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PRO P. XXI. PROBLEM.

To cut a given right line in extreme and mean proportion.

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Let AB be the given right line ; it is required to cut it in extreme and mean proportion.

Upon Ab describe the square Ac (II. 1.), and bisect the side Adin E (1. 10.); and join be.

In ea produced, take Ef equal to EB (I. 3.); and upon AF describe the square FH (II. 1.); then will AB be divided at the point H as was required.

For fince Dy is the sum of EB, ED, or EB, EA, and AF is their difference, the rectangle of DF, FA is equal to the difference of the squares of EB, EA (II. 13.)

But the rectangle of DF, FA is equal to DG, because Fa is equal to Fg; and the difference of the squares of EB, £ A is equal to the square of AB (II. 14. Cor.) ; whence DG is equal to AC.

And if from each of these equals, the part AK, which is common, be taken away, the remainder AG will be equal to the remainder HC.

But equal parallelograms have the sides about equal angles reciprocally proportional (VI. 15.) ; whence HK is to HG as HA tO HB.

And since Hk is equal to AD, or AB, and HG to HA, AB will be to HA AS Ha is to HB.

Q. E. D.

PROP. PRO P. XXII. PROBLEM.

To divide a given right line into two such parts, that their rectangle may be equal to a given square, the side of which is not greater than half the given line.

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Let AB be the given line, and c the side of the given square ; it is required to divide ab into two such parts that their rectangle may be equal to the square of c.

Upon as describe the semicircle bda, and make BF perpendicular to AB (I. 11.), and equal to c (I. 3.)

Through F draw parallel to AB (I. 27.); and from the point D where it cuts the circle, let fall the perpendicular de (I. 12.); and AB will be divided at È as was required.

For since BDA is a simicircle (by Const.), and de is perpendicular to the diameter AB (by Conft.), the rectangle of AE, EB will be equal to the square of ED (VI. 7. Car.)

But Ed is equal to FB (I. 30.) or c; whence the rectangle of AE, EB will be equal to the square of c as was to be thewn.

SCHOLIUM. When Br, or C, is equal to half AB, FD will be a tangent to the circle, and the rectangle of AE, zB will be the greatefi posible.

PROP. XXIII. PROBLEM.

To a given right line to add another right line such, that the rectangle of the whole and the part added shall be equal to a given square.

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Let all be the given line, and c the side of the given square ; it is required to add a line to AB such, that the rectangle of the whole and the part added Inall be equal to the fquare of C.

Make BE perpendicular to AB (I. 11.), and equal to c (I. 3.); also bisect AB in G (1. 10.), and join ge.

Then, if AB be produced, and Go be taken equal to GE (I. 3.), the part 'BD will be added to AB, as was required.

For on as describe the semicircle BFA, cutting Ge in F, and join FD.

Then, since the two sides GB, GE of the triangle GEB, are .equal to the two sides GF, GD, of the triangle GDF, and the angle G is common, the angle gbe will be equal to the angle GFD, and the fide Fd to the side BE (1. 4.)

But the angle gbe is a right angle (by Conft.); whence the angle GFD is also a righe angle ; and confequently FD is a tangent to the circle at F (III. 10.)

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And since DF is a tangent to the circle, and Da is drawn to the opposite part of the circumference, the rectangle of AD, DB will be equal to the square of DF (III. 29.)

But DF has been shewn to be equal to Be, or c; whence the rectangle of AD, DB will also be equal to the square of G:

Q. E. D.

PRO P. XXIV. THEOREM.

Angles at the centres or circumferences of equal circles, have the same ratio with the arcs on which they stand,

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: - Let ABC, D'EF be two equal circles, in which BGC, EHF are angles at the centre, and BAC, EDF angles at the circuinference;

then will the arc Bc be to the arc Ef aş the angle bgc is to the angle Ehf, or as the angle BAC to the angle EDF.

For on the circumference of the circle: ABC take any number of arcs whatever CK, KL each equal to BC; and on the circumference of the circle Def any number of arcs whatever FM, MN, each equal to Ef; and join GK, GL, HM, HN.

Then, because the arts BC, CK, KL are all equal to each other, the angles BG€, C'GK, KGL will also be equal to each other (III. 21.)

And, therefore, whatever multiple the arc BL is of the arc BC, the same multiple will the angle BGL be of the angle BGC.

For the same reason, whatever multiple the arc en is of the arc ef, the same multiple will the angle Ehn be of the angle EHF.

If, therefore, the arc Bl be equal to the arc En, the angle BGL will be equal to the angle EHN; and if equal, equal; and if less, less.

But el and bol are any equimultiples whatever of BC and BGC, and En and EHn of EF and EHF; whence the arc Bc is to the arc Ef as the angle bgc is to the angle EHF (V.5.)

And since the angle Bgc is double the angle BAC, and the angle ehf is double the angle EDF (III. 14.), the arc Bc will also be to the arc Ef as the angle BAC is to the angle edr (V. 13.)

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PRO P. XXV. THEOREM,

The rectangle of the two sides of any

triangle, is equal to the rectangle of the segments of the base, made by a line bisecting the verticle angle, together with the square of that line.

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Let ABC be a triangle, having the angle ACB bisected by the right line cd; then will the rectangle of AC, CB

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