Two Lines Meet at Unique Point

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Theorem

Let two straight line segments be constructed on a straight line segment from its endpoints so that they meet at a point.

Then there cannot be two other straight line segments equal to the former two respectively, constructed on the same straight line segment and on the same side of it, meeting at a different point.


In the words of Euclid:

Given two straight lines constructed on a straight line (from its extremities) and meeting in a point, there cannot constructed on the same straight line (from its extremities) and on the same side of it, two other straight lines meeting in another point and equal to the former two respectively, namely each to that which has the same extremity with it.

(The Elements: Book $\text{I}$: Proposition $7$)


Proof

Euclid-I-7.png

Let $AC$ and $CB$ be constructed on $AB$ meeting at $C$.

Let two other straight line segments $AD$ and $DB$ be constructed on $AB$, on the same side of it, meeting at $D$, such that $AC = AD$ and $CB = DB$.


Suppose, with a view to obtaining a contradiction, $C$ and $D$ are different points.

Let $CD$ be joined.


Since $AC = AD$ it follows that $\angle ACD = \angle ADC$.

Therefore $\angle ACD$ is greater than $\angle DCB$ because the whole is greater than the part.

Therefore $\angle CDB$ is much greater than $\angle DCB$.

Now since $CB = DB$, it follows that $\angle CDB = \angle DCB$.

But it was proved much greater than it.

From this contradiction it follows that $C$ and $D$ can not be different points.

Hence the result.

$\blacksquare$


Historical Note

This theorem is Proposition $7$ of Book $\text{I}$ of Euclid's The Elements.


Sources