Circumscribing Circle about Triangle

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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 $\triangle ABC$.

Euclid-IV-5a.png

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, it circumscribes $\triangle ABC$.

$\Box$


Euclid-IV-5b.png

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, it circumscribes $\triangle ABC$.

$\Box$


Euclid-IV-5c.png

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, it circumscribes $\triangle ABC$.

$\blacksquare$


Also see


Historical Note

This theorem is Proposition $5$ of Book $\text{IV}$ of Euclid's The Elements.


Sources