https://www.cut-the-knot.com/triangle/geoprobability.shtml

To save space, let E denote the event that AB is the shorter segment. Then we want to find P(C in AB | E). From the rules of probability, we know that this is P(C in AB and E) / P(E). Now the probability of E is just 50% since as you point out AB and BD are equally likely to be the shorter segment.

The probability that C is in AB and E occurs is the probability that B lies to the left of the midpoint of AD and so does C and C < B. Since B and C are chosen independently, the probability that both lie to the left of the midpoint of AD is just (1/2)(1/2) = 1/4. Furthermore, once we condition on B and C being left of the midpoint of AD, there is a 50% chance that B < C and 50% chance that C > B. So the total probability that C lies in AB and E occurs is (1/2)(1/2)(1/2) = 1/8. Thus P(C in AB | E) = (1/8) / (1/2) = 1/4.

Whew! Of course, that doesn't solve Amit's orginal problem, it just shows why your argument doesn't quite work.

To solve the orignal problem, first let me copy the Description of the problem from the CTK page https://www.cut-the-knot.com/triangle/geoprobability.shtml:
Two points are selected at random on a straight line segment of length 1. What is the probability that a triangle can be constructed out of thus obtained three segments?

To be precise, I'll mean that the two points are selected uniformly at random from 0 to 1. I'll call the two points B and C as you did. The only way that the resulting three pieces do not make a triangle is if one of the pieces has length at least 1/2 (this is equation (2) of the CTK Barycentric solution.) There are three ways this can happen. Either (B < 1/2 and C < 1/2) or (B > 1/2 and C > 1/2) or (|B - C| > 1/2). These are disjoint events (that is, only one can happen) and so to find the total probability, we add the probabilities of the three events. The probability that B < 1/2 and C < 1/2 is just (1/2)(1/2) = (1/4), and the probability that B> 1/2 and C > 1/2 is also (1/2)(1/2) = (1/4).

To find the probability of the last event, picture a square with B plotted on the x-axis and C plotted on the y-axis. Then for |B - C| > 1/2, we must have either B - C > 1/2 or C - B > 1/2. The event that (B - C > 1/2) is the lower right hand corner of the square, and (C - B > 1/2) is the upper left hand corner. Each of these are triangles, and so we can find their total area to be 1/4.

Hence there is a 1/4 + 1/4 + 1/4 = 3/4 chance that the resulting three piece DO NOT form a triangle. Hence the chance that they do is just 1 - 3/4 = 1/4, the same probability as found on the Barycentric page.

Mark