# Geometry for Elementary School/Bisecting a segment

 Geometry for Elementary School Bisecting an angle Bisecting a segment Congruence and similarity

In this chapter, we will learn how to bisect a segment. Given a segment ${\displaystyle {\overline {AB}}}$, we will divide it to two equal segments ${\displaystyle {\overline {AC}}}$ and ${\displaystyle {\overline {CB}}}$. The construction is based on book I, proposition 10.

## The construction

1. Construct the equilateral triangle ${\displaystyle \triangle ABD}$ on ${\displaystyle {\overline {AB}}}$.
2. Bisect an angle on ${\displaystyle \angle ADB}$ using the segment ${\displaystyle {\overline {DE}}}$.
3. Let C be the intersection point of ${\displaystyle {\overline {DE}}}$ and ${\displaystyle {\overline {AB}}}$.

## Claim

1. Both ${\displaystyle {\overline {AC}}}$ and ${\displaystyle {\overline {CB}}}$ are equal to half of ${\displaystyle {\overline {AB}}}$.

## The proof

1. ${\displaystyle {\overline {AD}}}$ and ${\displaystyle {\overline {BD}}}$ are sides of the equilateral triangle ${\displaystyle \triangle ABD}$.
2. Hence, ${\displaystyle {\overline {AD}}}$ equals ${\displaystyle {\overline {BD}}}$.
3. The segment ${\displaystyle {\overline {DC}}}$ equals to itself.
4. Due to the construction ${\displaystyle \angle ADE}$ and ${\displaystyle \angle EDB}$ are equal.
5. The segments ${\displaystyle {\overline {DE}}}$ and ${\displaystyle {\overline {DC}}}$ lie on each other.
6. Hence, ${\displaystyle \angle ADE}$ equals to ${\displaystyle \angle ADC}$ and ${\displaystyle \angle EDB}$ equals to ${\displaystyle \angle CDB}$.
7. Due to the Side-Angle-Side congruence theorem the triangles ${\displaystyle \triangle ADC}$ and ${\displaystyle \triangle CDB}$ congruent.
8. Hence, ${\displaystyle {\overline {AC}}}$ and ${\displaystyle {\overline {CB}}}$ are equal.
9. Since ${\displaystyle {\overline {AB}}}$ is the sum of ${\displaystyle {\overline {AC}}}$ and ${\displaystyle {\overline {CB}}}$, each of them equals to its half.