248.The diagonal BD of a parallelogram ABCD intersects the segment AE at the point F, where E is any point on the side BC. Prove that DF × EF = FB × FA.
In ∆AFD and ∆BFE, we have ∠1 = ∠2 [Vertically opposite angles] ∠3 = ∠4 [Alternate angles]
So, by AA-criterion of similarity, we have
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Short Answer Type
249.In a ∠ABC, BD and CE are the altitudes. Prove that ∠ADB and ∠AEC are similar. Is ∆CDB ~ ∆BEC?
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250.The perimeters of two similar triangles ABC and PQR are respectively 36 cm and 24 cm. If PQ = 10 cm, find AB.