Return to Euclid’s Elements of Geometry
To bisect a given rectilineal angle.
Let the angle BAC be the given rectilineal angle.
It is then required to bisect it.
Let a point D be taken at random on AB;
let AE be cut off from AC equal to AD; [I. 3]
let DE be joined, and on DE let the equilateral triangle DEF be constructed;
let AF be joined.
I say that the angle BAC has been bisected by the straight line AF.
For, 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. [I. 8]
Therefore the given rectilineal angle BAC has been bisected by the straight line AF.
Q.E.F.
Return to Euclid’s Elements of Geometry
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