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PROPERTIES OF CIRCLES. THEOREM XXXVI.—If any point D (fig. 56, p. II.), be taken in ACB, the diameter of a circle DB, that part of the diameter which passes through the centre, C, is greater than any other line, as DE, that can be drawn from D to the circumference of the circle; and D E is greater than any other line DF drawn from D to the circumference; but at a greater distance from D B.

For, join CE and C F, then as C E is equal to CB, DC and C E together are equal to DB; and, as the sum of DC and C E is greater than DE (Theorem XIII.), D B is greater than D E. Again as DC and CE are respectively equal to DC and CF, but the angle DCE is greater than the angle DCF, DE (Theorem XI.) is greater than DF.

THEOREM XXXVII.-An angle ACB (fig. 57, p. II.) at the centre of a circle is double the angle ADB at the circumference, when they both stand on the same arc A B.

For join D C, and produce it as to E; then (Theorem XIX.) the angle ACE is equal to the sum of the angles CAD, and CDA; but, as CA is equal to CD, the angles CAD and CDA are equal; the angle ACE is therefore double the angle ADC. For a like reason the angle BCE is double the angle BDC, consequently the angle AC B is double the angle A D B.

Cor. 1. As the angle AC B, at the centre, is measured by the arc A B on which it stands, the angle ADB at the circumference is measured by half the arc A B, on which it stands.

Cor. 2. All angles in the same segment of a circle, or standing on the same arc, are equal to each other.

Cor. 3. The angle in a semicircle is a right angle; for, when AC and C B become one straight line, the angle ACB becomes equal to two right angles, and the angle A D B, which is in that case an angle in a semicircle, is a right angle.

THEOREM XXXVIII.-The sum of any two opposite angles BAD, BCD, of a quadrilateral inscribed in a circle, is two right angles (fig. 58, p. II.)

For the three angles of the triangle A D B are together equal to two right angles (Theorem XIX. Cor. 1). But (Theorem XXXVII. Cor. 2), the angle A D B is equal to the angle AC B, and the angle ABD to the angle A CD, hence the angles BAD and BDC are together equal to two right angles.

Cor. If A B be produced, as to E, then (Theorem VI.) the angles D A B and DAE, together, make two right angles; hence the angle DAE is equal to the inward and opposite one BCD.

THEOREM XXXIX.-In any circle as ABCD (fig. 59, p. II.) parallel chords as A B, CD, intercept equal ones, AC and B D.

For, BC being joined, the angles ABC and BCD are equal (Theorem XV.), and hence (Theorem XXXVII. Cor. 1) half the arc A C is equal to half the arc BD, or the whole arc A C, to the arc BD.

Cor. If A B be conceived to move parallel to itself till it coincide with the tangent EG F, then the arcs CG and G D intercepted by a tangent and a chord parallel to it will appear to be equal.

THEOREM XL-If A B (fig. 60, p. II.) be a tangent to a circle, and AC a chord, drawn from the point of contact A, the angle DAC is equal to any angle in the alternate segment ADC.

For from C draw the chord DC parallel to AB, and join AD. Then, as the arcs AC and AD are equal (Theorem XXXIX. Cor.), the angles ACD and ADC are equal (Theorem XXXVII. Cor. 2). But (Theorem XV.) the angles ACD and BAC are equal; therefore the angles BAC and ACD are equal; and, as ADC is equal to any angle in the same segment, BAC is also equal to any angle in that segment.

THEOREM XLI.-The angle EAC (fig. 61, p. II.), formed by two lines cutting a circle, is measured by half the sum or half the difference of the intercepted arcs EC and BD, according as the point A is within or without the circle.

For join EB; then when A is within the circle, the angle EAC is the sum of the angles CBE and DEB; but when A is without the circle, the angle EAC is the difference of the angles C BE and D E B (Theorem XIX. Cors. 1 and 2). But (Theorem XXXI. Cor. 1) CBE is measured by half E C, and the angle D E B by half B D ; hence the truth of the proposition is manifest.

THEOREM XLII.-If A'B (fig. 62, p. II.), any chord of a circle, be bisected by a line C D drawn to the centre, CD will be perpendicular to AB; or if CD, drawn from the centre, be perpendicular to AB, A B will be bisected in D.

Draw the radii C A, C B, then A B is bisected in D, the triangles ACD and BCD will be identical (Theorem V.), and have the angles ADC and BDC, opposite to the equal sides A C and B C, equal to each other, and CD is therefore perpendicular to A B. Or again, if CD is perpendicular to A B, the angles CDA and CDB will be equal, and, because of the equal sides AC and CB, the angles CAD and CBD will be equal (Theorem III); hence the angles ACD and BCD will also be equal, and as CD is common to both triangles, the side A D will be equal to the side B D (Theorem II.), or A B is bisected in D.

