Exercise:
Construct a right triangle by the hypotenuse and the leg with a compass and a ruler without graduations.
The solution figures:
Note: The text book that I learn math with is on the Ukrainian language, that is the part of cyrillic. So lets rename the figures identifiers below from "б" to "b", from "в" to "c" and "a" is "a", it doesn't need to be changed.
The whole solution by the textbook:
Analysis:
We have two line segments of the length c and b, and b < c (fig. "a"). . Since any hypotenuse is greater than any leg, then our hypotenuse c is the bigger line segment, when the leg b - smaller. So, we need to construct a right triangle ABC in which ∠С = 90°, AB = c, AC = b.
Solution:
Let's draw two perpendicular lines m and n. Let C be their point of the intersection. On the straight line m, we put the segment CA, which is equal to the given leg b (fig. "b"). Let's make a circle with the center at the point A with a radius that is equal to the given hypothesis c. Let this circle cross the line n at two points B₁ and B₂ (fig. "c").
Each of the triangles ACB₁ and ACB₂ are solutions. And as the triangles ACB₁ and ACB₂ are congruent, this problem has a unique solution.
The problem has a unique solution in the sense that any two such triangles are congruent, independent of how they might have been constructed. That follows from the fact that they have the same three side lengths.
There are many such triangles, because there are many choices for the lines $m$ and $n$. In that sense there is no "unique solution".
I think the text is confusing when it calls attention to the fact that the proposed construction happens to find two instances of the required triangle.