I need help. I don’t know what the answer is.

I think it is B if not sorry

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)

secant-secant formula:

(whole secant) x (external part) =           (whole secant) x (external part)

(8+y-2) *8=                                           (y+4+6) *6

(y+6)*8 = (y+10) *6

distribute

8y+48 = 6y +60

subtract 6y from each side

2y +48 = 60

subtract 48 from each side

2y = 12

divide by 2

y = 6

sv =     (y+4+6)

sv   = 6+4+6

sv = 16