With various triangles to which the law of sines, the law of cosines, and Pythagoras' Theorem may be applied, there are no doubt many solutions to this problem.
Let BD = d, so that AD = d − 1.
Applying the law of sines (also known as the sine rule) to:
CAD, (sin C) / (d − 1) = (sin 45°) / 1 = 1 /,
ABD, (sin B) / (d − 1) = (sin 45°) / d = 1 / (d).
Note that sin B = AC/BC = cos C.
Squaring and adding both sides, we obtain (since sin2C + cos2C = 1),
1/(d − 1)2 = 1/2 + 1/(2d2).
Multiplying both sides of the equation by 2d2(d − 1)2, and simplifying, we get
d4 − 2d3 − 2d + 1 = 0.
Although the general quartic equation is difficult to solve, we can make a substitution that will halve the degree of this particular equation.
Notice that the sequence of coefficients, (1, −2, 0, −2, 1), forms a palindrome. Since d = 0 is not a root, we may divide by d2, yielding
d2 − 2d − 2/d + 1/d2 = 0.
If we substitute u = d + 1/d, then u2 = d2 + 1/d2 + 2. We thereby obtain
u2 − 2u − 2 = (u − 1)2 − 3 = 0.
Rejecting the negative root, as it would lead to non-real values of d, we have
u = d + 1/d = 1 + . (1)
Multiplying by d, and rearranging, we get
d2 − (1 + )d + 1 = 0.
Solving this quadratic equation, we get d = ½ (1 + ± ).
We reject the smaller root as it is less than 1. (The length of AD = d − 1 must be positive.)
Therefore the length of AD is ½ (−1 + + ).
Now let AC = x and AB = y.
Applying the law of sines to:
CAD, 1 / sin 45° = x / sin ADC,
ABD, d / sin 45° = y / sin BDA.
Note that sin BDA = sin (180° − ADC) = sin ADC.
Hence x = sin ADC / sin 45°, y = d · sin ADC / sin 45°, and so y = dx.
Applying Pythagoras' Theorem to ABC, we obtain
x2 + y2 = (d + 1)2.
Substituting y = dx, we get x2(d2 + 1) = (d + 1)2.
Hence x2 = (d2 + 2d + 1)/(d2 + 1) = 1 + 2d/(d2 + 1).
From (1), d + 1/d = (d2 + 1)/d = 1 + .
Therefore 2d/(d2 + 1) = 2/(1 + ).
Rationalizing the denominator of this fraction, we obtain
2d/(d2 + 1) = 2( − 1) / [(1 + )( − 1)] = − 1.
Hence x2 = 1 + 2d/(d2 + 1) = .
Therefore the length of AC is .
Having found AD, we could obtain the numerical value of AC by applying the law of cosines to ADC, leading to a quadratic equation in x. However, it would be difficult to obtain the exact solution from this equation.
Notice that, in proving AB/AC = BD/CD, we did not make use of the fact that ABC is a right triangle. Indeed, this is a general result, known as the Angle Bisector Theorem.
The polynomial p(x) = anxn + ... + a1x + a0, with real coefficients, is said to be reciprocal if ai = an−i for i = 0, ..., n. If r is a root of p(x) = 0, it is easy to show, by direct substitution, that 1/r is also a root.
It can be shown that any reciprocal polynomial p(x) of degree 2n can be written in the form p(x) = xnq(u), where u = x + 1/x, and q(u) is a polynomial of degree n Further, every reciprocal polynomial of odd degree is divisible by x + 1 (since f(−1) = 0), and the quotient is a reciprocal polynomial of even degree. Hence a reciprocal equation of degree 2n or 2n + 1 can always be reduced to one equation of degree n and one quadratic equation.