PROP. XIV. THEOREM.

An angle at the centre of a circle is double to that at the circumference, when both of them ftand upon the fame arc.

Let the angle BEC be an angle at the centre of the circle ABC, and BAC an angle at the circumference, both standing upon the fame arc BC; then will the angle BEC be double the angle BAC.

Firft, let E, the centre of the circle, be within the angle BAC, and draw AE, which produce to F.

Then, because EA is equal to EB, the angle EAB wilk be equal to the angle EBA (1.5.)

And, because AEB is a triangle, the outward angle BEF will be equal to the two inward oppofite angles EAB, EBA, taken together (I. 28.)

But fince the angles EAB, EBA, are equal to each other, they are, together, double the angle EAB; whence the angle BEF is alfo double the angle EAB.

And, in the fame manner it may be fhewn, that the angle FEC is double the angle EAC; confequently the whole angle BEC will also be double the whole angle

BAC.

Again, let E, the centre of the circle ABC, fall without the angle BAC, and join AE.

Then, fince the angle BF E, of the triangle EFB, is equal to the angle CFA of the triangle CAF (I. 15.) the remaining angles BEF, FBE of the one, are, together, equal to the remaining angles FAC, FCA of the other (I. 28.)

But the angle FBE is equal to the angle EAF (1.5.), and the angle FCA, or ECA, to the angle EAC (1.5.); therefore the angles BEF, EAF, are, together, equal to the angles FAC, EAC.

And, if the angle EAF, which is common, be taken away, the remaining angle BEF or BEC, will be equal to twice the angle FAC, or BAC, as was to be shewn,

PROP. XV. THEOREM.

All angles in the fame fegment of a circle, or which stand on the fame arc, are equal to each other.

B

Let ABCD be a circle, and BAC, BDC any two angles in the fame fegment BADC; then, will the angles BAC, BDC be equal to each other.

For, firft, let the fegment BADC be greater than a femicircle, and having found the centre E, join BE and EC,

Then,

Then, fince an angle at the centre of a circle is double to that at the circumference (III. 14.), the angle BAC will be half the angle BEC.

And, for the fame reason, the angle BDC will, alfo, be half the angle BEC.

But things which are halves of the fame thing are equal to each other; consequently the angle BAC is equal to the angle BDC.

Again, let the fegment BADC be not greater than a femicircle:

Then, fince the right lines BD, AC interfect each other in F, the angle BFA will be equal to the oppofite angle DFC (I. 15.)

And becaufe the fegment ABCD is greater than a femi circle, the angles ABD, ACD, which fland in that fegment, are equal to each other (III. 15.)

But fince the two angles BFA, ABF of the triangle IBA, are equal to the two angles DFC, FCD of the triangle DCF, the remaining angle BAF, or BAC, will alfo be equal to the remaining angle FDC, or BDC. ↑

PROP. XVI. THEOREM.

Q. E. D.

An angle in a femicircle is a right angle.

Let ABC be a femicirele; then will any angle ACB, in that femicircle, be a right angle.

For, find the centre of the circle E (III. 1.) and draw the diameter CED.

Then, because an angle at the centre of a circle is double to that at the circumference (III.14.) the angle AED will be double the angle ACD.

And, for the fame reason, the angle BED will be double the angle BCD.

The angles AED, BED, therefore, taken together, are double the whole angle ACB.

But the angles AED, BED, are, together, equal to two right angles (I. 13.); confequently the angle ACB will be equal to one right angle. Q. E. D. COROLL. The angle BAC, which stands in a segment greater than a femi-circle, is lefs than a right angle (1. 28.):

And the angle BCF, which stands in a segment less than a femi-circle, is greater than a right angle.

The oppofite angles of any quadrilateral figure, infcribed in a circle, are equal to two right angles.

Let ABCD be a quadrilateral, infcribed in the circle ADCB then will the oppofite angles BAD, BCD, taken together, be equal to two right angles.

For,

the remaining square of AC will also be equal to the remaining fquare of DF; or AC equal to Dr (II. 3.), and AB to DE (I. Ax. 6.)

Again, let the chord AB be equal to the chord DE; then will oc, or, or their distances from the centre, be equal to each other.

For the fquares of AC, CO are equal to the fquare of OA (II. 14.), and the squares of DF, FO to the fquare of OD.

But the square of OA is equal to the square of OD (II. 2.); therefore the squares of AC, co are equal to the fquares of DF, FO.

And fince AC is the half of AB (III. 3.), and DF is the half of DE (III. 3.), the square of AC is equal to the fquare of DF (II. 2.)

The remaining square of co is, therefore, equal to the remaining square of Fo; and confequently co is equal to FO (II. 3.), as was to be fhewn.

COROLL. If two right angled triangles, having equal hypotenuses, have two other fides alfo equal, the remaining fides will likewise be equal, and the triangles will be equal in all respects.