#### Question

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

#### Solution

We have,

ABCD is a trapezium with AB || DC

In ΔAOB and ΔCOD

∠AOB = ∠COD [Vertically opposite angles]

∠OAB = ∠OCD [Alternate interior angles]

Then, ΔAOB ~ ΔCOD [By AA similarity]

`therefore"OA"/"OC"="OB"/"OD"` [Corresponding parts of similar Δ are proportional]

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#### APPEARS IN

Solution Diagonals Ac and Bd of a Trapezium Abcd with Ab || Dc Intersect Each Other at the Point O. Using Similarity Criterion for Two Triangles, Show that `"Oa"/"Oc"="Ob"/"Od"` Concept: Triangles Examples and Solutions.