Circumscribing Circle about Triangle
Theorem
In the words of Euclid:
- About a given triangle to circumscribe a circle.
(The Elements: Book $\text{IV}$: Proposition $5$)
Construction
Let $\triangle ABC$ be the given triangle.
Let the straight lines $AB$ and $AC$ be bisected at $D$ and $E$.
Let $DF$ and $EF$ be drawn perpendicular to $AB$ and $AC$ respectively.
Let a circle be drawn with center $F$ and radius $AF$.
This is the circle required.
Proof
The point $F$ will be either inside, outside or on the edge $BC$ of $\triangle ABC$.
First suppose $F$ is inside $\triangle ABC$.
Join $FB$ and $FC$.
We have that $AD = DB$ and $DF$ is common and at right angles to $AB$.
So from Triangle Side-Angle-Side Congruence, $\triangle ADF = \triangle BDF$, and so $AF = BF$.
Similarly we can prove that $BF = CF$.
So $AF = BF = CF$.
Thus the circle with center $F$ and radius $AF$ also passes through $B$ and $C$.
That is, it circumscribes $\triangle ABC$.
$\Box$
Secondly, suppose $F$ lies on $BC$.
We have that $AD = DB$ and $DF$ is common and at right angles to $AB$.
So from Triangle Side-Angle-Side Congruence, $\triangle ADF = \triangle BDF$, and so $AF = BF$.
Similarly we can prove that $BF = CF$.
So $AF = BF = CF$.
Thus the circle with center $F$ and radius $AF$ also passes through $B$ and $C$.
That is, it circumscribes $\triangle ABC$.
$\Box$
Thirdly, suppose $F$ lies outside $\triangle ABC$.
Join $FB$ and $FC$.
We have that $AD = DB$ and $DF$ is common and at right angles to $AB$.
So from Triangle Side-Angle-Side Congruence, $\triangle ADF = \triangle BDF$, and so $AF = BF$.
Similarly we can prove that $BF = CF$.
So $AF = BF = CF$.
Thus the circle with center $F$ and radius $AF$ also passes through $B$ and $C$.
That is, it circumscribes $\triangle ABC$.
$\blacksquare$
Also see
Historical Note
This proof is Proposition $5$ of Book $\text{IV}$ of Euclid's The Elements.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 2 (2nd ed.) ... (previous) ... (next): Book $\text{IV}$. Propositions
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {IV}$. Pure Geometry: Plane Geometry: The circumcentre