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the greater equal to the greater, and the less to the less; and AD, DB are each of them less than a semicircle, because (III. 1. cor.) DC passes through the centre: wherefore the arc AD is equal to the arc DB: therefore the given arc is bisected in D: which was to be done.

PROP. XXXI. THEOR.

In a circle, (1.) the angle in a semicircle is a right angle; (2.) the angle in a segment greater than a semicircle is acute; and (3.) the angle in a segment less than a semicircle is obtuse.

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Let ABC be a circle, of which F is the centre, BC a diameter, and consequently BAC a semicircle ; and let the segment BAD be greater, and BAE less than a semicircle: the angle BAC in the semicircle is a right angle ; but the angle BAD in the segment greater than a semicircle is acute; and the angle BAE in the segment less than a semicircle is obtuse.

Join AF and produce it to G. Then (III. 20.) the angle BAG at the circumference, is half of BFG at the centre, both standing on the same arc BG; and for the same reason, GAC is half of GFC. Therefore the whole angle BAC is half of the angles BFG, GFC: and (I. 13.) these are together equal to two right angles ; therefore BAC is a right angle, and it is an angle in a semicircle. But (I. ax. 9.) the angle BAD is less, and BXE greater than the right angle BAC: therefore an angle in a segment greater than a semicircle is acute, and an angle in a segment less than a semicircle is obtuse.

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PROP. XXXII. THEOR.

If a straight line touch a circle, and from the point of contact a straight line be drawn dividing the circle into two segments; the angles made by this line with the tangent are equal to the angles which are in the alternate segments.

Let the straight line DE touch the circle BCG in the point B, and let the straight line BA be drawn dividing the circle into the segments AGB, ACB : the angle ABE is equal to any angle in the segment AGB, and the angle ABD to any angle in ACB.

If AB (fig. 1.) be perpendicular to DE, it passes (III. 19.) through the centre, and the segments being therefore semicircles,

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the angles in them are (III. 31.) right angles, and consequently equal to those which AB makes with DE.*

But if BA (fig. 2.) be not perpendicular to DE, draw BF perpendicular to it; join FA and produce it to E; join also CA, CB, C being any point in the arc AB. Then (III. 19.) BF is a diameter, and (III. 31. and I. 13.) the angles BAF, BAE are right angles. Therefore, in the triangles BAE, FBE, the angle E is common, and the angles BAE, FBE equal, being right angles ; wherefore (I. 32. cor. 5.) the remaining angle ABE is equal to the remaining angle F, which is an angle in the remote or alternate segment AGB.

Again, the two angles ABD, ABE are equal (I. 13.) to two right angles; and because ACBF is a quadrilateral in the circle, the opposite angles C and F are also equal (III. 22.) to two right angles : therefore (I. ax. 1.) the angles ABD, ABE are together equal to C and F. From these equals take away the angles ABE and F, which have been proved to be equal ; then (I. ax. 3.) the remaining angle ABD is equal to the remaining angle C, which is an angle in the remote segment ACB. If, therefore, a straight line, &c.t

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PROP. XXXIII. PROB.

UPON a given straight line to describe a segment of a circle, containing an angle equal to a given angle.

Let AB be the given straight line, and C the given angle; it is required to describe on AB a segment of a circle, containing an angle equal to C.

First, if C be a right angle, bisect (I. 10.) AB in F, and from

This case is wanting in most editions of Euclid. † This proposition, and the 21st and 22d, when viewed in a certain light, may be regarded as the same. To illustrate tbis, let AB be a fixed chord, and through B draw any other line DE cutting the circle in C; and join AC. Now (10. 21.) wherever C is taken in the arc ACB, the angle ACE is constantly of the same magnitude; and so also (I. 13.) is the exterior angle ACD. If C be now taken as coinciding with B, the straight line DE will become the tangent D'BE', AC will coincide with A B, and the angles ACE, ACD will become A BE', ABD'. If again C take the position C", the angles ACE, ACD will become A C''E", AC"D". Now, the equality of ABE', ACE, and of ABD', AC"D" is what is proved in the 32d proposition ; and from the equality of ACD and AC"D" the 22d follows by the addition of ACB.

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the centre F, at the distance FB, describe the semicircle AHB; therefore (III. 31.) any angle AHB in the semicircle is equal to the right angle C.

But, if C be not a right angle, make (I. 23.) the angle BAD equal to C, and (I. 11.) from A draw AE perpendicular to AD; bisect (I. 10. and 11.) AB by the perpendicular FG, and join GB. Then, because AF is equal to FB, FG common to the triangles AFG, BFG, and the angle AFG equal to BFG ; therefore (I. 4.) AG is equal to GB; and the circle described from the centre G, at the distance GA will pass through the point B: let this be the circle AHB. Then, because from the point A to the extremity of the diameter AE, AD is drawn at right angles to AE, therefore AD (III. 16. cor.) touches the circle ; and, (III. 32.) because AB, drawn from the point of contact A, cuts the circle, the angle DAB is equal to any angle in the alternate segment AHB: but DAB is equal to C; therefore also C is equal to the angle in the segment AHB: wherefore upon the given straight line AB the segment AHB is described, which contains an angle equal to C: which was to be done.*

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PROP. XXXIV. PROB.

