Naismax .
Let $D = BI \cap NK, E = BI \cap AC, F = AD \cap BC$.
So, $$\angle DEN = 180^{\circ} - \angle A - \dfrac{B}{2},$$
and
$$\angle END = 90^{\circ} - \dfrac{C}{2}.$$
Thus, $$\angle IDN = \angle DEN +\angle END = \left({180^{\circ} - \angle A - \dfrac{B}{2}}\right) + \left({90^{\circ} - \dfrac{C}{2}}\right) = 180^{\circ} - \dfrac{A}{2} = 180^{\circ} - \angle IAN$$$$\implies \angle IAN + \angle IDN = 180^{\circ} \implies IDNA \text{ is cyclic.}$$
So, $\angle IDA = \angle INA = 90^{\circ}$, then,
$$\angle BAF = 180^{\circ} - \angle BDA - \dfrac{\angle B}{2} = 90^{\circ} - \dfrac{\angle B}{2}$$
and $$\angle BFA = 180^{\circ} - \angle BDF - \dfrac{\angle B}{2} = 90^{\circ} - \dfrac{\angle B}{2} \implies \angle BAF = \angle BFA \implies BF = BA = c.$$
So $FC = BC - FB = a - c$, and we are done.
We constructed a total of three lines, namingly,
$BI$
$KN$
$AD$.