PROP. 46. THEOR. Of all the triangles (CDB CIB) standing on the same base (CB), on the same side of it, and having the equal vertical angles, the greatest is that (CDB) which

is isosceles. Fig. 47.

Describe a segt. of a circle on CB to pass through D, it must also pass through 1, draw to the point D & tangent DH, then the angle HDB is = to the angle DCB(prop.32.b.3.Elr.) and .. to the angle DBC, DH is parallel to CB, produce BI to meet it in H join HC, the triangles DCB,HCB are evidently=to one another. and ICB is a part of HCB.. &c.

PROP. 47. THEOR. If the hypothenuses of two dissimilar right angled

triangles [ABD ABI] be equal, the triangle (A1B] which has the smaller acute angle adjacent to its base,

shall have the greatest base. Fig. 6.

Apply to the base of either a triangle = to the other, let it lie towards the same part with the triangle on this base, describe on this base a semicircle ADB, then because the angle BAI is less than BAD, AI is nearer to AB than AD.. AI greater than AD, -. &c.

PROP. 48. THEOR. If from the centre [A] of a circle, a right line be drawn

to any point [C] in a chord, the square of that line together with the rectangle under the segments [GC Ci] of the chord, shall be equal to a square of the

radius. Fig. 51.

Draw any other chord BCF at right angles to AC, join AB.

Then since BC is = to CF, CB will be to the rectangle BCF, but the rect. BCF is to the rectangle GCI, add to both AC, then CA + BC X CFis = to AC2+ IC X CG; but CA + BC X CF is = to AB? .. &c.


PROP. 49. THEOR. The rectangle under the corresponding sides of equian. gular triangles [ABC DEF] taken alternately are equal, i. e.-if the angle A = D and B=E, then AB

X DF = AC X DE. Fig. 52.

Produce BA until AG is = to EF, then through the the points B C G describe a circle, produce CA to meet it in H, and join HG.

Then the angles H and B are = [prop 21.b.3.Elr.) and also the angles GAH,CAB. , the angles G and Care = .. the triangle AGH is in every way to DEF, .. AH is = to DE and AG to EF, but the rect. CAH is = to the rect. BAG .. the rect. under CA ED is = to the rect, under BA EF... &c.


The rectangle under two sides [AB BC] of any trian

gle is equal to the rectangle under the perpendicular (CD) to the base, and the diameter [EC] of the cire cumscribing circle.


T Fig. 53.

For, join BE. Then the angles A and E are = and also the angles ADC EBC are = being right ... the triangles EBC, ADC are similar ... AC, EC, CD, CB being corresponding sides opposite to = angles, the rectangle under AC, CB is = to the rectangle under EC, CD by the foregoing prop.


Through a given point(P) within a given circle to drare

a right line so that the parts of it between the given point and circumference shall have a given difference [F].

ANALYSIS. Suppose EI is the line required to be drawn, join the centre A with P, and draw AO at right angles to EI, it is evident that PO is = to lialf the difference of EP, PI and ... to half F.

Fig. 54.

Then join AP, upon it describe a semicircle POA inflect in it PO = to half F and produce it to meet the circumference in E and I, EI is the line required to be drawn.

For, join OA: then because POA is right El is bisected in' O..OP is half the difference between EP. and PI, .. &c.


If from any point[B] in the circumferenceof the greater

of two concentric circles, right lines [BO,BT]be drawn to the extremeties of any diameter [OT] of the less, the sum of the squares of those drawn lines shall be equal to double the squares of two radii of the given circles.

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Fig. 29.

For, draw BI: then in the triangle OB'T the ers of OB BT together are = to twice the O? of BI with twice [ of 10, [cor. prop. 13. b. 2. Elr.] but BI 10 are the semidiameters of the given circles, .. &c.



To describe a circle whose circumference shall pass

through a given point, [C] and touch a given circle

in another given point [B]. Fig. 55.

Suppose it done, and that CBD is the circle required : join their centres, OK must pass through the point of contact B, join BC bisect it in I, join KI it is at right angles to BC.

If therefore the given points [BC] be joined, BC bisected in 1, IK drawn at right angles to BC, the centre O of the given circle and the given point B joined, and OB be produced to meet IK, the point K in which they meet, must be the centre of the required circle.

PROP. 94: THEOR. If from a given point [Oj within a circle which is not

the centre, right lines [OC OD: 0A] be drawn to the circumference, making equal angles with each other, those[OD,DA] which are next that part of the diameter (OL) passing through the given point, shall cut off a greater circumference than the two which are more

remotes Fig. 56.

ANALYSIS. Suppose that the circumferenee DA is greater than CD, join CD DA; then the right line DA is greater than CD, draw OL through the centre R, then it is evideut that OC is less than DA, make 01 = to OC join DL then it is evident the triangles OCD OID are. = in every way, .. DI is = to DC, .. DA is greater than DI, if then this is proved the proposition is true.

Produee DO and AO to S and P, and join AS AC and CS.

Then the angle AID is = to the angles IOD, IDO to the angles DOC ODC, and the angle DOC is greater the angle OSC, and OSC is = to DAC', (prop.21.b.3.Elr.] .. DOC or DOT is greater than DAC, also the angle CDS is = to the angle CAS, :: CDO and CODIDO IOD are together greater than DAS, . DIA is greater than DAS and .:. than DAL · D4 is greater than Di, but DI is = to DC .. DA is greater than DC.


Through a given point[A], within a circle, which is not

the centre, to draw the least possible chord, Fig. 57.

Find the centre K, join KA and through A draw CP at right angles to KA it is less than any other chord DL.

For, draw KS at right angles to Dl: then since the angle KSA is right, KA is greater than KS .. CP is less than DL, in like manner it can be proved, that CO is less than any other right line drawn through A, and termicated by the circumference.

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Through either of the points of intersection [A] of

two given circles, which cut each other, to draw the greatest possible line, which shall be terminated both

ways by the two circumferences. Fig. 58.

Join the centres 0,S, and draw through A the right line CD, parallel to OS it is greater than any other right line, BF drawn through A.

For draw OL SI at right angles to CD, then LI is evidently half of CD; also draw OP SQ at right angles to BF and draw ST parallel to FB: it is also evident that ST is half of BF, and because ST is parallel to BF, STO is a right angle, .. SO is greater than ST, .'. LI is greater than PQ, and .:. CD is greater than BF, in like manner CD can be proved greater than any other right line drawn through A and terminated by the circumferences.


PROP. 58. THEOR. Of all the right lines [AF EC] that can be drawn to

cut off equal areas, from a given triangl [IDL) that (AF) is the least which makes the triangle (Led] cut op an isocecles one.

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