(a) Let $N$ be the midpoint of $KM$. Then triangles $KNE$ and $DHE$ are similar whence $\frac{KN}{KE}=\frac{DH}{DE}$ and so $KM=2KN=2\frac{DH \cdot KE}{DE}=2 \cdot \frac{BD \cdot BE}{DE}$. But this is symmetric in $D,E$ so the same formula holds for $LH$ and we are done. (b) Let $S$ be the point of intersection of that line through $B$ with $HK$. Since $SB$ is tangent to both circles we have $SB^2=SH \cdot SL=SK \cdot SM$. So $SH \cdot SL=SK \cdot SM$. Suppose the claim was false. Then w.l.o.g. we have $SH>SK$. Then from (a) we find that $SL>SM$ and hence $SH \cdot SL>SK \cdot SM$ which is a contradiction. Hence the claim.