From A as a centre with the radius AC, the greater of the two sides, describe the circle CFG: produce AB to meet the circumference in E and F, and CB to meet it in G. Then because AF=AC, BF=AB+AC, the sum of the sides; and since AE=AC, BE=AC-AB= the difference of the sides. Also, because AD drawn from the centre cuts GC at right angles, it bisects it; therefore, when the perpendicular falls within the triangle, BG=DG-DB=DC-DB= the difference of the segments of the base, and BC=BD+DC= the sum of the segments. But when AD falls without the triangle, BG=DG+DB=CD+DB= the sum of the segments of the base, and BC=CD-DB= the difference of the segments of the base. Now, in both cases, because B is the intersection of the two lines FE, GC, drawn in the circle, FB.BE=CB.BG; that is, as has been shewn, (AC+AB) (AC—AB)=(CD+DB) (CD—DB).

PROBLEMS

RELATING TO THE SIXTH BOOK.

PROP. L. PROBLEM.

To construct a square that shall be equivalent to a given rectilineal figure.

Let A be the given rectilineal figure; it is required to describe a square that shall be equivalent to A.

Describe (Prop. 45. 1.) the rectangular parallelogram BCDE equivalent to the rectilineal figure A; produce one of the sides BE, of this rectangle, and make EF= ED; bisect BF in G, and from the centre G, at the distance GB, or GF, describe the semicircle BHF,

and produce DE to H.

HE2=BEXEF, (13. 6.); therefore the be equivalent to the rectilineal figure A.

SCHOLIUM.

This problem may be considered as relating to the second Book: Thus, join GH, the rest of the construction being the same, as above; because the straight line BF is divided into two equal parts in the point G, and into two unequal in the point E, the rectangle BE.EF, together with the square of EG, is equal (5. 2.) to the square of GF: but GF is equal to GH;

therefore the rectangle BE, EF, together with the square of EG, is equal to the square of GH: But the squares of HE and EG, are equal (47. 1.) to the square of GH: Therefore also the rectangle BE.EF, together with the square of EG, is equal to the squares of HE and EG. Take away the square of EG, which is common to both, and the remaining rectangle BE.EF is equal to the square of EH: But BD is the rectangle contained by BE and EF, because EF is equal to ED; therefore BD is equal to the square of EH; and BD is also equal to the rectilineal figure A; therefore the rectilineal figure A is equal to the square of EH: Wherefore a square has been made equal to the given rectilineal figure A, viz. the square described upon EH.

Note. This operation is called squaring the rectilineal figure, or finding the quadrature of it.

PROP. M. PROB.

To construct a rectangle that shall be equivalent to a given square, and the difference of whose adjacent sides shall be equal to a given line.

Suppose C equal to the given square, and AB the difference of the sides.

Upon the given line AB as a diameter, describe a circle; at the extremity of the diameter draw the tangent AD equal to the side of the square C; through the point D, and the centre O, draw the secant DF; then will De and DF be the adjacent sides of the rectangle required.

First, the difference of their sides is equal to the diameter EF or AB; secondly, the rectangle DE.DF is equal to AD2 (36. 3.); hence that rectangle is equivalent to the given square C.

C

D

E

PROP. N. PROB.

To construct a rectangle equivalent to a given square, and having the sum of its adjacent sides equal to a given line.

T

Let C be the given square, and AB equal to the sum of the sides of the required triangle.

Upon AB as a diameter,

cuts the circumference, draw EF perpendicular to the diameter; AF and FB will be the sides of the rectangle required.

For their sum is equal to AB; and their rectangle AF.FB is equal to the square EF, or to the square AD; hence that rectangle is equivalent to the given square

C.

SCHOLIUM.

To render the problem possible, the distance AD must not exceed the radius; that is, the side of the square C inust not exceed the half of the line AB.

To construct a square that shall be to a given square as a given line to a given

line.

Upon the indefinite straight line GH take GK=E, and KH=F; describe on GH a semicircle, and draw the perpendicular KL. Through the points G, H, draw the straight lines LM, LN, making the former equal AB, the side of the given square, and through the point M, draw

E

F

D

L

MN parallel to GH, then will LN be the side of the square sought.

