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.

Answer 1

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Answer 2

OA/OC = OB/OD  proved

Step-by-step explanation:

Given: ABCD is a trapezium.

AB||CD

Diagonals AB and CD intersect at O

To prove: OA/OC = OB/OD

Proof:

In ΔOAB and ΔOCD

∠AOD = ∠DOC (vertically opposite)

∠ABO = ∠CDO (alternate angles)

∠BAO = ∠OCD (alternate angles)

ΔOAB ~ ΔOCD (AAA similarity)

When 2 triangles are similar, corresponding sides are proportionate.

So  OA/OC = OB/OD

Hence proved.