Proposition 10: To bisect a given finite straight line.


Proof.
Let this straight line be \(BC\).

Use proposition 1 to construct an equilateral triangle \(ABC\) using \(BC\) as a base.

Bisect the angle \(A\) using proposition 9.

We claim that \(BD = DC\). To see this, notice in the triangles \(ABD\) and \(ADC\) below that \(AB = AC\) by construction, \(AD\) is common to both triangles, and \(\angle ABC = \angle ACD\) by construction.

Therefore, we can conclude by proposition 4 that the triangles are equal in all respects and consequently we will have \(BD = DC\) as required.



Thoughts: pretty straight forward proof!



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