Proposition I.9 · Problem
To bisect a given rectilineal angle.
Drag the points circled in red. The figure changes, and every step of the proof still holds.
- Let the angle BAC be the given rectilineal angle. It is required to bisect it.
- Let a point D be taken at random on AB (you can drag it).
- From AC let AE be cut off equal to AD.Proposition I.3Proposition I.3: Given two unequal straight lines, to cut off from the greater a straight line equal to the less.
- Let DE be joined.Postulate 1Postulate 1: To draw a straight line from any point to any point.
- On DE let the equilateral triangle DEF be constructed, with F on the side away from A.Proposition I.1Proposition I.1: On a given finite straight line to construct an equilateral triangle.
- Let AF be joined. The claim: the angle BAC has been bisected by the straight line AF.Postulate 1Postulate 1: To draw a straight line from any point to any point.
- Since AD is equal to AE, and AF is common, the two sides DA, AF are equal to the two sides EA, AF respectively; and the base DF is equal to the base EF; therefore the angle DAF is equal to the angle EAF. So the given rectilineal angle BAC has been bisected by the straight line AF.Proposition I.8Proposition I.8: If two triangles have the two sides equal to two sides respectively, and have also the base equal to the base, they will also have the angles equal which are contained by the equal straight lines.
Being what it was required to do.