Let AB be the stick. WLOG we may assume AB=1(Since the probability won't depend on the length of AB). Let the points at which the stick is broken be P and Q.

AP=x,PQ=y and QB=z.

Since $0\leq AP,PQ,QB \leq 1$ we need to consider all the points inside the $1\times 1\times 1 $ cube. Futhermore the points lie on the x+y+z=1 plane.

x+y+z=1 plane(click on the link to see the image of the plane)

On applying the triangle inequalities (i.e $x+y>z,y+z>x\text{ and }x+z>y$) we find that the net area of points satisfying the condition of forming a triangle is the shaded potion.

Shaded Area(Click on the link to see the shaded area)

Since the points $J,K,I$ are the midpoints of the sides of the of the triangle ACE.The probability = $\dfrac{\text{Area of }\Delta JKI}{\text{Area of }\Delta ACE}=\dfrac{1}{4}$

Let AB be the stick. WLOG we may assume AB=1(Since the probability won't depend on the length of AB). Let the points at which the stick is broken be P and Q.

$AP=x$, $PQ=y$ and $QB=z$.

Since $0\leq AP,PQ,QB \leq 1$ we need to consider all the points inside the $1\times 1\times 1 $ cube. Furthermore the points lie on the x+y+z=1 plane.

x+y+z=1 plane(click on the link to see the image of the plane)

On applying the triangle inequalities (i.e $x+y>z,y+z>x\text{ and }x+z>y$) we find that the net area of points satisfying the condition of forming a triangle is the shaded potion.

Shaded Area(Click on the link to see the shaded area)

Since the points $J,K,I$ are the midpoints of the sides of the of the triangle ACE.The probability = $\dfrac{\text{Area of }\Delta JKI}{\text{Area of }\Delta ACE}=\dfrac{1}{4}$