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 OAOC=OBOD.
It is given that trapezium $A B C D$ with $A B \| D C . O$ is the point of intersection of $A C$ and $B D$.
We have to prove that $\frac{O A}{O C}=\frac{O B}{O D}$
Now, in $\triangle A O B$ and $\triangle C O D$
$\angle A O B=\angle C O D \quad$ (Vertically opposite angles)
$\angle \mathrm{OAB}=\angle \mathrm{OCD}$ (Alternate angles)
$\therefore \triangle A O B \sim \triangle C O D \quad$ (AASimilarity)
Hence, $\frac{O A}{O C}=\frac{O B}{O D}$ (Corresponding sides are proportional)
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