IXL geometry help please!

Work Shown:

Inscribed angle theorem

arc measure = 2*(inscribed angle)

arc BCD = 2*(angle A)

arc BCD = 2*(97)

arc BCD = 194 degrees

Break arc BCD into its smaller pieces

arc BCD = (minor arc BC) + (minor arc CD)

194 = (90) + (minor arc CD)

194-90 = minor arc CD

104 = minor arc CD

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.

add 5 subtract 1

step-by-step explanation: