The circle with diameter DE is the A-Apollonius circle of $\triangle ABC,$ the locus of points F with constant ratio of distances $\frac{FB}{FC}= \frac{AB}{AC}$ from the vertices B, C. The pedal triangle $\triangle KLM,\ K \in BC, \L \in CA,\ M \in AB$ of any point F on the A-Apollonius circle is isosceles with KL = KM. For example, quadrilaterals FKCL, FKMB are both cyclic and FC, FB are diameters of their circumcircles. Therefore, $KL = FC \sin C,\ KM = FB \sin B.$ Then $\frac{KL}{KM}= \frac{FC}{FB}\cdot \frac{\sin C}{\sin B}= \frac{AC}{AB}\cdot \frac{\sin C}{\sin B}= 1$ by the sine theorem for $\triangle ABC.$

An interesting remark. $m(\widehat{LKM})=\left|A-m(\widehat{BFC})\right|$ and the points $F,K,L,M$ never are concyclically.

In fact, it is the solution of IMO 1996/2. IMO 2010/4 is also similar to the problem above. All uses Apollonius circle for angle bisectors.