Question 86
In a triangle $\triangle ABC$ with $\angle ABC \lt \angle BCA$ we define $K$ as the centre of the excircle opposite $B$ (the excircle tangent to side $AC$). The lines $AK$ and $BC$ intersect at a point $D$. Let $E$ be the centre of the circumcircle of $\triangle BKC$. Prove that
$$
\frac{1}{|KA|} = \frac{1}{|KD|} + \frac{1}{|KE|}
$$
$$
\frac{1}{|KA|} = \frac{1}{|KD|} + \frac{1}{|KE|}
$$