Exactly one plane that contains noncollinear points a, b, and c. true or no justify answer

The answers 5 so there’s the you need trust me it’s right

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

ac ≅ ab (given)< cax ≅ < bax   (given)ax ≅ ax (common side)

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

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

in δacy and δaby,

ac ≅ ab (given)< cay ≅ < bay (since < cax ≅ < bax is given and xy is the extension of line ax)ay ≅ ay (common side)

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)

Chope this have a nice day btw the other person deserves brainlest

3,843,000 is how that would be read