Construction of Triangle from Given Lengths
Contents |
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.
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 $K$, WLOG the top one in the accompanying diagram.
proof that the circles intersect
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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 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 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 is Proposition 22 of Book 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 poof 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.
