Page:The Elements of Euclid for the Use of Schools and Colleges - 1872.djvu/129

Rh Therefore the angle ABF is equal to the angles BAD, ABD. From each of these equals take away the common angle ABD; therefore the remaining angle DBF is equal to the remaining angle BAD, [Axiom 3. which is in the alternate segment of the circle.

And because ABCD is a quadrilateral figure in a circle, the opposite angles BAD, BCD are together equal to two right angles. [III. 22. But the angles DBF, DBE are together equal to two right angles. [I. 13. Therefore the angles DBF, DBE are together equal to the angles BAD, BCD. And the angle DBF has been shewn equal to the angle BAD; therefore the remaining angle DBE is equal to the remaining angle BCD, [Axiom 3. which is in the alternate segment of the circle.

Wherefore, if a straight line &c.

PROPOSITION 33. PROBLEM.

On a given straight line to describe a segment of a circle, containing an angle equal to a given rectilineal angle.

Let AB be the given straight line, and C the given rectilineal angle: it is required to describe, on the given straight line AB, a segment of a circle containing an angle equal to the angle C.

First, let the angle G be a right angle. Bisect AB at F, [1. 10. and from the centre F, at the distance FB, describe the semicircle AHB.

Then the angle AHB in a semicircle is equal to the right angle C. [III. 31.