Problem. To bisect a given arc ACB.
Sol.—Draw the chord AB; bisect it in D; erect DC at right angles to AB, meeting the arc in C; then the arc is bisected in C.
Dem.—Join AC, BC. Then the triangles ADC, BDC have the side AD equal to DB (const.), and DC common to both, and the angle ADC equal to the angle BDC, each being right. Hence the base AC is equal to the base BC. Therefore [xxviii.] the arc AC is equal to the arc BC. Hence the arc AB is bisected in C.