Geometry for elementary academy Bisecting a articulation

 06 August 03:32   

    In this chapter, we will apprentice how to bifurcate a segment. Accustomed a articulation $overline$, we will bisect it to two according segments $overline$ and $overline$. The architecture is based on [http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI10.html book I, hypothesis 10].

    # $riangle\; ABD$ on $overline$.

    # on $angle\; ADB$ using the articulation $overline$.

    # Let C be the circle point of $overline$ and $overline$.

    # Both $overline$ and $overline$ are according to bisected of $overline$.

    # $overline$ and $overline$ are abandon of the boxlike triangle $riangle\; ABD$.

    # Hence, $overline$ equals $overline$.

    # The articulation $overline$ equals to itself.

    # Due to the architecture $angle\; ADE$ and $angle\; EDB$ are equal.

    # The segments $overline$ and $overline$ lie on anniversary other.

    # Hence, $angle\; ADE$ equals to $angle\; ADC$ and $angle\; EDB$ equals to $angle\; CDB$.

    # Due to the triangles $riangle\; ADC$ and $riangle\; CDB$ congruent.

    # Hence, $overline$ and $overline$ are equal.

    # Back $overline$ is the sum of $overline$ and $overline$, anniversary of them equals to its half.