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In the figure above,  at point B. If AE = 16, how many times greater is the area of ΔDBE than the area of ΔABC?
    Correct answer is '9'. Can you explain this answer?
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    In the figure above, at point B. If AE = 16, how many times greater is...
    Because AC and DE are parallel, we can establish three pairs of congruent angles:
    ∠ABC ≌ ∠DBE (vertical angles)
    ∠ACD ≌∠CDE (alternate interior angles)
    ∠DEA ≌ ∠EAC (alternate interior angles)
    Therefore the two triangles are similar, and all pairs of corresponding sides are proportional. If the ratio of corresponding sides in two similar figures is m : n, the ratio of areas of the two figures must by m2 : n2'. If AE = 16, then BE = 16 - 4 = 12, and therefore BE : AB = 12 : 4 = 3 : 1. Since the sides of ΔDBE are 3 times greater than the corresponding sides of ΔABC, the area of ΔDBE must be 32 = 9 times greater than the area of ΔABC.
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    In the figure above, at point B. If AE = 16, how many times greater is the area of ΔDBE than the area of ΔABC?Correct answer is '9'. Can you explain this answer?
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