Problem 2002

Let $\alpha$ be the size of the angle $\angle CAD$. Because $|AB| = |AC| = |CD|$, we have $\angle ABC = \angle BCA = \angle CAD + \angle CDA = 2\angle CAD = 2\alpha$. Therefore, $\angle BAD = \angle DBA$ if and only if $\angle BAC = \alpha$. On the other hand, because $\angle BAC = 180^\circ - 4\alpha$, we see that $\angle BAC = \alpha$ if and only if $\alpha = 36^\circ$. Therefore, it suffices to show that
$$\frac{1}{|CD|} - \frac{1}{|BD|} = \frac{1}{|CD| + |BD|} \quad \text{is equivalent to} \quad \angle BAD = \angle DBA.$$

Because $|AB| = |AC| = |CD|$ and $C$ is between $B$ and $D$, the equation
$$\frac{1}{|CD|} - \frac{1}{|BD|} = \frac{1}{|CD| + |BD|} \quad \text{is equivalent to}$$
$$\frac{1}{|AB|} - \frac{1}{|BC| + |AB|} = \frac{1}{|AB| + |BD|}, \quad \text{or}$$
$$\frac{|BC|}{|AB|( |BC| + |AB|)} = \frac{1}{|AB| + |BD|}, \quad \text{which simplifies to}$$
$$|BC| \cdot |BD| = |AB|^2, \text{ or equivalently, } \frac{|AB|}{|BD|} = \frac{|BC|}{|AB|}.$$
Because the triangles $ABC$ and $ABD$ have a common angle at $B$, this last equation is equivalent to these two triangles being similar such that sides $AB$ and $BD$ in triangle $ABD$ correspond to sides $BC$ and $AB$ in triangle $ABC$. Because $\triangle ABC$ is isosceles, this is equivalent to triangle $BDA$ being isosceles with $\angle BAD = \angle DBA$, which, as seen above, is equivalent to $\angle BAC = 36^\circ$.

As usual, we let $\alpha = \angle BAC$, $\beta = \angle ABC$, $a = |BC|$ and $b = |AC| = |AB| = c$. Because $|CD| = b$ and $C$ is between $B$ and $D$, the equation
$$\frac{1}{|CD|} - \frac{1}{|BD|} = \frac{1}{|CD| + |BD|} \quad \text{is equivalent to}$$
$$\frac{1}{b} - \frac{1}{a+b} = \frac{1}{a+2b} \quad \text{or} \quad \frac{a}{b(a+b)} = \frac{1}{a+2b}, \quad \text{which can be written as}$$
$$a(a+2b) = b(a+b), \text{ i.e. } a^2 + ab = b^2. \text{ Introducing } x = a/b, \text{ we see now that}$$
$$\frac{1}{|CD|} - \frac{1}{|BD|} = \frac{1}{|CD| + |BD|} \quad \text{is equivalent to} \quad x^2 + x - 1 = 0.$$

Because $A$ is not on $BC$, the triangle inequality implies $0 < a < 2b$, hence $0 < x < 2$. Therefore, $0 \neq (x-2)(x+1)$ and so $x^2+x-1=0$ if and only if
$$0 = (x^2+x-1)(x-2)(x+1) = x^4 - 4x^2 - x + 2.$$

The Cosine Rule gives $a^2 = 2b^2 - 2b^2 \cos(\alpha)$ and $b^2 = a^2 + b^2 - 2ab \cos(\beta)$, from which we obtain $2 \cos(\alpha) = 2 - x^2$ and $2 \cos(\beta) = x$. Therefore, the equation $(x^2 - 2)^2 - 2 = x$ is equivalent to $4 \cos^2(\alpha) - 2 = 2 \cos(\beta)$. Because $2 \cos^2(\alpha) - 1 = \cos(2\alpha)$, the above is equivalent to $\cos(2\alpha) = \cos(\beta)$.
We would like to conclude that $\beta = 2\alpha$. This follows as soon as we exclude that $360^\circ - 2\alpha = \beta$. But we know that $\alpha + 2\beta = 180^\circ$, hence $360^\circ - 2\alpha = 4\beta$. Therefore, $360^\circ - 2\alpha = \beta$ can only happen if $\beta = 0$, which is excluded by the assumption that $A$ is not on $BC$. Therefore, we have shown that
$$\frac{1}{|CD|} - \frac{1}{|BD|} = \frac{1}{|CD| + |BD|} \quad \text{is equivalent to} \quad \beta = 2\alpha.$$
From $\alpha + 2\beta = 180^\circ$ we easily obtain that $\beta = 2\alpha$ if and only if $\alpha = 36^\circ$.