Two Lines Meet at Unique Point

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

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 $CB = DB$.

Aiming for a contradiction, suppose $C$ and $D$ are different points.

Let $CD$ be joined.

We have by hypothesis that:

$AC = AD$

Hence from Proposition $5$: Isosceles Triangle has Two Equal Angles:

$\angle ACD = \angle ADC$

From Common Notion $5$:

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

Therefore:

$\angle CDB > \angle DCB$

We have by hypothesis:

$CB = DB$

Hence it follows that:

$\angle CDB = \angle DCB$

But this contradicts:

$\angle CDB > \angle DCB$

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

Hence the result.

$\blacksquare$

Historical Note

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

Sources

• 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 1 (2nd ed.) ... (previous) ... (next): Book $\text{I}$. Propositions