The construction
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Construct the equilateral triangle △ A B D {\displaystyle \triangle ABD} on A B ¯ {\displaystyle {\overline {AB}}} .
Bisect an angle on ∠ A D B {\displaystyle \angle ADB} using the segment D E ¯ {\displaystyle {\overline {DE}}} .
Let C be the intersection point of D E ¯ {\displaystyle {\overline {DE}}} and A B ¯ {\displaystyle {\overline {AB}}} .
Both A C ¯ {\displaystyle {\overline {AC}}} and C B ¯ {\displaystyle {\overline {CB}}} are equal to half of A B ¯ {\displaystyle {\overline {AB}}} . The proof
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A D ¯ {\displaystyle {\overline {AD}}} and B D ¯ {\displaystyle {\overline {BD}}} are sides of the equilateral triangle △ A B D {\displaystyle \triangle ABD} .
Hence, A D ¯ {\displaystyle {\overline {AD}}} equals B D ¯ {\displaystyle {\overline {BD}}} .
The segment D C ¯ {\displaystyle {\overline {DC}}} equals to itself.
Due to the construction ∠ A D E {\displaystyle \angle ADE} and ∠ E D B {\displaystyle \angle EDB} are equal.
The segments D E ¯ {\displaystyle {\overline {DE}}} and D C ¯ {\displaystyle {\overline {DC}}} lie on each other.
Hence, ∠ A D E {\displaystyle \angle ADE} equals to ∠ A D C {\displaystyle \angle ADC} and ∠ E D B {\displaystyle \angle EDB} equals to ∠ C D B {\displaystyle \angle CDB} .
Due to the Side-Angle-Side congruence theorem the triangles △ A D C {\displaystyle \triangle ADC} and △ C D B {\displaystyle \triangle CDB} congruent.
Hence, A C ¯ {\displaystyle {\overline {AC}}} and C B ¯ {\displaystyle {\overline {CB}}} are equal.
Since A B ¯ {\displaystyle {\overline {AB}}} is the sum of A C ¯ {\displaystyle {\overline {AC}}} and C B ¯ {\displaystyle {\overline {CB}}} , each of them equals to its half.