Look at the figure below and the proof.

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 ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

answer: a

step-by-step explanation: the answer is a due to the rule of symmetry; that a line of symmetry must divide a figure into 2 pieces that are the exact same, only reflected over a line of symmetry. b and c do not have lines of symmetry and would not divide the figure evenly, thus making the answer a.

yeah i know the problem you need. i am taking it right now. the answer is step 3

~mickey