Cor. The angles ACD and B C D are equal, therefore the arcs A E and B E are equal.

THEOREM XLIII.-If on A B, CD (fig. 63, p. II.), two chords in a circle, the perpendiculars EG and GF be drawn; then if the perpendiculars are equal, the chords are equal; and if the chords are equal, the perpendiculars are equal.

For join AG and GC, then (Theorem XLII.) the chords A B and C D are bisected in E and F, if therefore A B is equal to C D, A E is equal to CF. But (Theorem XXXI. Cor.) A G = AE,2+ EG2; and GC' (or A C2) C'F' + GF. Hence A E2 + EG2 = C F2 + GF2; and if from these equals the equal squares A E2 and CF be taken, we have EGG F, or EGGF. In the same way it may be shown that, when EG CF, AE = C F, and consequently A B = CD.

THEOREM XLIV.-A perpendicular as BC (fig. 64, p. II.), at the extremity B of the radius of a circle, is a tangent to the circle.

For from the centre A draw AC to any point C in the line BC; then as ABC is a right angle, ACB is less than a right angle; hence

(Theorem X. AC is greater than AB, or the point C is without the circle. In the same way it may be shown that BC meets the circle in the point B only, and it is therefore a tangent to the circle.

OF PROPORTION.

Proportion is the numerical relation which one quantity bears to another. Quantities between which proportion can exist must be of the same kind, as a line and a line, a surface and a surface, a solid and a solid, an angle and an angle.

A greater quantity is said to be a multiple of a less, when it contains the less a certain number of times without any remainder; and quantities so related are said to have the same relation to each other that unity has to the number which indicates how often the less is contained in the greater.

If a quantity, as A (fig. 66, p. II.), be contained exactly a certain number of times in another quantity B, the quantity A is said to measure the quantity B; and, if the same quantity A be contained exactly a certain number of times in another quantity C, A is also said to be a measure of C; and it is called a common measure of the quantities B and C. The quantities B and C will evidently bear the same relation to each other, that the numbers do which represent the multiple that each quantity is of the common measure A.

Again, if a quantity, D, be contained as often in another quantity, E, as A is contained in B, and as often in another quantity F, as A is contained in C, then the proportion that E has to F will be the same that B has to C, and the quantities B, C, E, and F, are said to be proportional quantities, a relation which is commonly expressed thus B: C:: E: F.

THEOREM XLV.-Any two quantities as A B, CD (fig. 67, p. II.), have the same proportion that their like multiples have.

Let A B be to CD as any number (say 3) to any other number (say 4); or let A B contain three such equal parts as those of which CD contains 4; and let Ef, fg, g F, be any like multiples A a, a b, and b B; and Gh, hi, ik, and k H, the same multiples of Cc, cd, de, and eD; then EF is the same multiple of A B that GH is of CD; and the same that each part of the one is of the corresponding part of the other. And, as the parts of AB and CD are equal, the like multiples of those parts which constitute the parts of EF and GH are also equal. Hence EF is to GH as 3 is to 4, the same proportion that A B has to CD. In the same way may the property be proved, whatever numerical relation AB may have to CD.

Cor. Quantities of the same kind are to each other as their like parts.

THEOREM XLVI.-In any four quantities AB, BC, DE, and EF (fig. 68, p. II.) of the same kind, if AB BC: DE: EF, then also alternately A B : DE:: BC: EF. If A B contain any number of such equal parts A a (say 4), as those of which BC contains any other number (say 3), Bb, then DE also will contain four such equal parts D d as those of which EF con

tains three, Ee. Then (Theorem XLV. Cor.)
AB will be to DE as Aa to Dd, and BC to
EF as Bb to Ee; or as Aa to Dd; hence
AB: DE:: BC: EF.

Cor. If AB: BC:: DE:EF; then BC:
AB::FE: ED.

THEOREM XLVII.-In any four quantities AB, BC, D E, and E F (fig. 68, p. II.), of the same kind, if AB: BC :: DE: EF; then AB: AB+ BC:: DE: DE+ EF.

For let A B BC, or DE: EF::m: n; then AB: AB BC, or DE: DE EF:m: mn. In the same way it may be shown that AB+BC: AB BC:: DE+EF: DE EF.

THEOREM XLVIII.-Triangles, as ABC, DEF (fig. 69, p. II.), between the same parallels AE, CF, or that have equal altitudes, are to each other as their bases, A B and D E.

