A triangle's angle bisectors of $\alpha$ and $\beta$ intersect the circumcircle of the triangle at points $D$ and $E$. What angles does the segment $DE$ form with the angle bisector of the triangle's $\gamma$ angle?
I. solution. Let the intersection of the angle bisector from $C$ and $DE$ be $F$, and the intersection of the angle bisectors be $O$. (Fig. 2. - We will use the theorem that the internal angle bisectors intersect at a single point.)

![](https://cdn.mathpix.com/cropped/2024_05_02_c3eb8eb4b60d11c9ab26g-1.jpg?height=518&width=548&top_left_y=245&top_left_x=775)

Fig. 2

Then, as the external angle of $\triangle AOC$,

$$
\angle COD \equiv \angle FOD = \frac{\alpha}{2} + \frac{\gamma}{2}.
$$

By the inscribed angle theorem,

$$
\angle ADE \equiv \angle ODF = \frac{\beta}{2}.
$$

Thus, from $\triangle DFO$,

$$
\angle DFO = 180^\circ - (\angle ODF + \angle FOD) = 180^\circ - \frac{\alpha + \beta + \gamma}{2} = 90^\circ.
$$

Therefore, $DE$ is perpendicular to the angle bisector from $C$.

II. solution: The points $D$ and $E$ bisect the arcs of the circumcircle over the sides $BC$ and $AC$, respectively. Thus, due to the equality of inscribed angles subtended by equal arcs, in the quadrilateral $CDOE$, the diagonal $DE$ bisects the angles at its endpoints (Fig. 3), making it the axis of symmetry of the quadrilateral. (The quadrilateral is a kite.)

![](https://cdn.mathpix.com/cropped/2024_05_02_c3eb8eb4b60d11c9ab26g-1.jpg?height=569&width=548&top_left_y=1450&top_left_x=775)

Since $C$ and $O$ are reflections of each other across $DE$, it follows that

$$
DE \perp CO
$$

but $CO$ is the angle bisector from $C$, as $O$ must lie on the third angle bisector as well.

III. solution: The fact that the quadrilateral $CDOE$ is a kite can also be proven as follows:

If we fix points $B$ and $C$ and let $A$ move along the arc $\widehat{BEC}$, then $O$ moves along the arc $BC$ of a circle whose center is $D$ (Fig. 3 - see the first problem of the A. D. competition, first round, in the K. M. L. October 1954 issue). Similarly, if we fix points $A$ and $C$ and let $B$ move, $O$ moves along the arc $AC$ of a circle whose center is $E$. Therefore,

$$
DC = DO \quad \text{and} \quad EC = EO
$$

which was to be proven.

The other solutions do not use the theorem about the intersection of angle bisectors.

IV. solution. Let $G$ be the intersection of the angle bisector from $C$ with the circle (Fig. 4).

![](https://cdn.mathpix.com/cropped/2024_05_02_c3eb8eb4b60d11c9ab26g-2.jpg?height=613&width=550&top_left_y=122&top_left_x=755)

$D$, $E$, and $G$ bisect the arcs over the sides of the triangle. Draw a line through $D$ parallel to $CG$. Let its second intersection with the circle be $H$. Then

$$
\widehat{CH} = \widehat{CD}
$$

so the arc $\widehat{EAGH}$ is composed of half-arcs over the three sides, making it a semicircle.

This implies that

$$
\angle EDH = 90^\circ,
$$

and thus, due to the parallelism $DH \parallel CG$,

$$
DE \perp CG
$$

Remark: Essentially, we proved a general theorem for the specific case here, that the angle between two chords is equal to the inscribed angle subtended by the sum of the arcs opposite the angle and its vertex angle. The proof used here can be extended to arbitrary chords.

Several other solutions are essentially proofs of this theorem using different auxiliary lines. (We do not include these here.)

V. solution: Let the line $DE$ intersect sides $AC$ and $BC$ at $K$ and $L$, respectively. Several contestants showed that the triangle $CKL$ is isosceles. This implies that $DE$ is perpendicular to the angle bisector from $C$, because in an isosceles triangle, the altitude and the angle bisector at the vertex are perpendicular to each other.

![](https://cdn.mathpix.com/cropped/2024_05_02_c3eb8eb4b60d11c9ab26g-2.jpg?height=520&width=550&top_left_y=1575&top_left_x=771)

Fig. 5

The isosceles property can be shown simply as follows: Since $E$ bisects the arc $AC$ and $D$ bisects the arc $BC$ (Fig. 5), it follows that

$$
\angle CED = \angle BCD \quad \text{and} \quad \angle ACE = \angle CDE,
$$

as inscribed angles subtended by equal arcs. This implies that the third angles of $\triangle CKE$ and $\triangle DLC$ are equal, which are also the exterior angles of $\triangle CKL$ at $K$ and $L$.

Thus, $\triangle CKL$ is isosceles.

VI. solution: Let $C_0$ be the midpoint of the arc $AB$ that contains $C$. The angle bisector from $C_0$ in the isosceles triangle $ABC_0$ is the axis of symmetry of the figure, and thus the line connecting the intersection points of the other two angle bisectors, $D_0$ and $E_0$, is perpendicular to it (Fig. 6).

![](https://cdn.mathpix.com/cropped/2024_05_02_c3eb8eb4b60d11c9ab26g-3.jpg?height=574&width=548&top_left_y=125&top_left_x=775)

Fig. 6

If $C_0$ moves along the circle to $C$, then $D_0$ and $E_0$ move in the same direction by half the arc. Let the intersection of $D_0E_0$ and $DE$ be $H$. According to the above,

$$
\angle D_0E_0D = \angle EDE_0 = \frac{1}{2} \angle C_0GC.
$$

Thus, as the external angle of $\triangle E_0DH$,

$$
\angle E_0HE = \angle HE_0D + \angle HDE_0 = \angle C_0GC,
$$

which means that the line $DE$ has turned by the same angle as the angle bisector from $C$, maintaining their perpendicularity.