In a triangle ABC, \(AB = 8\text{ cm}\) and D and E are points on AB and AC respectively such that DE is parallel to BC. If \(BC = 5DE\), then what is \(AD \times BD\) equal to?
\(10.24\text{ cm}^2\)
Since DE ∥ BC, \(\triangle ADE \sim \triangle ABC\), so \(\dfrac{AD}{AB} = \dfrac{DE}{BC} = \dfrac15\) (as \(BC = 5DE\)). Thus \(AD = \dfrac{8}{5} = 1.6\text{ cm}\) and \(BD = AB - AD = 8 - 1.6 = 6.4\text{ cm}\). Hence \(AD\times BD = 1.6\times 6.4 = 10.24\text{ cm}^2\).
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