For let A B be to DE as any number, (3, for example), to any other number, (as 4); that is, let AB contain three such equal parts, Aa, ab, b B, as those of which DE contains four, Dc, ce, ef, fE; and join Ca, Cb, Fc, Fe, and Ff.

Then the triangles CA a, Cab, CbB, F D c, Fce, &c., are all (Theorem XXV. Cor. 1) equal; therefore the triangle ABC contains three such equal parts as those of which the triangle D E F contains four. Hence the triangle ABC is to the triangle DEF as the base AB is to the base D E.

Cor. Parallelograms and rectangles between the same parallels, or that have equal altitudes, are to each other as their bases; for the parallelograms are doubles of their respective triangles.

THEOREM XLIX.-If two triangles, as A B C, DEF (fig. 70, p. II.), stand on equal bases, AB, DE, the triangles are to each other as their altitudes CH, FI.

Let BP be perpendicular to A B and equal to CH; in BP take BQ equal to F I, and join AP, AQ, and CP. Then the triangle AP B is equal to the triangle ABC, and the triangle ABQ to the triangle D F E. But ABP: ABQ:: BP: BQ (Theorem XLVIII.), therefore ABC: DEF::HC: FI.

Cor. Parallelograms and rectangles on equal bases are to each other as their altitudes.

THEOREM L.-If four lines, as A, B, C, D (fig. 71, p. II.), be proportional, the rectangle under the extremes A and D will be equal to the rectangle under the means B and C.

For AD: BD::A: B (Theorem XLIX. Cor.), and BC: B D:: C: D; consequently as A: B:: C: D, A D: BD:: B.C: BD; and therefore A D and B C are equal because they bear the same proportion to BD.

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Cor. If the means are equal, the rectangle of the extremes will be equal to the square of the

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THEOREM LI.—If of four lines, as A, B, C, D · (fig. 71, p. II.), the rectangle A D of two of them, be equal to the rectangle B C of the other two, then the sides of those rectangles will be inversely proportional; viz. A: B::C:D; or B: A:: D: C, or A: C:: B : D, &c.

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THEOREM LII.-If a line, as B E (fig. 72, p. II.), be drawn parallel to CD, one of the sides of a triangle ACD, it divides the other two sides, AC and AD, in the same proportion, or so that AB: BC::AE: ED.

For let CE and BD be joined, then the triangle BCE and BDE are equal (Theorem XXIV. Cor.), and therefore the triangle ABE bears the same proportion to BCE that it does to BDE. But (Theorem XLVIII.) ABE: BCE::AB: BC; and A BE:BDE:: AE : ED; hence AB: BC::AE: ED.

Cor. AB: AB+ BC, or AC::AE:AE + ED or AD (Theorem XLVIII.); or alternately AB AE:: AC:AD; and the triangles ABE, ACD, being equiangular or similar, we may hence infer that similar triangles have the sides about their like angles in the same proportion, the homologous sides being opposite to the equal angles.

THEOREM LIII.-If two triangles as ABC, DEF (fig. 74, p. II.), have one angle, as A, in the one, equal to an angle, as E, in the other, then if A B: BC:: DE: EF, the triangles are similar.

For, if they are not, make the angle EDG equal to the angle BA C, then the angle AGB is also equal to the angle D G E, and the triangles A B C and DEG are consequently similar. Hence A B : BC::DE: EĠ (Theorem LI. Cor.); but AB: BC:: DE: EF; therefore E F and EG are equal, which is impossible. The triangle ABC and DEF are therefore similar.

THEOREM LIV.-If CD (fig. 73, p. II.), be a perpendicular from C, the right angle of a rightangled triangle ABC, on the hypothenuse AB; then AD AC:: AC: AB; BD: BC:: BC: AB, and AD: DC :: DC: DB.

For the triangles ABC, ADC, having the common angle A, and the right angles ADC, and AC B, right angles, are similar; and for a like reason AC B, and DCB, are similar, as also are ADC and DCB. Hence (Theorem II. Cor.) AB AC:: AC: AD, AB: BC: BCBD and AD: DC :: DC: DB.

Cor: 1. AB, ADAC, AB · BDBC and ADDB=DC.

Cor. 2. ABAD+AB BD=AC+ BC. But (Theorem XXVII. Cor. 2), AB AD+AB BDA B2; hence A B A C + BC3, another demonstration of the important property deduced at Theorem XXXI. Cor.

