Question 143
In a triangle $ABC$ with $\hat{B} \lt \hat{C}$, let $K$ be the center of the excircle that is tangent to the side $[AC]$. The lines $AK$ and $BC$ intersect at $D$, and $E$ is the center of the circumcircle of $BKC$. Prove that
$$
\frac{1}{|KA|} = \frac{1}{|KD|} + \frac{1}{|KE|}.
$$
$$
\frac{1}{|KA|} = \frac{1}{|KD|} + \frac{1}{|KE|}.
$$