Take the ruler and compass to a middle (grade) school. Trade them for a calculator. You may need to throw some paste into the bargain to get the calculator, or you can resort to schoolyard fisticuffs (remember you have a sharp compass if you're losing). After binding your wounds, punch in length N and hit the √ key.

The math behind it

BD = (AB)1/2 iff BC = 1

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

2. ABD
3. DBC

Triangle ADC is inscribed in a semi-circle. The angle ADC intersects an arc of 180o, therefore angle ADC is a right angle (90o. 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:

1. AD2 + CD2 = AC2
2. AB2 + BD2 = AD2
3. BC2 + BD2 = CD2

Substitute equations 2 and 3 into equation 1. This gives:
AB2 + BD2 + BC2 + BD2 = AC2

BC = 1; AB = N; AC = N+1
N2 + BD2 + BD2 + 1 = (N+1)2

Expanding out:
N2 + 2(BD2) + 1 = N2 + 2N + 1

Cancel terms common to both sides of the equation:
N2 + 2(BD2) + 1 = N2 + 2N + 1
2(BD2) = 2N
Divide each sides by 2:
BD2 = N
Take the square root of each side:
BD = N1/2

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