Euclid's Elements Book 1: Proposition 2

Proposition 2: “From a given point, to draw a straight line equal to a given finite straight line”. In other words, given a point $A$ and another finite straight line $CB$, draw a line equal to $CB$ at $A$.

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Proof.
Let the given point be $A$ and the given finite line be $CB$.

Use postule 1 to draw a line between $A$ and one of the end points say $B$.

Next, use proposition 1 to draw an equilateral triangle from the drawn line $AB$.

Next, describe a circle with center $B$ and radius $CB$ using postulate 3

Extend the line DB all the way till intersects the circle at $E$ using postulate 2.

Now describe another circle with center $D$ and radius $DE$ (postulate 3).

Extend the line DA until it intersects the circle at $F$.

The claim is that $AF=BC$ which is what we want. To see this, we know that $DF=DE$ by definition 15. We know that $DA=DB$ by construction. Therefore, $BE=AF$ by common notion 3. But $CB=BE$ by definition 15 and so $CB=AF$ (common notion 3) as required.

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References:

• Oliver Byrne's Elements of Euclid