Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that `(AO)/(OC) = (OB)/(OD)`

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#### Solution

In ΔDOC and ΔBOA,

∠CDO = ∠ABO [Alternate interior angles as AB || CD]

∠DCO = ∠BAO [Alternate interior angles as AB || CD]

∠DOC = ∠BOA [Vertically opposite angles]

∴ ΔDOC ∼ ΔBOA [AAA similarity criterion]

∴ `(DO)/(BO) = (OC)/(OA)`... [Corresponding sides are proportional]

`⇒ (OA)/(OC) = (OB)/(OD)`

Concept: Criteria for Similarity of Triangles

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