Geometry for Elementary School/Bisecting a segment

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

The construction

1. [[../Constructing equilateral triangle|Construct the equilateral triangle]] $\mathrm{△}ABD$ on $\overline{AB}$.
2. [[../ Bisecting an angle |Bisect an angle]] on $\mathrm{\angle }ADB$ using the segment $\overline{DE}$.
3. Let C be the intersection point of $\overline{DE}$ and $\overline{AB}$.

Claim

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

The proof

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