THEOREM LV.-If an angle C, of a triangle ABC (fig. 75 p. I.), be bisected by CD, meeting the base in D, then A C; CB::AD: DB. Let BE, parallel to CD, meet AC produced in E; then the angles DCB and CBE are equal (Theorem XV.), and the angles ACD and CEB are equal (Theorem XVII.); hence, as ACB and DC B are equal, CBE and CEB are equal, and therefore Theorem IV.), CE is equal to C B. But (Theorem LII., Cor.), AC: CE::AD: DB; hence AC:CB::AD: D B.

THEOREM LVI.-If two chords, CE and DB, intersect each other in any point A, within a circle DCDE (fig. 76, p. II.), the rectangle CA AE, of the segments of the one, is equal to

the rectangle, BA AD, the segments of the other.

For join C, B, and D, E, then_(Theorem XXXVII., Cor. 2), the angles BCE, BDE, standing on the same arc BE, are equal, and the angles CBA and DEA are equal for a like reason, and the vertical angles BAC and DAE being equal, the triangles ABC and AED are similar, and consequently (Theorem LII., Cor.), A D:AE::AC:AB; therefore (Theorem L.), AD AB AE A Č.

In like manner, if, as in fig. 77, the chords CE and BD meet when produced without the circle, the rectangles ACA E and AB-AD are equal. For the angles ABC and AED are equal, as are also the angles ACB and ADE (Theorem XXXVIJI., Cor.); and, the angle A being_common to both the triangles ABC and AED, those triangles are similar, and therefore (Theorem LII., Cor.), AC: A B ::AD: AE; whence (Theorem L.) AD AB = AE·ÁC.

Cor. If A B (fig. 77, p. II.) revolve round A till the points B and D meet, then the rectangle AC AE will be equal to the square of the tangent drawn from A; and hence all tangents drawn to the same circle from the same point are equal.

THEOREM LVII.-If in a triangle as ACB (fig. 78, p. II.) the vertical angle be bisected, as by C D, then AC.C =AË·EB + EC.3

For let CE produced meet the circumscribing circle in D, and join D B, then, as the angles A Č E and DCB are equal, and the angles CA E and CDB are equal (Theorem XXXVII., Cor. 2), the angles A ÉC and DBC are equal; whence the triangles AC E and DCB are similar, and consequently (Theorem LII., Cor.) AC:CE:: DC: CB; whence (Theorem L.) AC • C B = CE CD. But (Theorem XXVÍ. Cor.), CE CD=CE+CÈ·ED; and (Theorem LVI.) CE ED AE EB; therefore AC CB AE EB+EC.'

THEOREM LVIII.-If ABCD (fig. 79, p. 11.) be a quadrilateral inscribed in a circle, the rectangle ACBD is equal to the sum of the two rectangles AD BC and A B C D.

From C draw C E, making the angle DCE equal to the angle ACB; then, as the angles CDE, CA B, are equal (Theorem XXXVII., Cor. 2), the remaining angles D E C and A BC are equal, and the triangles A B C and DEC are similar; therefore AB: AC::DE: DC (Theorem LII. Cor.); whence (Theorem L.) A B⚫DC= AD DE. Again, if from the equal angles DCE and AC B, the common angle ACE be taken, the remaining angles DCA and ECB will be equal; and, as the angles DAC and EBC are equal (Theorem XXXVII., Cor. 2.), the remaining angles ADC and BEC are equal; whence the triangles ADC and BEC are similar; and therefore AD: AC::BE: BC; or AD BC=AC·E B. Hence AB. DC+AD BC=AC.DE+AC EB, or =AC BD (Theorem XXVII.).

THEOREM LIX.-If ABCDE (fig. 80, p. II.) be an equilateral polygon inscribed in a circle whose centre is M, and FGHIK an equilateral polygon of the same number of sides in

scribed in a circle whose centre is L, the circumference of the polygon ABCDE is to the circumference of the polygon FG HIK as the radius A M to the radius F L.

For join M and L to the angles of the polygons, then the triangles AM B, BMC, &c., being mutually identical, and the triangles FLG, GLH, &c., being also mutually identical (The orem V.), the angles AM B, FLG, are equal, being like parts of four right angles; and therefore, as the triangles are isosceles, they are similar. Hence A M: FL::AB:FC (Theorem LII., Cor.); or perimeter ABCDEA: perimeter FGHIKF (Theorem XLV.), since the perimeters are like multiples of AB and FC. Cor. If we conceive the sides of the polygons to be equal in number, and indefinitely small, their sides will coincide with the circumferences of the circumscribing circles. Hence the circumferences of circles are to each other as their radii.

THEOREM LX. Similar triangles, as A B C, DEF (fig. 81. p. II.), are to each other as the squares of their like sides.

