We need to prove that $\frac{|KA|}{|KD|}+\frac{|KA|}{|KE|}=1$. Simple Angle-Chasing gives us $\angle ADC=\frac{C-B}2$. Hence, $$\frac{|KA|}{|KD|}=\frac{|AC|}{|CD|}=\frac{\sin{ADC}}{\sin{CAD}}=\frac{\sin{(\frac{C-B}2)}}{\cos{(\frac A2)}}$$ Also, $$\frac{|KA|}{|KE|}=\frac{2|KA|}{2R_{BKC}}=\frac{2|KA|\cdot\sin{(\frac B2)}}{|KC|}=\frac{2\cos{(\frac C2)\cdot \sin{(\frac B2)}}}{\cos{(\frac A2)}}$$ Therefore, it suffices to show that $$1=\frac{\sin{(\frac{C-B}2)}}{\cos{(\frac A2)}}+\frac{2\cos{(\frac C2)\cdot \sin{(\frac B2)}}}{\cos{(\frac A2)}}=\frac{\sin{(\frac{C+B}2)}}{\cos{(\frac A2)}}\Leftrightarrow \sin{\left(\frac{C+B}2\right)}=\cos{\left(\frac A2\right)}$$ , which is correct. Done.

we know that if $(A,B;C,D)=-1$ then $\frac{2}{AB}=\frac{1}{CB}+\frac{1}{DB}$ ( with signed distance) just consider $T$ the symmetric of $K$ in $E$ which is on $(E)$... My regards RH HAS