Question 62
The three distinct points $B$, $C$, $D$ are collinear with $C$ between $B$ and $D$. Another point $A$ not on the line $BD$ is such that $|AB| = |AC| = |CD|$.
Prove that $\angle BAC = 36^\circ$ if and only if $\frac{1}{|CD|} - \frac{1}{|BD|} = \frac{1}{|CD| + |BD|}$.
Prove that $\angle BAC = 36^\circ$ if and only if $\frac{1}{|CD|} - \frac{1}{|BD|} = \frac{1}{|CD| + |BD|}$.