For, since MN is parallel to GH, LM: LN: LG:

LH; consequently, LM2: LN2 :: LG2 : LH2 (22. 6.); but, since the triangle LGH is right angled, we have LG2 : LH2:: GK: KH; hence LM2 : IN2 :: GK: KH; but, by construction GK=E, and KH-F, also LM =AB; therefore, the square described on AB is to that described on LN, as the line E is to the line F.

PROP. P. PROB.

To divide a triangle into two parts by a line from the vertex of one of its angles, so that the parts may be to each other as a straight line M to another straighs line N.

Divide BC into parts BD, DC proportional to M, N; draw the line AD, and the triangle ABC will be divided as required.

For, since the triangles of the same altitude are to each other as their bases,

we have ABD: ADC:: BD: DC :: B D M: N.

SCHOLIUM.

A

A triangle may evidently be divided into any number of parts proportional to given lines, by dividing the base in the same proportion.

PROP. Q. PROB.

To divide a triangle into two parts by a line drawn parallel to one of its sides, so that these parts may be to each other as two straight lines M, N.

As M+N: N, so make AB to AD2 (Prob. 4.); Draw DE parallel to BC, and the triangle is divided as required.

For the triangles ABC, ADE being similar, ABC: ADE:: AB2: AD2; but M+NN:: AB2: AD2; therefore ABC ADE M+N: N; consequently BDECADE:: M: N.

A

B

E

To divide a triangle into two parts, by a line drawn from a given point in one of its sides, so that the parts may be to each other as two given lines M, N.

Let ABC be the given triangle, and P the given point; draw PC, and divide AB in D, so that AD is to DB as M is to N; draw DE parallel to PC, join PE, and the triangle will be divided by the line PE into the proposed parts.

For join DC; then because PC, DE are parallel, the triangles PDE, CDE are equal; to each add the triangle DEB, then PEB= DCB; and consequently, by taking each from the triangle ABC, there results the quadrilateral ACEP equivalent to the triangle ACD.

B

D

P

Now, ACD: DCB:: AD: DB:: M: N; consequently,

The above operation suggests the method of dividing a triangle into any number of equal parts by lines drawn from a given point in one of its sides; for if AB be divided into equal parts, and lines be drawn from the points of equal division, parallel to PC, they will intersect BC, and AC; and from these several points of intersection if lines be drawn to P, they will divide the triangle into equal parts.

PROP. S. PROB.

To divide a triangle into three equivalent parts by lines drawn from the vertices of the angles to the same point within the triangle.

the

Make BD equal to a third part of BC, and draw DE parallel to BA, side to which BD is adjacent. From F, the middle of DE, draw the straight lines FA, FB, FC, and they will divide the triangle as required.

A

E

For, draw DA; then since BD is one third of BC, the triangle ABD is one third of the triangle ABC; but ABD= ABF (37. 1.); therefore ABF is one third of ABC; also, since DF=FE, BDF = AFE; likewise CFD = = CFE, consequently the whole triangle FBC is equal to the whole triangle FCA; and FBA has been shown to be equal to a third part of the whole triangle ABC; consequently the triangles FBA, FBC, FCA, are each equal to a third part of ABC.

B

F

D

PROP. T. PROB.

To divide a triangle into three equivalent parts, by lines drawn from a given point within it.

Divide BC into three equal parts in the points D, E, and draw PD, PE; draw also AF parallel to PD, and AG parallel to PE; then if the lines PF, PG, PA be drawn, the trian

gle ABC will be divided by them into three equivalent parts.

A

Λ

For, join AD, AE; then because AF, PD are parallel, the triangle AFP is equivalent to the triangle AFD; consequently, if to each of these there be added the triangle ABF, there will result the quadrilateral ABFP equivalent to the triangle ABD; but since BD is a third part of BC, the triangle ABD E G is a third part of the triangle ABC; consequently the quadrilateral ABEP is a third part of the triangle ABC. Again, because AG, PE are parallel, the triangle AGP is equivalent to the triangle ACE; and this triangle is one third of ABC; hence the quadrilateral ACGP is one third of the triangle ABC: consequently, the spaces ABFP, ACPG, PFG are each equal to a third part of the triangle ABC.

F D

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