triangle.

a) All three segments turned out to be equal. Is it true that the triangle is equilateral?

b) Two segments turned out to be equal. Is it true that the triangle is isosceles?

c) Can the lengths of the segments be 4, 4, and 3?
a) Let in triangle $ABC \quad \angle A < \angle B \leq \angle C$. Then the perpendicular bisectors of sides $AC$ and $BC$ intersect side $AB$. The segments of these perpendicular bisectors lying inside the triangle have equal projections on lines perpendicular to $AB$, but they form different angles with these lines. Therefore, they are not equal.

Now let $\angle A \leq \angle B < \angle C$. Then the perpendicular bisectors of $AB$ and $AC$ intersect $AC$ and $AB$ respectively, and thus cut off similar but not equal triangles from triangle $ABC$. The segments of the perpendicular bisectors lying inside the triangle are corresponding sides of these triangles and, therefore, are not equal.

b) For example, consider a triangle with angles $A = \pi / 8, B = \pi / 4, C = 5 \pi / 8$ and a unit radius of the circumscribed circle. In this triangle, the perpendicular bisector of $AB$ intersects side $AC$ and the segment of this bisector lying inside the triangle is equal to $1 / 2 \mathrm{AB} \operatorname{tg} \angle A = \sin 5 \pi / 8 \operatorname{tg} \pi / 8 = \cos \pi / 8 \operatorname{tg} \pi / 8 = \sin \pi / 8$. The perpendicular bisector of $BC$ intersects $AB$ and the length of the corresponding segment is $1 / 2 BC \operatorname{tg} \angle B = \sin \pi / 8$. Thus, these segments are equal.

c) If the triangle is isosceles, then as shown in a), the segments of the perpendicular bisectors to the legs are shorter than the height.

Let $\angle A < \angle B < \angle C$. Then from the condition it follows that $\cos \angle A \operatorname{tg} \angle B = \sin \angle C$. Therefore, the ratio $t$ of the segments of the perpendicular bisectors to the largest and middle sides of the triangle is equal to the ratio of these sides, that is, $t = \frac{\sin \angle C}{\sin \angle B} = \frac{\cos \angle A}{\cos \angle B}$.

From the equality $\sin \angle B \cdot \frac{\cos \angle A}{\cos \angle B} = \sin \angle C = \sin (\angle A + \angle B) = \sin \angle A \cos \angle B + \sin \angle B \cos \angle A$, we get $\sin^2 \angle A \cos^2 \angle B = \sin^2 \angle B (1 - \cos \angle B)^2 t^2$, or $\left(1 - t^2 \cos^2 \angle B\right) \cos^2 \angle B = \left(1 - \cos^2 \angle B\right) (1 - \cos \angle B)^2 t^2$, from which $t = \frac{\cos \angle B}{\sqrt{1 - 2 \cos \angle B + 2 \cos^3 \angle B}}$.

By investigating the function $\frac{x}{\sqrt{1 - 2x + 2x^3}}$, we can see that its maximum is less than $4 / 3$.

## Answer

a) Correct; b) Incorrect; c) Cannot.