From a given circle, to cut off a segment which shall contain an angle equal to a given angle.

Let ABC be a given circle, and D a given angle ; it is required to cut a segment from ABC containing an angle equal to D.

Draw (III. 17.) the straight line EF touching the circle in B, and (I. 23.) make the angle FBC equal to D. Then, because EF touches the circle, and BC is drawn from the point of contact B, the angle FBC is equal (111. 32.) to any angle in the alternate segment BAC: but FBC is equal to D; therefore any angle in the segment BAC is equal to D : wherefore the segment BAC is cut off from the given circle, containing an angle equal to the given angle D: which was to be done.

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It is evident there may be two segments answering the conditions of the problem, one on each side of the given line. It is also plain, that when C is an acute angle, and the segment is to be above AB, G is above AB, but when obtuse, it is below it. It is likewise plain, that the angle BAE is the complement of the given angle C; that is, the difference between it and a right angle.

PROP. XXXV. THEOR.

Ir two chords of a circle cut one another, the rectangle contained by the segments of one of them is equal to the rectangle contained by the segments of the other.

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In the circle ABCD, let the two chords AC, BD cut one another in the point E: the rectangle AE.EC is equal to the rectangle BE.ED.

IfAC, B D both pass through the centre, so that E is the centre ; it is evident, that (I. def. 30.) AE, EC, BE, ED, being all equal, the rectangle A E.EC is equal to the rectangle BE.ED.

But let one of them BD pass through the centre, and cut the other AB which does not pass through the centre, at right angles, in the point E. Then if BD be bisected in F, F is the centre. Join AF; and, because BD, which passes through the centre, is perpendicular to AC, AE, EC are (III. 3.) equal to one another. Now, because BD is divided equally in F, and unequally in E, the rectangle BE.ED, and the square of EF are equal (II. 5.) to the square of FB; that is, (I. 46. cor. 2.) to the square

of FA. But (I. 47.) the squares of AE, EF are equal to the square of FA; therefore the rectangle BE.ED, and the square of EF are equal to the squares of AE, EF. Take away the common square of EF, and the remaining rectangle BE.ED is equal to the remaining square of AE; that is, to the rectangle AE.EC.

Next, let BD pass through the centre, and cut AC, which does not pass through the centre, in E, but not at right angles. Then, as before, if BD be bisected in F, F is the centre of the circle. Join AF, and (I. 12.) draw FG perpendicular to AC; therefore (III. 3.) AG is equal to GC; wherefore (II. 5.) the rectangle AE.EC and the square of EG are equal to the square of AG. To each of these equals add the square of GF: therefore the rectangle A E.EC and the squares of EG, GF are equal to the squares of AG, GF: but (I. 47.) the squares of EG, GF are equal to the square of EF ; and the squares of AG, GF are equal to the square of AF: therefore the rectangle AE.EC and the square of EF are equal to the square of AF; that is, (I. 46. cor. 2.) to the square of FB. But (11. 5.) the square of FB is equal to the rectangle BE.ED, together with the square of EF; therefore the rectangle AE.EC and the square of EF are equal to the rectangle BE.ED and the square of EF: take

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away the common square of EF, and the remaining rectangle A E.EC is equal to the remaining rectangle BE.ED.*

Lastly, let neither of the lines AC, BD pass through the centre. Take the centre F, and through E draw the diameter GEFH. Then, because the rectangle AE.EC is equal, as has been shown, to the rectangle GE.EH; and, for the same reason, the rectangle BE.ED is equal to the same rectangle GE.EH; therefore (I, ax. 1.) the rectangle AE.EC is equal to the rectangle BE.ED. If, therefore, two chords of a circle, &c.

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PROP. XXXVI. THEOR.

Ir from any point without a circle two straight lines be drawn, one of which cuts the circle, and the other touches it; the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, is equal to the square of the line which touches it.

Let D be any point without the circle ABC, and DCA, DB two straight lines drawn from it, of which DCA cuts the circle in A and C, and DB touches it in B : the rectangle AD.DC is equal to the square of DB.

Either DCA passes through the centre, or it does not : first, let it pass through the centre E, and join EB. Therefore (III. 18.) the angle EBD is a right angle : and (II. 6.) because the straight line AC is bisected in E, and produced to D, the rectangle AD.DC and the square of EC are together equal to the square of ED; and CE is equal to EB: therefore the rectangle AD.DC and the square of EB are equal to the square of ED. But (I. 47.) the square of ED is equal to the squares of EB, BD, because EBD is a right angle: therefore the rectangle AD.DC and the square of EB are equal to the squares of EB, BD: take away the mon square of EB; and the remaining rectangle AD.DC is equal to the square of the tangent DB.

But if DCA do not pass through the centre, take (III. 1.)

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* The second and third cases may be thus demonstrated in one :

Join FC. Then the rectangle AE.EC is equal (II. 5. cor. 5.) to the difference of the squares of AF and FE, or of DF and FE, or (II. 5. cor. 1.) to the rectangle DE.EB.

Proportion however, affords much the easiest method of demonstrating both this proposition and the following. See the Notes at the end of the volume.

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