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9514 1404 393

Step-by-step explanation:

Congruent sides are the same length.

  AB = DF

  3(2x +10) = 12x +12

  6x +30 = 12x +12 . . . . eliminate parentheses

  18 = 6x . . . . . . . . . . . . subtract 6x+12

  3 = x . . . . . . . . . . . . . divide by 6

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  BC = FG

  2y +12 = 2(2y -3)

  y +6 = 2y -3 . . . . . . . divide by 2

  9 = y . . . . . . . . . . . . subtract y-3

< 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)

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step-by-step explanation:

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