Arc-midpoint (perpendicular-bisector ∧ angle-bisector) lemma – often nick-named “Fact 5"

The statement

Arc-midpoint (perpendicular-bisector ∧ angle-bisector) lemma – often nick-named “Fact 5”.
In any triangle $ABC$, the intersection $E$ of

• the internal (or external) bisector of $\angle C$, and

• the perpendicular bisector of $AB$
  lies on the circumcircle $(ABC)$.
  In fact, $E$ is the midpoint of the arc $AB$:

• with the internal bisector it is the midpoint of the arc $AB$ not containing $C$;

• with the external bisector it is the midpoint of the arc $AB$ containing $C$.

Proof:

---

1. Mid-arc argument (the slick one)

1. Draw the circumcircle $\omega=(ABC)$ and let $M$ be the midpoint of the minor arc $AB$ (the arc that does not contain $C$).

2. Why $M$ is on the perpendicular bisector of $AB$.
   Equal arcs subtend equal chords, so $MA=MB$. Hence $M$ lies on the perpendicular bisector of $AB$. Because perpendicular bisector of AB contains all the points equidistant from A,B.

3. Why $M$ is on the internal angle bisector of $\angle C$.
   Arcs $AM$ and $MB$ are equal, so their subtended angles at $C$ are equal:

   $$\angle ACM=\angle MCB.$$

   Thus $M$ lies on the bisector of $\angle C$.

4. The perpendicular bisector of $AB$ and the angle bisector of $\angle C$ meet in only one point.
   Since $M$ satisfies both properties and $E$ was defined as that unique intersection, we must have $E=M$.

Hence $E$ is literally the midpoint of the arc $AB$, i.e. it lies on $\omega$.

(This little fact is ubiquitous in olympiad geometry hand-outs; many authors simply call it “Fact 5”.)

---

So above we have proved that internal angle bisector of A and the perpendicular bisector of BC meet at the circumcircle.
Let's call that point M.
Let N be the second point where the perpendicular bisector of BC meets the circumcircle.
Since MN is the perpendicular bisector of BC and in a circle a chord is bisected by the perpendicular passing through the center, it means that MN goes through center.
Hence MN is a diameter.
Hence Angle MAN = 90.
Since AM is the internal angle bisector of A and it meets AN at 90 degree, it means that AN is the external angle bisector of A.
Hence proved.