For let AK, DM, be squares on the like sides AB and D E, BI and EL, the diagonals of these squares, and CG, F B, perpendiculars from C and F upon A B and D E. Then as the angles CAG and FDH are equal, and the right angles AGC and DHF are also equal, the triangles ACG and DF H, as well as the triangles ABC, D FE, are similar. Hence (Theorem LII., Cor.), AC: DF::CG: FH, and AC: DF:: A B: DE or AI: DL; therefore CG: FE ::AI: DI; or CG: AI:: FH: DI (Theorem XLVI.). But (Theorem XLIX.) the triangle ABC is to the triangle ABI as CG is to AI; and the triangle D E F is to the triangle DLE as FH is to DL, therefore the triangle ABC is to the triangle ABI as the triangle DFE to the triangle DLE; or the triangle ABC is to the triangle DFE as twice the triangle ABI is to twice the triangle DLE; or as AK to DM; that is as the square of A B is to the square of D E.

THEOREM LXI.-All similar rectilineal figures, as ABCDE, FGHIK (fig. 81, p. II.), are to each other as the squares of their like sides..

For draw BE, BD, G K, GI; then the two figures being similar they are equiangular, and have their like sides proportional. Hence, as the angles A and F are equal, the triangles A E B and KF G are similar (Theorem LIV.); and for a like reason the triangles DBC and ICHI are similar. And if from the equal angles AED and KFI the equal angles A E B and FKG be taken, the remaining angles EDB and IKC are equal; similarly the angles E DB and KIG are equal; therefore the triangles EBD and KIG are similar. Hence the triangles A E B and FKG are to each other as A E2 to K F,2 and EBD to KGI as ED2: KI, and DBC to IGH as DC: I H2; but the corresponding sides of the polygons, and consequently their squares, are proportional; therefore, each of the triangles that compose A B C D E is to the corresponding one in FGHIK, as the square of a side of the former is to a square of a like side of the latter, and consequently the whole polygons are to each other in the same proportion.

Cor. The areas of circles are to each other as the squares of their radii; for (fig. 80, p. II.) ABCF:AM: FL; or A B2: C F2:: AM2: FL2; but the areas of the polygons are by this theorem as A B2: G F2; and therefore also A M2: FL2; which, when the polygons coincide with their circumscribing circles, is the proportion that those circles have to each other.

OF PLANES, AND THEIR INTERSECTIONS.
DEFINITIONS.

1. A straight line is perpendicular to a plane when it is at right angles with every line which it meets in that plane.

2. If two planes cut each other, and, from any point in the line of their common section, two straight lines be drawn at right angles to that line, one in the one plane and the other in the other plane, the angle contained by these two lines is the angle made by the planes.

3. Two planes are perpendicular to each other when any straight line, drawn in one of the planes perpendicular to their line of common section, is perpendicular to the other plane.

4. A straight line is parallel to a plane, when it does not meet the plane, though produced ever so far.

5. Planes are parallel to each other when they do not meet, though produced ever so far.

6. A solid angle is formed by the meeting, in one point, of three or more plane angles, which are not in the same plane with each other.

THEOREM LXII.-Any three lines as AB, CD, CB (fig. 82, p. II.), which meet each other, not in the same point, are in the same plane.

For conceive a plane, passing through the line AB, to revolve round that line till it passes through the point C, then as the points E and Care in that plane, the line C D is in it; and for a similar reason the line CB is in the same plane; therefore all the three lines are in the same plane.

THEOREM LXIII.-If two planes as A B, BC (fig. 83, p. II.), cut each other, their common section BD is a straight line.

For join B, D, by a straight line; then, as the points B and D are in both the planes, the straight line BD, which joins them, is in both the planes; and A, therefore, is their line of common section.

THEOREM LXIV.-If a line, as AB (fig. 84, p. II.), be at right angles to each of two other straight lines, AD) and A C, at A, their point of meeting; A B is also at right angles to the plane passing through AC and A D.

For through A, in the plane passing through A C and A D, draw any line A E; and through any point E in that line draw EF parallel to AD, meeting AC in F. In A F produced take FC equal to A F; join C E and produce A to D. Then (Theorem LII.) AF: FC :: DE: EC; and consequently as AF is equal to FC, DE is equal to AC. Hence (Theorem XXXV.) AD2+ AC2 2 D Ea + 2 A E2; and therefore, by adding 2 A B2 to each of these equals, we have AB+ A D'+ A B2 + A C2 = 2 A B2 + 2 AE+ 2 ED2; or B D2+B C2 = 2 A B2 + 2A

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