# Construction of Triangle from Given Lengths

## Theorem

Given three straight lines such that the sum of the lengths of any two of the lines is greater than the length of the third line, it is possible to construct a triangle having the lengths of these lines as its side lengths.

In the words of Euclid:

*Out of three straight lines, which are equal to three given straight lines, to construct a triangle; thus it is necessary that two of the straight lines taken together in any manner should be greater than the remaining one.*

(*The Elements*: Book $\text{I}$: Proposition $22$)

## Construction

Let the three straight lines from which we are going to construct the triangle be $a$, $b$, and $c$.

Let $D$ and $E$ be any distinct points.

Construct $DE$ and extend it beyond $E$.

We cut off a length $DF$ on $DE$ equal to $a$.

Similarly, we cut off $FG = b$ and $GH = c$ on $DE$.

Now we can construct a circle centered at $F$ with radius $DF$.

Similarly, we can construct a circle centered at $G$ with radius $GH$.

Call one of the intersections of the two circles $K$.

Without loss of generality, let this be the top one in the accompanying diagram.

This needs considerable tedious hard slog to complete it.In particular: proof that the circles intersectTo discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Finish}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

Finally, we can construct the segment $FK$.

$\triangle FGK$ is the required triangle.

## Proof

Since $F$ is the center of the circle with radius $FD$, it follows from Book $\text{I}$ Definition $15$: Circle that $DF = KF$, so $a = KF$ by Euclid's first common notion.

Since $G$ is the center of the circle with radius $GH$, it follows from Book $\text{I}$ Definition $15$: Circle that $GH = GK$, so $c = GK$ by Euclid's first common notion.

$FG = b$ by construction.

Therefore the lines $FK$, $FG$, and $GK$ are, respectively, equal to the lines $a$, $b$, and $c$, so $\triangle FGK$ is indeed the required triangle.

$\blacksquare$

## Historical Note

This proof is Proposition $22$ of Book $\text{I}$ of Euclid's *The Elements*.

Note that the condition required of the lengths of the segments is the equality shown in Proposition $20$: Sum of Two Sides of Triangle Greater than Third Side. Thus, this is a necessary condition for the construction of a triangle.

When Euclid first wrote the proof of this proposition in *The Elements*, he neglected to prove that the two circles described in the construction actually do intersect, just as he did in Proposition $1$: Construction of Equilateral Triangle.

## Sources

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