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For ΔABC, ∠A = 4x - 4, ∠B = 6x - 1, and ∠C = 8x - 13. If ΔABC undergoes a dilation by a scale factor of 2 to create ΔA'B'C' with ∠A' = 51 - x, ∠B' = 4x + 21, and ∠C' = 6x + 9, which confirms that ΔABC∼ΔA'B'C by the AA criterion?

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  1. 26 May, 14:56
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    4x-4+6x-1+8x-13=180 (angle sum of triangle)

    18x-18=180

    x=11

    ∠A=4*11-4=40°

    ∠B=6*11-1=65°

    ∠A' = 51-11=40°

    ∠B'=4*11+21=65°

    since ∠A = ∠A' = 40°, ∠B = ∠B'=65°

    ΔABC∼ΔA'B'C (AA)
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