Circumscribing a Circle about a Triangle

Theorem

About any given triangle it is possible to circumscribe a circle.

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 Equality, $\triangle ADF = \triangle BDF$, 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, is 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 Equality, $\triangle ADF = \triangle BDF$, 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, is 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 Equality, $\triangle ADF = \triangle BDF$, 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, is circumscribes $\triangle ABC$.

$\Box$

Historical Note

This is Proposition 5 of Book IV of Euclid's The Elements.