First let's draw a line from 0 to 1, and define two uniform random variables $x$ and $y$ as positions at which the breaks occur. We can simplify the calculation by imposing that $x \leq y$ - if $x$ happens to be greater than $y$ then we can simply relabel them to maintain this condition without changing the result. $$ $$ We now use the Triangle Inequality Theorem, which states that if we have a triangle with the two shortest sides denoted $s_1$ and $s_2$, and the hypotenuse denoted $s_3$, then $$ s_3 \leq s_1 + s_2 $$ must be true. Now, let us denote the lengths of our three pieces formed by the breaks, which we will call $A$, $B$ and $C$, $$ A = x, $$ $$ B = y - x, $$ $$ C = 1 - y. $$ We apply the Triangle Inequality Theorem to the three scenarios of $A$, $B$ or $C$ being the hypotenuse to get $$ A \leq B + C $$ $$ B \leq A + C $$ $$ C \leq A + B $$ and we get the following inequalities $$ y \leq \frac{1}{2}, $$ $$ y - x \leq \frac{1}{2}, $$ $$ x \leq \frac{1}{2}. $$ Now, if we draw out a unit square and draw the boundaries of each inequality (including the condition we imposed that $x \leq y$), we see that a triangle which occupies a quarter of the area of our sample space satisfies all three conditions. Consequently, the probability of the three pieces constructing a triangle is $$ P(\text{triangle}) = \frac{1}{4}. $$