equal angles, are equal. In like manner AD and BC are equal; therefore the opposite sides of the parallelogram are equal.

Again, from the equality of the triangles, it follows, that the angle A is equal to the angle C; and it has been shown that the angles ADB, BDC, are respectively equal to the angles CBD, DBA; therefore the whole angle ADC is equal to the whole angle ABC, and thus the opposite angles are equal.

85. Corollary.—Two parallels, AB, CD, comprehended between two other parallels, AD, BC, are equal.

Theorem 29.

86. If the opposite sides of a quadrilateral be equal, the figure is a parallelogram.

For, drawing the diagonal BD, (as above,) the triangles ABD, BDC, have the three sides equal, each to each; therefore the angle ADB, opposite to the side AB, is equal to the angle CBD, opposite to the side CD {theorem 10); hence the side AD is parallel to BC (Theorem 19). For the like reason AB is parallel to CD: therefore the quadrilateral, ABCD, is a parallelogram.

Theorem M.

87. A straight line, BD, drawn perpendicular to the extremity of a radius, CA, is a tangent to the circumference.

For every oblique line, CE, is longer than the perpendicu

B A ED

lar CA {theorem 15); therefore the point E must be without the circle; and since this is true of every point in the line BD, except the point A, the line BD is a tangent {def.9,p. 11).

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THEOREM 31.

88. Only one tangent can be drawn from a point, A, in the circumference of a circle.

Let BD be a tangent at A, in the circumference, de- Ah. —^ »

scribed with the radius CA; and let AG be another tangent, if possible; then, as CA would not be perpendicular to AG, another line, CF, would be perpendicular to AG, and so CF would be less than CA (theorem 14); therefore F would fall within the circle, and AF, if produced, would cut the circumference.

Theorem 32.

89. If two circumferences cut each other, the straight line which passes through their centres shall be perpendicular to the chord which joins the points of intersection, and shall divide it into two equal parts.

For the line AB, which joins the points of intersection, being a common chord to the two circles; if, through the middle of this chord, a perpendicular be drawn, it will pass through the points C, D, the centres of the two circles. But only one line can be drawn through two given points; therefore the straight line which passes through the centres is a perpendicular to the middle of the common chord.

Theorem M.

90. In the same circle, or in equal circles, equal angles ACB, DCE, at the centre, intercept equal arcs, AB, DE, on the circumference; and conversely, if the arcs AB, DE, be equal, the angles ACB and DCE are also equal.

If the angle ACB be equal to DCE, these two angles may be placed on each other; and, as their sides are equal, the point A will fall on D, and the point B on E; but then the arc AB must also fall on DE; for, if the two arcs did not coincide, there would be, in one or the other, points unequally distant from the centre; therefore the arc AB is equal to DE.

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Next, if the arc AB be equal to DE, the angle ACB shall be equal to DCE; for, if they are not equal, let ACB be the greater, and take ACI equal to DCE; then, by what has been demonstrated, AI is equal to DE; but, by hypothesis, the arc AB is equal to DE; therefore the arc AI is equal to AB, which is impossible: therefore the angle ACB is equal to DCE.

Theorem M.

91. An angle, ACB, at the centre of a circle, is double of the angle at the cirumference, upon the same arc, AB.

Draw DC, Mg. 1,) and produce it to E. First, let the angle at the centre be within the angle at the circumference, then the angle ACE is equal to the sum of the angles CAD, CDA (Theorem 27); but, because CA is equal to CD, the angle CAD is equal to CDA {theorem 11); therefore the angle ACE is equal to twice the angle CDA. By the same reason the angle BCE is equal to twice the angle CDB; therefore the whole angle ACB is double the whole angle ADB.

Next, let the angle at the centre {fig. 2,) be without the angle at the circumference. It may be demonstrated, as in the first case, that the angle ECB is equal to twice the angle EDB, and that the angle ECA, a part of the first, is equal to twice EDA, a part of the second; therefore, the remainder, ACB, is double the remainder ADB.

Theorem M.

92. The angles, ADB, AEB, in the same segment, AEB, of a circle, are equal to one another.

Let C {fig. 1) be the centre of the circle; and, first, let the segment AEB be greater than a semi-circle. Draw CA, CB, to the ends of the base of the segment; then each of the angles, ADB, AEB, will be half of the angle ACB {theorem 34); therefore the angles ADB and AEB are equal.

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straight line which meets another, each of these angles is a right angle; therefore each of the angles ABD and DBF is a right angle; and, consequently, the angle ABD in a semi-circle is a right angle.

Again, because in the triangle ABD, the angle ABD is a right angle; therefore BAD, which is manifestly in a segment less than a semi-circle, is less than a right angle: and, lastly, because ABED is a quadrilateral in a circle, the sum of the two angles A, E, is equal to two right angles; but the angle A is less than a right angle; therefore E, which is in a segment less than a semi-circle, is greater than a right angle.

Theorem 38.

95. The angle BAE, contained by a tangent AE to a circle, and a chord AB, drawn from the point of contact, is equal to the angle AGB in the alternate segment.

Let the diameter ACF be drawn, and GF be joined; and, because the angles FGA, FAE, are right-angles (theorems 37,30), and of these FGB, a part of the one, is equal to FAB, a part of the other, [theorem 35,) the remainders BAE, BAG, Tr are equal.

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ALGEBRA.

96. ALGEBRA is a method of demonstrating propositions, and resolving questions, by means of the letters of the alphabet used as symbols.

Letters are employed to denote angles, lines, surfaces, or solids; each being considered as a multitude of units of the kind to which it belongs; and, consequently, as number, abstracted from figure. Thus, a letter may denote the number 3, or a line of five equal parts of any measure; as inches, yards, miles, &c. and so on.

Letters which are thus generally employed are called quantities; they being the representation of quantities.

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