A triangle has corners at points A, B, and C. Side AB has a length of #15 #. The distance between the intersection of point A's angle bisector with side BC and point B is #6 #. If side AC has a length of #15 #, what is the length of side BC?

Let the point where the angle bisector intersects with side BC be D

As BD = DC, it’s an isosceles triangle & BC = 2*BD

Using the angle bisector theorem, we can find the length of side BC. The theorem states that in a triangle, if an angle bisector intersects the opposite side, it divides that side into segments proportional to the lengths of the other two sides.

Given that the distance between the intersection of the angle bisector of angle A with side BC and point B is 6 units, and side AC has a length of 15 units, we can set up a proportion:

(Length of side BC) / (Length of side AC) = (Distance from point B to the intersection of angle bisector of angle A) / (Distance from point C to the intersection of angle bisector of angle A)

Let's denote the length of side BC as x. So, the proportion becomes:

x / 15 = 6 / (15 - x)

Solving this proportion for x will give us the length of side BC.

Cross multiplying, we get:

x(15 - x) = 6 * 15

Expanding and simplifying:

15x - x^2 = 90

Rearranging terms and setting the equation equal to zero:

x^2 - 15x + 90 = 0

We can solve this quadratic equation using the quadratic formula:

x = [ -b ± √(b^2 - 4ac) ] / (2a)

Where a = 1, b = -15, and c = 90.

Plugging in the values:

x = [ -(-15) ± √((-15)^2 - 4190) ] / (2*1)

x = [ 15 ± √(225 - 360) ] / 2

x = [ 15 ± √(-135) ] / 2

Since the term inside the square root is negative, the solution will be imaginary, meaning the triangle cannot be formed with these given conditions. Therefore, there is no real value for the length of side BC that satisfies the given conditions.