For, draw the diagonals AC, BD, and produce the fide

BA to E.

Then, because the outward angle of any triangle, is equal to the two inward oppofite angles taken together (I. 28.), the angle EAD will be equal to the angles ABD,

ADB.

And, because all angles in the fame fegment of a circle are equal to each other (III. 15.), the angle ABD will be equal to the angle ACD, and the angle ADB to the angle ACB.

The angle EAD, therefore, which is equal to the angles ABD, ADB, taken together, will alfo be equal to the angles ACD, ACB, taken together, or to the whole angle BCD.

But the angles EAD, BAD, taken together, are equal to two right angles (I. 13.); confequently the angles BCD, BAD, taken together, will alfo be equal to two right angles. Q. E. D. COROLL. If any fide AB, of the quadrilateral ABCD, be produced, the outward angle EAD will be equal to the inward oppofite angle BCD.

PROP. XVIII. PROBLEM.

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

B

1

Let A, B, C, be any three points, not fituated in the fame right line; it is required to defcribe the circum-ference of a circle through thofe 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 less 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 also 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 pass through the points A, B and C, as was to be fhewn.

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 quadriz lateral.

D

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

For fince the circumference of a circle may be described 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 pafs through the fourth point A, let it pafs, if poffible, through some 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 angles 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 impoffible (I. 16.)

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

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 fegments be equal to each other.

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

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.

って

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 fuppofe to fall at G within the circle ACB; and draw the lines AGC, BC and BG.

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 alfo equal to the angle DFE, ber cause the fegments 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 stand upon equal chords, and contain equal angles, have equal circumferences.