Irish Mathematical Olympiad

Because $|B'B| = |BA| = |AC| = |CC'|$, we have $|B'X| = |AB| + |BX|$ and $|BC'| = |BC| + |AB|$. For the first equation we used that $B$ is between $B'$ and $X$. To see this, we have to exclude that $X$ is between $B$ and $B'$ (by assumption, $B'$ is not between $X$ and $B$). If $|AX| = |BX|$, $X$ cannot be between $B$ and $B'$, because the angle $\angle CBA$ is acute. If $\frac{1}{|AB|} = \frac{1}{|B'X|} + \frac{1}{|BC'|}$, we have $|B'X| > |AB| = |B'B|$, hence $X$ is not between $B$ and $B'$.

We see now that the equation $\frac{1}{|AB|} = \frac{1}{|B'X|} + \frac{1}{|BC'|}$ is equivalent to
$$\frac{1}{|AB|} = \frac{1}{|AB| + |BX|} + \frac{1}{|BC| + |AB|}$$ which rewrites as
$$(|AB| + |BX|)(|BC| + |AB|) = |AB|(|AB| + |BX|) + |AB|(|BC| + |AB|)$$
which simplifies to $|BC| \cdot |BX| = |AB|^2$, or equivalently $\frac{|AB|}{|BC|} = \frac{|BX|}{|AB|}$. Because the triangles $ABX$ and $ABC$ have a common angle at $B$, this last equality is equivalent to these two triangles being similar with sides $AB$ and $BX$ in triangle $ABX$ corresponding to sides $BC$ and $AB$ in triangle $ABC$. Because $\triangle ABC$ is isosceles, the same is then true for the triangle $ABX$, hence $|AX| = |BX|$. On the other hand, if $|AX| = |BX|$, triangle $ABX$ is isosceles and because it shares an angle with $\triangle ABC$ at $B$, both triangles are similar. This finishes the proof of the desired equivalence.