# Proof: By Euclid

(related to Proposition: 6.03: Angle Bisector Theorem)

• For let $CE$ have been drawn through (point) $C$ parallel to $DA$.
• And, $BA$ being drawn through, let it meet ($CE$) at (point) $E$.1
• And since the straight line $AC$ falls across the parallel (straight lines) $AD$ and $EC$, angle $ACE$ is thus equal to $CAD$ [Prop. 1.29].
• But, (angle) $CAD$ is assumed (to be) equal to $BAD$.
• Thus, (angle) $BAD$ is also equal to $ACE$.
• Again, since the straight line $BAE$ falls across the parallel (straight lines) $AD$ and $EC$, the external angle $BAD$ is equal to the internal (angle) $AEC$ [Prop. 1.29].
• And (angle) $ACE$ was also shown (to be) equal to $BAD$.
• Thus, angle $ACE$ is also equal to $AEC$.
• And, hence, side $AE$ is equal to side $AC$ [Prop. 1.6].
• And since $AD$ has been drawn parallel to one of the sides $EC$ of triangle $BCE$, thus, proportionally, as $BD$ is to $DC$, so $BA$ (is) to $AE$ [Prop. 6.2].
• And $AE$ (is) equal to $AC$.
• Thus, as $BD$ (is) to $DC$, so $BA$ (is) to $AC$.
• And so, let $BD$ be to $DC$, as $BA$ (is) to $AC$.
• And let $AD$ have been joined.
• I say that angle $BAC$ has been cut in half by the straight line $AD$.
• For, by the same construction, since as $BD$ is to $DC$, so $BA$ (is) to $AC$, then also as $BD$ (is) to $DC$, so $BA$ is to $AE$.
• For $AD$ has been drawn parallel to one (of the sides) $EC$ of triangle $BCE$ [Prop. 6.2].
• Thus, also, as $BA$ (is) to $AC$, so $BA$ (is) to $AE$ [Prop. 5.11].
• Thus, $AC$ (is) equal to $AE$ [Prop. 5.9].
• And, hence, angle $AEC$ is equal to $ACE$ [Prop. 1.5].
• But, $AEC$ [is] equal to the external (angle) $BAD$, and $ACE$ is equal to the alternate (angle) $CAD$ [Prop. 1.29].
• Thus, (angle) $BAD$ is also equal to $CAD$.
• Thus, angle $BAC$ has been cut in half by the straight line $AD$.
• Thus, if an angle of a triangle is cut in half, and the straight line cutting the angle also cuts the base, then the segments of the base will have the same ratio as the remaining sides of the triangle.
• And if the segments of the base have the same ratio as the remaining sides of the triangle, then the straight line joining the vertex to the cutting (point) will cut the angle of the triangle in half.
• (Which is) the very thing it was required to show.

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### References

1. The fact that the two straight lines meet follows because the sum of $ACE$ and $CAE$ is less than two right angles, as can easily be demonstrated. See [Post. 5] (translator's note).