can you add an image ?
option a, c,d are correct.
from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have
therefore, by rhs rule,δtzb ≅δszb
by cpctc, sz≅tz
also, from δctz and δasz,
by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz
also,from δasz and δzsb,
by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb
hence, option a, c,d are correct.