ABCD is a trapezium in which $ A B \| C D $. The diagonals $ A C $ and $ B D $ intersect at $ O . $ Prove that $ \triangle A O B \sim \Delta C O D $.

Given:

ABCD is a trapezium in which \( A B \| C D \). The diagonals \( A C \) and \( B D \) intersect at \( O . \)

To do:

We have to prove that \( \triangle A O B \sim \Delta C O D \).

Solution:

$AB \parallel CD$

In $\triangle AOB$ and $\triangle COD$,

$\angle AOB=\angle COD$   (Vertically opposite angles)

$\angle BAO=\angle DCO$    (Alternate angles)

Therefore,

$\triangle AOB \sim\ \triangle COD$   (By AA similarity)

Hence proved.

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Updated on: 10-Oct-2022

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