Plate I.

Cor. 3. Hence an Angle in a Segment greater than a Semicircle is lefs than a right Angle; thus ADB is measured by half the Arc AB, but as the Arc AB is less than a Semicircle, therefore half the Arc AB, or the Angle ADB is less than half a Semicircle, and confequently less than a right Angle. Fig. 26.

Cor. 4. An Angle in a Segment lefs than a Semicircle is greater than a right Angle, for since the Arc AEC is greater than a Semicircle, its half, which is the Measure of the Angle ABC, must be greater than half a Semicircle, that is greater than a right Angle. Fig. 27.

Cor. 5. An Angle in a Semicircle is a right Angle, for the Measure of the Angle ABD, is half of a Semicircle AED, and therefore a right Angle.

THE O. VIII.

If from the Center C of a Circle ABE, there be let fall the Perpendicular CD on the Chord AB, it will bifect it in the Point D. Fig. 29.

Let the Lines CA and CB be drawn from the Center to the Extremities of the Chord, then fince CA=CB, the Angle CAB=CBA (by the Lemma). But the Triangles ADC, BDC are right angled ones, fince the Line CD is a Perpendicular, and fo the Angle ACD=DCB; (by Cor. 2. Theo. 5.) then have we AC, CD, and the Angle ACD in one Triangle; feverally equal to CB, CD, and the Angle BCD in the other: Therefore (by Theo. 6.) AD DB. Q, E. D.

Cor.

Plate I.

Cor. 3. Hence an Angle in a Segment greater than a Semicircle is lefs than a right Angle, thus ADB is measured by half the Arc AB, but as the Arc AB is lefs than a Semicircle, therefore haif the Arc AB, or the Angle ADB is less than half a Semicircle, and confequently lefs than a right Angle. Fig. 26.

Cor. 4. An Angle in a Segment less than a Semicircle is greater than a right Angle, for fince the Arc AEC is greater than a Semicircle, its half, which is the Measure of the Angle ABC, must be greater than half a Semicircle, that is greater than a right Angle. Fig. 27.

Cor. 5. An Angle in a Semicircle is a right Angle, for the Measure of the Angle ABD, is half of a Semicircle AED, and therefore a right Angle.

THEO. VIII.

If from the Center C of a Circle ABE, there be let fall the Perpendicular CD on the Chord AB, it will bifet it in the Point D. Fig. 29.

Let the Lines CA and CB be drawn from the Center to the Extremities of the Chord, then fince CA=CB, the Angle CAB=CBA (by the Lemma). But the Triangles ADC, BDC are right angled ones, fince the Line CD is a Perpendicular, and fo the Angle ACD=DCB; (by Cor. 2. Theo. 5.) then have we AC, CD, and the Angle ACD in one Triangle; feverally equal to CB, CD, and the Angle BCD in the other: Therefore (by Theo. 6.) AD DB. Q. E. D.

Cor.