PROP. XXV. PROBLEM.

Upon a given right line to defcribe a feg ment of a circle, that shall contain an angle equal to a given rectilineal angle,

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Let AB be the given right line, and c the given recti、 lineal angle; it is required to describe a segment of a cir cle upon the line AB that shall contain an angle equal to c.

Make the angle BAD equal to C (I. 20.), and, from the point A, draw AE at right angles to AD (I. 11.) and make the angle ABF equal to the angle FAB (I. 20,)

Then, fince the angles ABF, FAB are equal to each other, and less than two right angles, the fides AF, FB will meet, and be equal to each other (I, 25 Cor. and I. 6.)

From the point F, therefore, at the distance FA, or FB, defcribe the circle AEB, and ABEA will be the segment required.

For let AF be produced to cut the circle in E; and join the points E, B.

Then, because AD is perpendicular to the diameter AE, it will be a tangent to the circle at the point A (III. 10.) And, because AD touches the circle, and AB is a chord drawn from the point of contact, the angle BAD

will be equal to the angle AEB in the alternate segment (III. 24.)

But the angle BAD is equal to the angle. c, by con. ftruction; confequently the angle AEB is also equal to the angle c. Q.E.I. SCHOLIUM. When the given angle is a right angle, a femi-circle defcribed upon the given line will be the feg. ment required (III. 16.)

To cut off a fegment from a given circle, that fhall contain an angle equal to a given rectilineal angle,

Let ABC be a given circle, and D a given rectilineal angle; it is required to cut off a segment from the circle ABC, that shall contain an angle equal to D.

Draw the right line EF, to touch the circle ABC in the point A (III. 10.), and make the angle FAB equal to the angle D (I. 20.); then will ABCA be the fegment required.

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For, fince EF is a tangent to the circle, and AB is a chord drawn from the point of contact, the angle FAB will be equal to the angle ACB in the alternate segment (III. 24.)

But

But the angle FAB is equal to the angle D, by cơnftruction; confequently the angle ACB, in the fegment ABCA, is alfo equal to the angle D.

Q.E.I.

PROP. XXVII.

THEOREM.

If two right lines in a circle interfect each other, the rectangle contained under the segments of the one, will be equal to the rectangle contained under the fegments of the other.

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Let AB, CD be any two right lines, in the circle ACBD, interfecting each other in the point F; then will the rect. angle contained under the parts AF, FB of the one, be équal to the rectangle contained under the parts CF, FD of the other.

For, through the point F, draw the diameter HI (III. 1.); and, from the centre E, draw EG at right angles to AB (I. 12.), and join AE:

Then, fince AEF is a triangle, and the perpendicular EG divides the chord AB into two equal parts (III. 3.); the line FB will be equal to the difference of the segments ÁG, GF.

And, because E is the centre of the circle, and AE is equal to EI or EH, the line FI will be equal to the fum of

the fides AE and EF; and FH will be equal to their difference.

But the rectangle under the fum and difference of the two fides of any triangle, is equal to the rectangle under the base and the difference of the fegments of the base (Cor. II. 16.); whence the rectangle of IF, FH is equal to the rectangle of AF, FB.

And, in the fame manner, it may be proved, that the rectangle of IF, FH, is equal to the rectangle of DF, FC confequently the rectangle of AF, FB is alfo equal to the rectangle of DF, FC.

Q.E.D. SCHOLIUM. When the two lines interfect each other in the centre of the circle, the rectangles of their segments will manifeftly be equal, because the fegments themselves are all equal,

PROP.

PROP. XXVIII. THEOREM.

If two right lines be drawn from any point without a circle, to the oppofite part of the circumference, the rectangle of the whole and external part of the one, will be equal to the rectangle of the whole and external part of the other.

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Let ADFB be a circle, and AC, BC any two right lines, drawn from the point c, to the oppofite part of the cir. cumference; then will the rectangle of AC, CD be equal to the rectangle of BC, CF.

For, through the centre E, and the point c, draw the right line CH; and, from the point E, draw EG at right angles to AC (I. 12.) and join AE.

Then, fince AEC is a triangle, and the perpendicular EG divides the chord AD into two equal parts (III. 3.), the line DC will be equal to the difference of the fegments AG, GC.

And, because E is the centre of the circle, and AE is equal to EH, or E1, the line HC will be equal to the fum of the fides AE, EC, and IC will be equal to their difference.