option a, c,d are correct.
step-by-step explanation:
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
tz=sz(given)
bz=zb(common)
therefore, by rhs rule,δtzb â
δszb
by cpctc, szâ
tz
also, from δctz and δasz,
tz=sz(given)
â tcz=â saz(90°)
by rhs rule, δctz â
δasz, therefore by cpctc, â ctzâ
â asz
also,from δasz and δzsb,
zs=sz(common)
â zbs=â saz=90°
by rhs rule, δasz â
δzsb, therefore, by cpctc, â aszâ
â zsb
hence, option a, c,d are correct.