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PROP. 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.

B

Let AB be the given line, and c the fide of the given quare; 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 fquare of c.

Make BE perpendicular to AB (I. 11.), and equal to c (I. 3.); also bisect 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, GE 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 Conft.); 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 drawn 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 fhewn to be equal to BE, or c; whence the rectangle of AD, DB will also be equal to the square 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.

N

H

Let ABC, DEF be two equal circles, in which BGC, EHF 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 also 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 fame 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 lefs, lefs.

But EL and BGL 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 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. 13.)

PROP. XXV. THEOREM.

The rectangle of the two fides of any triangle, is equal to the rectangle of the segments of the bafe, made by a line bifecting the verticle angle, together with the fquare

of that line.

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

be

be equal to the rectangle of AD, DB, together with the fquare of CD.

For, about the triangle ABC, defcribe the circle AEC (IV. 5.), cutting CD, produced, in E; and join EB.

Then, because the angle ACD is equal to the angle ECD (by Hyp.), and the angle CAD to the angle CEB (III. 15), the remaining angle ADC will be equal to the remaining angle CBE (I. 28. Cor.)

The triangles CAD, CEB being, therefore, equiangular, CA will be to CD as CE to CB (VI. 5.); and confequently the rectangle of CA, CB is equal to the rectangle of CE, CD (VI. 12.)

But the rectangle of CE, CD is equal to the rectangle of ED, DC, together with the fquare of CD (II. 10.); whence the rectangle of CA, CB is alfo equal to the rectangle of ED, DC, together with the fquare of CD.

And fince the rectangle of ED, pc is equal to the rect angle of AD, DB (III. 27.), the rectangle of AC, CB is also equal to the rectangle of AD, DB, together with the fquare of CD.

Q. E. D.

PROP. XXVI. THEOREM.

The rectangle of the two fides of any triangle, is equal to the rectangle of the perpendicular, drawn from the vertical angle to the base, and the diameter of the circumfcribing circle.

Let ABC be a triangle, having CD perpendicular to AB; then will the rectangle of AC, CB be equal to the rectangle of CD and the diameter of the circumfcribing circle.

For, about the triangle ABC, defcribe the circle AEC (IV. 5.); in which draw the diameter CE; and join EB.

Then, fince the angle CAD is equal to the angle CEB (III. 15.) and the angle ADC to the angle EBC, being each of them right angles (Conft. and III. 16.), the remaining angle ACD will be equal to the remaining angle ECB (I. 28. Cor.)

The triangles ACD, ECB are, therefore, equiangular; whence AC is to CD as CE is to CB (VI. 5.); and confequently the rectangle of AC, CB is equal to the rectangle of CD, CE (VI. 12.) Q. E. D.

SCHOLIUM. When ABC is an obtufe angle, the perpendicular CD falls without the circle; but the fame demonstration will hold.

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