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9. 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
9. 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
CBSE | Class 10 | Exercise 4.5 | Maths | RD Sharma | Triangles
9. 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:

Given:

OC is the point of intersection of AC and BD in the trapezium ABCD, with AB ∥ DC.

To prove: 

OA/ OC = OB/ OD

Proof :

In ΔAOB and ΔCOD

∠AOB = ∠COD (Vertically Opposite Angles)

∠OAB = ∠OCD    (Alternate angles)

ΔAOB ∼ ΔCOD

Therefore, OA/ OC = OB/ OD   (Corresponding sides are proportional)

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