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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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