We use the notation $\angle C A B=2 \alpha, \angle A B C=2 \beta$ and $\angle B C A=2 \gamma$. Note that $\alpha+\beta+\gamma=\frac{1}{2}(\angle C A B+\angle A B C+\angle B C A)=\frac{1}{2} \cdot 180^{\circ}=90^{\circ}$.

We first prove that $K, A$ and $E$ are collinear. Since $K$ lies on the angle bisector of $\angle A B C$ and the external angle bisector of $\angle C A B$, we find that $\angle B K A=180^{\circ}-\angle K A B-\angle A B K=180^{\circ}-\left(90^{\circ}+\alpha\right)-\beta=90^{\circ}-\alpha-\beta=\gamma$. On the other hand, we know that $\angle B C K=90^{\circ}+\gamma>90^{\circ}$ is obtuse. Since $E$ is the circumcenter of $\triangle B K C$, $E$ must lie on the opposite side of $B K$ from $C$, and by the central angle theorem, we know that $\angle K E B=2 \cdot\left(180^{\circ}-\angle B C K\right)=2 \cdot\left(90^{\circ}-\gamma\right)=180^{\circ}-2 \gamma$. Since $E$ is the circumcenter of $\triangle B K C$, we also know that $\triangle B E K$ is isosceles with vertex angle $E$. Thus, $\angle B K E=\frac{1}{2}\left(180^{\circ}-\angle K E B\right)=\frac{1}{2} \cdot 2 \gamma=\gamma$. We conclude that $\angle B K A=\gamma=\angle B K E$, which implies that $K, A$ and $E$ are collinear.

Due to this collinearity, we find that $\angle A E B=\angle K E B=180^{\circ}-2 \gamma=180^{\circ}-\angle B C A$, so $A C B E$ is a cyclic quadrilateral. In particular, it follows that $\angle A E C=\angle A B C=\angle A B D$. Since also $\angle C A E=90^{\circ}+\alpha=\angle K A B=\angle D A B$, we find by (AA) that $\triangle A E C \sim \triangle A B D$.

The last two observations we need are that $|E C|=|E K|$ because $\triangle C E K$ is also isosceles with vertex angle $E$, and that $B K$ is a bisector of $\triangle A B D$. Now it follows that

$$1-\frac{|K A|}{|K E|}=\frac{|K E|-|K A|}{|K E|}=\frac{|A E|}{|K E|}=\frac{|A E|}{|C E|}=\frac{|A B|}{|D B|}=\frac{|A K|}{|D K|}$$

where we have used, respectively: the difference of fractions, that $K$, $A$ and $E$ are collinear, that $|E C|=|E K|$, that $\triangle A E C \sim \triangle A B D$, and the angle bisector theorem on $B K$ in $\triangle A B D$. The desired result now follows easily by moving $\frac{|K A|}{|K E|}$ to the other side and dividing by $|K A|$.