Let $\alpha = \angle CAI = \angle IAB$, $\beta = \angle ABI = \angle IBC$, $\gamma = \angle BCI = \angle ICA$. Then $\alpha + \beta + \gamma = 90^\circ$ and $\angle CBD = \angle CAD = \alpha = \angle DAB = \angle DCB$, yielding
$$\angle KBD = \angle IBD = \alpha + \beta = 90^\circ - \gamma, \\ \angle DCL = \angle DCI = \alpha + \gamma = 90^\circ - \beta.$$
Depending on the location of the point $K$ (Figures 50 and 51), we have either $\angle DKB = \angle DKI = \angle DCI$ or $\angle DKB = 180^\circ - \angle IKD = \angle DCI$; in each case $\angle DKB = 90^\circ - \beta$. Analogously, we obtain $\angle CLD = 90^\circ - \gamma$. Hence
$$\angle BDK = 180^\circ - (90^\circ - \gamma) - (90^\circ - \beta) = \beta + \gamma = 90^\circ - \alpha, \\ \angle LDC = 180^\circ - (90^\circ - \beta) - (90^\circ - \gamma) = \beta + \gamma = 90^\circ - \alpha.$$
On the other hand, we have $\angle BDC = 180^\circ - 2\alpha = 2(90^\circ - \alpha)$, meaning that both $DK$ and $DL$ bisect the angle $BDC$. Consequently, points $D, K$ and $L$ lie on a line. As the bisector of the vertex angle of the isosceles triangle $BCD$ is also the perpendicular bisector of the line segment $BC$, symmetry yields $\angle DCK = \angle KBD = 90^\circ - \gamma$ and $\angle LBD = \angle DCL = 90^\circ - \beta$.

a. If the point $K$ lies between points $C$ and $I$ then
$$\angle DKI = \angle DKB = 90^\circ - \beta = \angle LBD = \angle LID.$$
Hence the line $DI$ is tangent to the circumcircle of the triangle $IKL$ (as points $D, K, L$ lie on a line). If the point $K$ lies between points $I$ and $D$ then
$$\angle ILD = \angle CLD = 90^\circ - \gamma = \angle DCK = \angle DIK.$$
Analogously to the previous case, the line $DI$ must be tangent to the circumcircle of the triangle $IKL$.

b. Since $\angle DKB = 90^\circ - \beta = \angle LBD$ and points $D, K, L$ lie on a line, the line $DB$ is tangent to the circumcircle of the triangle $BKL$. On the other hand, we have $\angle DAB = \alpha = \angle CBD = \angle EBD$, implying that $DB$ is also tangent to the circumcircle of the triangle $BEA$. Using the power of the point $D$ w.r.t. to these circles, we get $DB^2 = DK \cdot DL$ and $DB^2 = DA \cdot DE$, respectively. Altogether, we obtain $DA \cdot DE = DK \cdot DL$. Hence points $A, K, L, E$ are concyclic.