Prove that each angle in an equiangular (not equilateral) triangle is 60 degrees. Must be a two-column proof with statements and reasons!

I need this for tomorrow so please only put legitimate answers.

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.

-362

step-by-step explanation:

(12)(4)+2(12)-34(12)+2(12)-3

  \/         \/         \/         \/

  48 +   24 -   408 + 24 - 3

      \   /               \   /

          72     -     432 - 3 = -363

not sure but it may be b, as i said i'm not sure.

80° is the measure of gh

step-by-step explanation:

circle f is congruent to circle j

we have two circle with center f and j.

both circles are congruent. the radius of the circle would be equal.

see the attachment for the circle.

we are given ∠efd ≅ ∠gjh

therefore, arc (ed) = arc (gh)     (∴ circle f is congruent to circle j)

but ∠dfe=80°=arc (ed)

therefore, arc (gh) = 80°

thus, 80° is the measure of gh