Through any three points, not fituate in the fame right line, to defcribe the circumference of a circle.

Let A, B, C, be any three points, not fituate in the fame right line; it is required to defcribe the circumference of a circle through those points.

Draw the right lines AB, BC, and bifect them with the perpendiculars DH, EG (I. 10 and 11.); and join DE.* Then, because the angles FED, FDE are lefs than two right angles, the lines DH, EG will meet each other, in fome point F (Cor. I. 25.); and that point will be the centre of the circle required.

For, draw the lines FA, FB and FC,

Then, fince AD is equal to DB, DF common, and the angle ADF equal to the angle FDB (I. 8.), the fide FA will alfo be equal to the fide FB (I. 4.)

And, in the fame manner, it may be fhewn, that the fide FC is alfo equal to the fide FB.

The lines FA, FB and FC, are, therefore, all equal to each other; and confequently F is the centre of a circle which will pafs through the points A, B and C, as was to be fhewn.

SCHO.

SCHO. If the fegment of a circle be given, and any three points be taken in the circumference, the centre of the circle may be found, as above.

PROP. XIX. THEOREM.

If the oppofite angles of a quadrilateral, taken together, be equal to two right angles, a circle may be described about that quadri

lateral.

D

Let ABCD be a quadrilateral, whofe oppofite angles DCB, DAB are, together, equal to two right angles: thenmay a circle be defcribed about that quadrilateral.

For fince the circumference of a circle may be defcribed through any three points (III. 18.), let E be the centre of a circle which paffes through the points D, C, B ; and draw the indefinite right line EFA.

And if the circle does not pass through the fourth point A, let it pass, if poffible, through fome other point F, in the line EA, and draw the lines DF, FB, and BD.

Then, fince the oppofite angles BFD, DCB are, together, equal to two right angles (III. 17.), and the angles BAD, DCB are alfo equal to two right angles (by Hyp.), the angles BFD, DCB will be equal to the angle BAD, DCB.

And if, from each of these equals, there be taken the angle DCB, which is common to both, the remaining angle BAD will be equal to the remaining angle BFD; or, which is the fame thing, the two angles DFE, EFB will *be equal to the two angles DAE, EAB, which is impof fible (I. 16.)

The circumference of the circle, therefore, cannot pass through the point F; and the fame may be demonftrated of any other point in the line EA, except the point A ; whence a circle may be described about the quadrilateral ABCD, as was to be shewn.

PRO P. XX. THEOREM.

Segments of circles, which ftand upon equal chords, and contain equal angles, are equal to each other.

Let ACB, DFE be two segments of circles, which stand upon the equal chords AB, DE, and contain equal angles; then will those segments be equal to each other.

For let the segment DFE be applied to the segment ACB, fo that the point D may fall upon the point A, and the line the line AB.

DE upon

Then, fince DE is equal to AB (by Hyp.), the point E will fall upon the point B, and the two fegments will coincide with each other.

For if they do not, there must be fome point, in the circumference of one of them, which will fall either within or without the other.

Let the point F, in the circumference of the circle DFE, be that point; which suppose to fall at G within the circle and draw the lines AGC, BC and BG.

ACB;

Then, fince the outward angle AGB, of the triangle BCG, is greater than the inward oppofite angle GCB, it will also be greater than the angle DFE, which is equal to GCB, or ACB (by Hyp.)

But the angle AGB is also equal to the angle DFE, because the segments in which they ftand are identical; whence they are equal and unequal at the fame time, which is abfurd.

The point F, therefore, cannot fall within the circle ACB; and in the fame manner it may be fhewn that it cannot fall without it; confequently the fegments must coincide, and be equal to each other. Q.E.D.

COROLL. Segments of circles, which ftand upon equal chords, and contain equal angles, have equal circumferences.

PROP. XXI. THEOREM.

In equal circles, equal angles ftand upon equal arcs, whether they be at the centres or circumferences; and if the arcs be equal, the angles will be equal.

B
K

Let ABC, DEF be two equal circles, having the angles AGB, DHE, at their centres, equal to each other, as alfo the angles ACB, DFE, at their circumferences; then will the arc AKB be equal to the arc DLE.

For, join the points AB, DE: then, fince the circles are equal to each other (by Hyp.), their radii and circumferences will also be equal (III. 5.)

And, fince the two fides AG, GB of the triangle ABG, are equal to the two fides DH, HE of the triangle DEH, and the angle AGB to the angle DHE (by Hyp.), their bafes AB, DE will likewife be equal (I. 4.)

The chord AB, therefore, being equal to the chord DE, and the angle ACB to the angle DFE (by Hyp.), the arc BCA will also be equal to the arc EFD (Cor. III. 20.)

But fince the whole circumference of the circle ABC, is equal to the whole circumference of the circle DEF, and the arc BCA to the arc EFD, the arc AKB will also be equal to the arc DLE.