Let ABC = t and AC = x.
Then ADC = t + 15° and DCA = t - 15°.
Applying the law of sines (also known as the sine rule) to ADC, we obtain:
AD/AC = sin(t - 15°)/sin(t + 15°). (1)
Dropping a perpendicular from A to BC, we have:
cos t = ½BC/AC = /(2x).
Hence AD/AC = (2 cos t)/(2)
By inspection, if t = 45°, then:
sin(t - 15°)/sin(t + 15°) = ½ / (½) = 1/.
(2 cos t)/ = / = 1/.
Hence t = 45° is one solution. To show it is the only solution, we first expand the right-hand side of (1) using standard trigonometric identities, and then divide by sin t sin 15°:
|sin(t - 15°)/sin(t + 15°)||= (sin t cos 15° - cos t sin 15°)/(sin t cos 15° - cos t sin 15°)|
|= (cot 15° - cot t)/(cot 15° + cot t)|
Clearly, 15° < t < 90°.
For this range of values of t, f(t) = (2 cos t)/ is a strictly decreasing function.
As cot t is strictly decreasing, g(t) = (cot 15° - cot t)/(cot 15° + cot t) is a strictly increasing function.
Since both f(t) and g(t) are continuous functions, it follows that t = 45° is the only solution.
Therefore, CAB is a right angle.
More generally, if BC/AD = r and BCD = u, then (2 cos t)/r = (cot u - cot t)/(cot u + cot t).
Setting x = cot t, we get cos t = x/(1 + x2)1/2.
Then, setting v = cot u, squaring, simplifying and collecting terms, we obtain a quartic equation in x:
(r2 - 4)x4 - 2v(r2 + 4)x3 + (v2(r2 - 4) + r2)x2 - 2vr2x + v2r2 = 0.
For given r and u, the (unique) positive root of this equation that is less than cot u (hence t > u) yields t, and hence CAB.