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 defcribe the fquare AC (II. 1.), and bisest the fide AD in E (I. 10.) and join, BE.

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

For fince DF is the fum of EB, ED, or EB, EA, and. AF is their difference, the rectangle of DF, FA is equal to the difference of the fquares 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 fquares of EB, EA is equal to the fquare of AB (II. 14. Cor.); whence DG is equal to AC,

And if from each of thefe 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 fides about equal angles reciprocally proportional (VI. 15.); whence нK is to HG as HA to HB.

And fince 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.

PROP. XXII. PROBLEM.

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

Let AB be the given line, and c the fide of the given fquare; it is required to divide AB into two fuch parts that their rectangle may be equal to the fquare of c.

Upon AB describe the femicircle BDA, and make BF perpendicular to AB (I. 11.), and equal to c. (I. 3.)

Through F draw FD parallel to AB (I. 27.); and from the point D where it cuts the circle, let fall the perpendicular DE (1. 12.); and AB will be divided at E as was required.

For fince BDA is a femicircle (by Conft.), and DE is perpendicular to the diameter AB (by Conft.), the rectangle of AE, EB will be equal to the fquare of ED (VI. 7. Cor.)

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 fhewn.

SCHOLIUM. When BF, or C, is equal to half AB, FD will be a tangent to the circle, and the rectangle of ae, EB will be the greatest poffible.

PRO P. XXIII. PROBLEM.

To a given right line to add another right line fuch, that the rectangle of the whole and the part added fhall be equal to a given fquare.

Let AB be the given line, and c the fide of the given fquare; it is required to add a line to AB fuch, that the rectangle of the whole and the part added fhall be equal to the square of c.

Make BE perpendicular to AB (I. 11.), and equal to c (I. 3.); alfo bifect AB in G (I. 10.), and join GE.

Then, if AB be produced, and GD be taken equal to GE. (I. 3.), the part BD will be added to AB, as was required. For on AB describe the femicircle BFA, cutting GE in F, and join FD.

Then, fince the two fides GB, CE of the triangle GEB, are equal to the two fides 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 fide BE (I. 4.). But the angle GBE is a right angle (by Const.); whence the angle GFD is also a right angle; and confequently FD is a tangent to the circle at F (III. 10.)

And

And fince DF is a tangent to the circle, and DA is drawni to the oppofite part of the circumference, the rectangle of AD, DB will be equal to the fquare 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 fquare of c.

Q.E.D.

PROP. XXIV. THEOREM.

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

Let ABC, DEF be two equal circles, in which BGC, EHP are angles at the centre, and BAC, EDF angles at the circumference; then will the arc BC be to the arc EF as 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 arcs BC, CK, KL are all equal to each other, the angles BGC, CGK, KGL will alfo be equal to each other (III. 21.)

And, therefore, whatever multiple the arc BL is of the arc BC, the fame 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 fame 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 BL and BGL are any equimultiples whatever of BC and BGC, and EN and EHN of EF and EHN; whence the arc BC is to the arc EF as the angle BGC is to the angle EHF (V.5.)

And fince, 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 EDF (V. 18.) Q.E. D.

PROP. XXV. THEOREM.

The rectangle of the two fides of any triangle, is equal to the rectangle of the fegments of the base, made by a line bifecting the vertical angle, together with the fquare 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

be