The math behind it

BD = (AB)^{1/2} iff BC = 1

First off, realize that there are several triangles described above:

- ADC
- ABD
- DBC

Triangle ADC is inscribed in a semi-circle. The angle ADC intersects an arc of 180^{o}, therefore angle ADC is a right angle (90^{o}. Therefore, triangle ADC is a right triangle. Furthermore, BD is perpendicular to AC. Therefore, the traingle ABD and the triangle DBC are both right triangles. From this we set up the equations from the Pythagorean Theorem:

- AD
^{2} + CD^{2} = AC^{2}
- AB
^{2} + BD^{2} = AD^{2}
- BC
^{2} + BD^{2} = CD^{2}

Substitute equations 2 and 3 into equation 1. This gives:

AB^{2} + BD^{2} + BC^{2} + BD^{2} = AC^{2}

BC = 1; AB = N; AC = N+1

N^{2} + BD^{2} + BD^{2} + 1 = (N+1)^{2}

Expanding out:

N^{2} + 2(BD^{2}) + 1 = N^{2} + 2N + 1

Cancel terms common to both sides of the equation:

~~N~~^{2} + 2(BD^{2}) + ~~1~~ = ~~N~~^{2} + 2N + ~~1~~

2(BD^{2}) = 2N

Divide each sides by 2:

BD^{2} = N

Take the square root of each side:

BD = N^{1/2}